A course in

projective geometry

Projektif geometri dersi

David PierceMatematik Boumlluumlmuuml

Mimar Sinan Guumlzel Sanatlar Uumlniversitesimatmsgsuedutr~dpierce

dpiercemsgsuedutr

September ndash Last edited January

Nesin Matematik KoumlyuŞirince Selccediluk İzmir

Preface

This is about a two-week course called Geometriler at theNesin Mathematics Village In the first week we read in myTurkish translation some of the lemmas for Euclidrsquos Porisms

in Book VII of the Collection of Pappus Students presentedthe lemmas at the board and I lectured on modern develop-ments due to Desargues and others In the following weekwe read Lobachevskirsquos ldquoGeometrical Researches on the Theoryof Parallelsrdquo in Halsteadrsquos English translation Again studentspresented propositions at the board but curved space turnedout to be a more difficult concept than points at infinity Thatweek is covered only briefly here in the Introduction

I had given a similar Geometriler course in my departmentat Mimar Sinan That course met two hours a week for the weeks of the fall semester of ndash I kept a record likethe present one Class at the Math Village met two hoursa day the first week an hour and a half the second Mondaythrough Sunday except Thursday In the second week I endedon Friday so that I could participate in the Thales Buluşmasıin Milet the next day

I thank the staff of the Village notably Aslı Can Korkmazfor having Pappus and Lobachevski printed and coil-boundfor distribution to students I thank all of the workers of theVillage from Ali Nesin on down for making it an excellentplace to continue and develop the thoughts begun more than years ago elsewhere in Ionia by Thales

Contents

Preface

Introduction

Overview Origins Additions

Monday

Quadrangle Theorem statement Desarguesrsquos Theorem statement Pappusrsquos Theorem statement Equality and sameness Thalesrsquos Theorem statement Thales himself Non-Euclidean planes Proportions

Tuesday

Pappusrsquos Theorem parallel case Quadrangle Theorem Pappusrsquos proof

Wednesday

Thalesrsquos Theorem Euclidrsquos proof A simple definition of proportion Cross ratio in Pappus

Friday

Addition and multiplication A non-commutative field Pappusrsquos Theorem intersecting case Pappusrsquos Theorem a second proof

Saturday

Cartesian coordinates Barycentric coordinates Projective coordinates Pappusrsquos Theorem third proof

Sunday

The projective plane over a field The Fano Plane Desarguesrsquos Theorem a proof Duality Quadrangle Theorem second proof

Bibliography

Contents

List of Figures

Pappusrsquos Hexagon Theorem The Quadrangle Theorem Desarguesrsquos Theorem

Degenerate case of Desarguesrsquos Theorem Points in involution An involution A five-line locus Solution of the five-line locus problem A diagram for Lemma I of Pappus Pappusrsquos Theorem in modern notation

The Quadrangle Theorem set up The Quadrangle Theorem ldquoCompleterdquo figures Desarguesrsquos Theorem Pappusrsquos Hexagon Theorem Triangles on the same base

Equal parallelograms Thalesrsquos Theorem Dates of some ancient writers and thinkers The angle in a semicircle Straight lines in the hyperbolic plane For a definition of proportion A circle of implications

Cases of Pappusrsquos Theorem A case of Pappusrsquos Theorem Lemma IV

Lemma from Lemma IV The Quadrangle Theorem

Thalesrsquos Theorem A proportion of lengths and areas A definition of proportion Lemma III Invariance of cross ratio in Pappus Lemma X Lemma III reconfigured Lemma X two halves of the proof

Addition of line segments Sum of two points with respect to a third Descartesrsquos definition of multiplication Commutativity of Descartesrsquos multiplication Lemma XI Lemma XI variant Lemma XII Steps of Lemma XII Pappusrsquos Theorem by projection

Line segments in the ratio of rectangles Alternation of ratio Associativity of addition Cevarsquos Theorem Pappusrsquos Hexagon Theorem

The Fano Plane Desarguesrsquos Theorem Proof of Desarguesrsquos Theorem The dual of Pappusrsquos Theorem The Quadrangle Theorem

List of Figures

Introduction

Overview

Of Pappusrsquos lemmas for Euclidrsquos Porisms students presentedsix VIII IV III X XI and XII in that order Lemmas VIIIXII and XIII are cases of what is now known as PappusrsquosTheorem while Lemmas III X and XI are needed to proveXII and XIII We skipped Lemma XIII in class its proof beingsimilar to that of XII Lemma IV is effectively what I shallcall the Quadrangle Theorem although Coxeter gives it noname [ p ] There is a related theorem calledDesarguesrsquos Involution Theorem by Field and Gray [ p ]Coxeter describes this as ldquothe theorem of the quadrangular setrdquo[ p ]

Two students volunteered to present the first two (VIII andIV) of Pappusrsquos lemmas above For the next two lemmas (IIIand X) volunteers were not forthcoming and so I picked twomore students When they had fulfilled their assignments onlytwo more students were still in class I asked them to preparethe last two lemmas (XI and XII) for the next day Class metat am a difficult time for many Nonetheless some absentstudents did return the next day

Every presenter of a lemma came more or less prepared forthe job though sometimes needing help from the audienceClass was mostly in Turkish except on the last day or twowhen only I was speaking with the remaining studentsrsquo per-mission I switched mostly to English

In the following week class was in the afternoon as one re-turning student had begged for it to be Most students in thesecond week were new With a couple of notable exceptionsthey did not prepare their presentations well Some of themleft the Village early earlier than I did without telling meand having accepted assignments for the day (Friday) whenthey would be gone I have doubts about how well even theremaining students understood Lobachevskirsquos non-Euclideanconception of parallelism they seemed to persist in their Eu-clidean notions

On the first day of that second week I reviewed the Eu-clidean geometry not requiring the Fifth Postulate that Loba-chevski summarizes in his Theorems ndash This is the geome-try of Propositions ndash of Book I of the Elements There isalso some solid geometry from Book XI though I did not gointo this I do not know how much the review of Euclidean ge-ometry meant to students who unlike those at Mimar Sinanhad not read Euclid in the first place I gave away the plotby describing the Poincareacute half-plane model for Lobachevskiangeometry but given the quality of later student presentationsI have doubts that the model made much sense I shall notsay more about that second week except that if that part ofthe course is repeated it should probably be coupled with areading of Euclid and then it would need a full week if nottwo

Details

One may refer to Pappusrsquos Theorem more precisely as Pap-pusrsquos Hexagon Theorem It is the theorem that if the ver-tices of a hexagon lie alternately on two straight lines thenthe intersection points of the three pairs of opposite sides of

Introduction

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

the hexagon lie on a straight line See for example Figure which is the same as Figure on page Pappusrsquos theo-rem remains true if some of the opposite sides of the hexagonare parallel provided one allows that parallel lines meet onthe ldquoline at infinityrdquo See Figure on page and Figure on page In any case the two straight lines holding thesix vertices between them can be considered as a degenerateconic section Pascal generalized to an arbitrary conic sectionthough without proof [ ]

The Quadrangle Theorem is that if a straight line intersectssix straight lines each of which passes through two of fourgiven points no three of the four being collinear then five ofthe intersection points determine the sixth regardless of thechoice of the four given points Thus in Figure which is thesame as Figure on page the points A B C D and Eon the same straight line determine the point L on that lineprovided it is understood that L is found by first picking apoint F not on the line AB then picking a point G on thestraight line AF letting GC and GE intersect FB and FD

Geometries

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

respectively at H and K and letting HK intersect the originalline AB

In class after we saw Pappusrsquos treatment of the Quadrangleand Hexagon Theorems I sketched a proof of all cases of thelatter by passing to a third dimension and projecting I gaveanother proof by introducing projective coordinates

Desarguesrsquos Theorem is that if the straight lines throughcorresponding vertices of two triangles intersect at one pointthen the intersection points of the corresponding sides of thetriangles lie on one straight line For example if the trianglesare ABC and DEF as in Figure which is the same as Fig-ure on page then HKL is straight I gave the proofattributed to Hessenberg from using three applicationsof Pappusrsquos Theorem as sketched in Figure on page Strictly this proof assumes that all points named in the dia-gram are distinct in the other case Cronheim gave in [] the argument sketched in Figure where by applying

Introduction

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

Pappusrsquos Theorem to hexagons GCELBA GAELBC andSRDCAF in turn we have that FHS DKR and finally LKHare straight

In class using Desarguersquos Theorem and its dual which is itsconverse I proved again the Quadrangle Theorem

Pappusrsquos proofs rely heavily on the proposition known inmodern times in some countries as Thalesrsquos Theorem []a straight line cutting two sides of a triangle cuts them pro-portionally if and only if it is parallel to the base This isldquoequationrdquo () on page with respect to Figure Euclidproves this theorem at the beginning of Book VI of the Ele-

ments using the theory of proportion developed in Book VThis theory relies on the so-called Archimedean Axiom thatof two line segments either can be multiplied so as to ex-ceed the other In fact one does not need this axiom but onecan develop a theory of proportion sufficient for establishing

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Preface

This is about a two-week course called Geometriler at theNesin Mathematics Village In the first week we read in myTurkish translation some of the lemmas for Euclidrsquos Porisms

in Book VII of the Collection of Pappus Students presentedthe lemmas at the board and I lectured on modern develop-ments due to Desargues and others In the following weekwe read Lobachevskirsquos ldquoGeometrical Researches on the Theoryof Parallelsrdquo in Halsteadrsquos English translation Again studentspresented propositions at the board but curved space turnedout to be a more difficult concept than points at infinity Thatweek is covered only briefly here in the Introduction

I had given a similar Geometriler course in my departmentat Mimar Sinan That course met two hours a week for the weeks of the fall semester of ndash I kept a record likethe present one Class at the Math Village met two hoursa day the first week an hour and a half the second Mondaythrough Sunday except Thursday In the second week I endedon Friday so that I could participate in the Thales Buluşmasıin Milet the next day

I thank the staff of the Village notably Aslı Can Korkmazfor having Pappus and Lobachevski printed and coil-boundfor distribution to students I thank all of the workers of theVillage from Ali Nesin on down for making it an excellentplace to continue and develop the thoughts begun more than years ago elsewhere in Ionia by Thales

Contents

Preface

Introduction

Overview Origins Additions

Monday

Quadrangle Theorem statement Desarguesrsquos Theorem statement Pappusrsquos Theorem statement Equality and sameness Thalesrsquos Theorem statement Thales himself Non-Euclidean planes Proportions

Tuesday

Pappusrsquos Theorem parallel case Quadrangle Theorem Pappusrsquos proof

Wednesday

Thalesrsquos Theorem Euclidrsquos proof A simple definition of proportion Cross ratio in Pappus

Friday

Addition and multiplication A non-commutative field Pappusrsquos Theorem intersecting case Pappusrsquos Theorem a second proof

Saturday

Cartesian coordinates Barycentric coordinates Projective coordinates Pappusrsquos Theorem third proof

Sunday

The projective plane over a field The Fano Plane Desarguesrsquos Theorem a proof Duality Quadrangle Theorem second proof

Bibliography

Contents

List of Figures

Pappusrsquos Hexagon Theorem The Quadrangle Theorem Desarguesrsquos Theorem

Degenerate case of Desarguesrsquos Theorem Points in involution An involution A five-line locus Solution of the five-line locus problem A diagram for Lemma I of Pappus Pappusrsquos Theorem in modern notation

The Quadrangle Theorem set up The Quadrangle Theorem ldquoCompleterdquo figures Desarguesrsquos Theorem Pappusrsquos Hexagon Theorem Triangles on the same base

Equal parallelograms Thalesrsquos Theorem Dates of some ancient writers and thinkers The angle in a semicircle Straight lines in the hyperbolic plane For a definition of proportion A circle of implications

Cases of Pappusrsquos Theorem A case of Pappusrsquos Theorem Lemma IV

Lemma from Lemma IV The Quadrangle Theorem

Thalesrsquos Theorem A proportion of lengths and areas A definition of proportion Lemma III Invariance of cross ratio in Pappus Lemma X Lemma III reconfigured Lemma X two halves of the proof

Addition of line segments Sum of two points with respect to a third Descartesrsquos definition of multiplication Commutativity of Descartesrsquos multiplication Lemma XI Lemma XI variant Lemma XII Steps of Lemma XII Pappusrsquos Theorem by projection

Line segments in the ratio of rectangles Alternation of ratio Associativity of addition Cevarsquos Theorem Pappusrsquos Hexagon Theorem

The Fano Plane Desarguesrsquos Theorem Proof of Desarguesrsquos Theorem The dual of Pappusrsquos Theorem The Quadrangle Theorem

List of Figures

Introduction

Overview

Of Pappusrsquos lemmas for Euclidrsquos Porisms students presentedsix VIII IV III X XI and XII in that order Lemmas VIIIXII and XIII are cases of what is now known as PappusrsquosTheorem while Lemmas III X and XI are needed to proveXII and XIII We skipped Lemma XIII in class its proof beingsimilar to that of XII Lemma IV is effectively what I shallcall the Quadrangle Theorem although Coxeter gives it noname [ p ] There is a related theorem calledDesarguesrsquos Involution Theorem by Field and Gray [ p ]Coxeter describes this as ldquothe theorem of the quadrangular setrdquo[ p ]

Two students volunteered to present the first two (VIII andIV) of Pappusrsquos lemmas above For the next two lemmas (IIIand X) volunteers were not forthcoming and so I picked twomore students When they had fulfilled their assignments onlytwo more students were still in class I asked them to preparethe last two lemmas (XI and XII) for the next day Class metat am a difficult time for many Nonetheless some absentstudents did return the next day

Every presenter of a lemma came more or less prepared forthe job though sometimes needing help from the audienceClass was mostly in Turkish except on the last day or twowhen only I was speaking with the remaining studentsrsquo per-mission I switched mostly to English

In the following week class was in the afternoon as one re-turning student had begged for it to be Most students in thesecond week were new With a couple of notable exceptionsthey did not prepare their presentations well Some of themleft the Village early earlier than I did without telling meand having accepted assignments for the day (Friday) whenthey would be gone I have doubts about how well even theremaining students understood Lobachevskirsquos non-Euclideanconception of parallelism they seemed to persist in their Eu-clidean notions

On the first day of that second week I reviewed the Eu-clidean geometry not requiring the Fifth Postulate that Loba-chevski summarizes in his Theorems ndash This is the geome-try of Propositions ndash of Book I of the Elements There isalso some solid geometry from Book XI though I did not gointo this I do not know how much the review of Euclidean ge-ometry meant to students who unlike those at Mimar Sinanhad not read Euclid in the first place I gave away the plotby describing the Poincareacute half-plane model for Lobachevskiangeometry but given the quality of later student presentationsI have doubts that the model made much sense I shall notsay more about that second week except that if that part ofthe course is repeated it should probably be coupled with areading of Euclid and then it would need a full week if nottwo

Details

One may refer to Pappusrsquos Theorem more precisely as Pap-pusrsquos Hexagon Theorem It is the theorem that if the ver-tices of a hexagon lie alternately on two straight lines thenthe intersection points of the three pairs of opposite sides of

Introduction

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

the hexagon lie on a straight line See for example Figure which is the same as Figure on page Pappusrsquos theo-rem remains true if some of the opposite sides of the hexagonare parallel provided one allows that parallel lines meet onthe ldquoline at infinityrdquo See Figure on page and Figure on page In any case the two straight lines holding thesix vertices between them can be considered as a degenerateconic section Pascal generalized to an arbitrary conic sectionthough without proof [ ]

The Quadrangle Theorem is that if a straight line intersectssix straight lines each of which passes through two of fourgiven points no three of the four being collinear then five ofthe intersection points determine the sixth regardless of thechoice of the four given points Thus in Figure which is thesame as Figure on page the points A B C D and Eon the same straight line determine the point L on that lineprovided it is understood that L is found by first picking apoint F not on the line AB then picking a point G on thestraight line AF letting GC and GE intersect FB and FD

Geometries

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

respectively at H and K and letting HK intersect the originalline AB

In class after we saw Pappusrsquos treatment of the Quadrangleand Hexagon Theorems I sketched a proof of all cases of thelatter by passing to a third dimension and projecting I gaveanother proof by introducing projective coordinates

Desarguesrsquos Theorem is that if the straight lines throughcorresponding vertices of two triangles intersect at one pointthen the intersection points of the corresponding sides of thetriangles lie on one straight line For example if the trianglesare ABC and DEF as in Figure which is the same as Fig-ure on page then HKL is straight I gave the proofattributed to Hessenberg from using three applicationsof Pappusrsquos Theorem as sketched in Figure on page Strictly this proof assumes that all points named in the dia-gram are distinct in the other case Cronheim gave in [] the argument sketched in Figure where by applying

Introduction

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

Pappusrsquos Theorem to hexagons GCELBA GAELBC andSRDCAF in turn we have that FHS DKR and finally LKHare straight

In class using Desarguersquos Theorem and its dual which is itsconverse I proved again the Quadrangle Theorem

Pappusrsquos proofs rely heavily on the proposition known inmodern times in some countries as Thalesrsquos Theorem []a straight line cutting two sides of a triangle cuts them pro-portionally if and only if it is parallel to the base This isldquoequationrdquo () on page with respect to Figure Euclidproves this theorem at the beginning of Book VI of the Ele-

ments using the theory of proportion developed in Book VThis theory relies on the so-called Archimedean Axiom thatof two line segments either can be multiplied so as to ex-ceed the other In fact one does not need this axiom but onecan develop a theory of proportion sufficient for establishing

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Contents

Preface

Introduction

Overview Origins Additions

Monday

Quadrangle Theorem statement Desarguesrsquos Theorem statement Pappusrsquos Theorem statement Equality and sameness Thalesrsquos Theorem statement Thales himself Non-Euclidean planes Proportions

Tuesday

Pappusrsquos Theorem parallel case Quadrangle Theorem Pappusrsquos proof

Wednesday

Thalesrsquos Theorem Euclidrsquos proof A simple definition of proportion Cross ratio in Pappus

Friday

Addition and multiplication A non-commutative field Pappusrsquos Theorem intersecting case Pappusrsquos Theorem a second proof

Saturday

Cartesian coordinates Barycentric coordinates Projective coordinates Pappusrsquos Theorem third proof

Sunday

The projective plane over a field The Fano Plane Desarguesrsquos Theorem a proof Duality Quadrangle Theorem second proof

Bibliography

Contents

List of Figures

Pappusrsquos Hexagon Theorem The Quadrangle Theorem Desarguesrsquos Theorem

Degenerate case of Desarguesrsquos Theorem Points in involution An involution A five-line locus Solution of the five-line locus problem A diagram for Lemma I of Pappus Pappusrsquos Theorem in modern notation

The Quadrangle Theorem set up The Quadrangle Theorem ldquoCompleterdquo figures Desarguesrsquos Theorem Pappusrsquos Hexagon Theorem Triangles on the same base

Equal parallelograms Thalesrsquos Theorem Dates of some ancient writers and thinkers The angle in a semicircle Straight lines in the hyperbolic plane For a definition of proportion A circle of implications

Cases of Pappusrsquos Theorem A case of Pappusrsquos Theorem Lemma IV

Lemma from Lemma IV The Quadrangle Theorem

Thalesrsquos Theorem A proportion of lengths and areas A definition of proportion Lemma III Invariance of cross ratio in Pappus Lemma X Lemma III reconfigured Lemma X two halves of the proof

Addition of line segments Sum of two points with respect to a third Descartesrsquos definition of multiplication Commutativity of Descartesrsquos multiplication Lemma XI Lemma XI variant Lemma XII Steps of Lemma XII Pappusrsquos Theorem by projection

Line segments in the ratio of rectangles Alternation of ratio Associativity of addition Cevarsquos Theorem Pappusrsquos Hexagon Theorem

The Fano Plane Desarguesrsquos Theorem Proof of Desarguesrsquos Theorem The dual of Pappusrsquos Theorem The Quadrangle Theorem

List of Figures

Introduction

Overview

Of Pappusrsquos lemmas for Euclidrsquos Porisms students presentedsix VIII IV III X XI and XII in that order Lemmas VIIIXII and XIII are cases of what is now known as PappusrsquosTheorem while Lemmas III X and XI are needed to proveXII and XIII We skipped Lemma XIII in class its proof beingsimilar to that of XII Lemma IV is effectively what I shallcall the Quadrangle Theorem although Coxeter gives it noname [ p ] There is a related theorem calledDesarguesrsquos Involution Theorem by Field and Gray [ p ]Coxeter describes this as ldquothe theorem of the quadrangular setrdquo[ p ]

Two students volunteered to present the first two (VIII andIV) of Pappusrsquos lemmas above For the next two lemmas (IIIand X) volunteers were not forthcoming and so I picked twomore students When they had fulfilled their assignments onlytwo more students were still in class I asked them to preparethe last two lemmas (XI and XII) for the next day Class metat am a difficult time for many Nonetheless some absentstudents did return the next day

Every presenter of a lemma came more or less prepared forthe job though sometimes needing help from the audienceClass was mostly in Turkish except on the last day or twowhen only I was speaking with the remaining studentsrsquo per-mission I switched mostly to English

In the following week class was in the afternoon as one re-turning student had begged for it to be Most students in thesecond week were new With a couple of notable exceptionsthey did not prepare their presentations well Some of themleft the Village early earlier than I did without telling meand having accepted assignments for the day (Friday) whenthey would be gone I have doubts about how well even theremaining students understood Lobachevskirsquos non-Euclideanconception of parallelism they seemed to persist in their Eu-clidean notions

On the first day of that second week I reviewed the Eu-clidean geometry not requiring the Fifth Postulate that Loba-chevski summarizes in his Theorems ndash This is the geome-try of Propositions ndash of Book I of the Elements There isalso some solid geometry from Book XI though I did not gointo this I do not know how much the review of Euclidean ge-ometry meant to students who unlike those at Mimar Sinanhad not read Euclid in the first place I gave away the plotby describing the Poincareacute half-plane model for Lobachevskiangeometry but given the quality of later student presentationsI have doubts that the model made much sense I shall notsay more about that second week except that if that part ofthe course is repeated it should probably be coupled with areading of Euclid and then it would need a full week if nottwo

Details

One may refer to Pappusrsquos Theorem more precisely as Pap-pusrsquos Hexagon Theorem It is the theorem that if the ver-tices of a hexagon lie alternately on two straight lines thenthe intersection points of the three pairs of opposite sides of

Introduction

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

the hexagon lie on a straight line See for example Figure which is the same as Figure on page Pappusrsquos theo-rem remains true if some of the opposite sides of the hexagonare parallel provided one allows that parallel lines meet onthe ldquoline at infinityrdquo See Figure on page and Figure on page In any case the two straight lines holding thesix vertices between them can be considered as a degenerateconic section Pascal generalized to an arbitrary conic sectionthough without proof [ ]

The Quadrangle Theorem is that if a straight line intersectssix straight lines each of which passes through two of fourgiven points no three of the four being collinear then five ofthe intersection points determine the sixth regardless of thechoice of the four given points Thus in Figure which is thesame as Figure on page the points A B C D and Eon the same straight line determine the point L on that lineprovided it is understood that L is found by first picking apoint F not on the line AB then picking a point G on thestraight line AF letting GC and GE intersect FB and FD

Geometries

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

respectively at H and K and letting HK intersect the originalline AB

In class after we saw Pappusrsquos treatment of the Quadrangleand Hexagon Theorems I sketched a proof of all cases of thelatter by passing to a third dimension and projecting I gaveanother proof by introducing projective coordinates

Desarguesrsquos Theorem is that if the straight lines throughcorresponding vertices of two triangles intersect at one pointthen the intersection points of the corresponding sides of thetriangles lie on one straight line For example if the trianglesare ABC and DEF as in Figure which is the same as Fig-ure on page then HKL is straight I gave the proofattributed to Hessenberg from using three applicationsof Pappusrsquos Theorem as sketched in Figure on page Strictly this proof assumes that all points named in the dia-gram are distinct in the other case Cronheim gave in [] the argument sketched in Figure where by applying

Introduction

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

Pappusrsquos Theorem to hexagons GCELBA GAELBC andSRDCAF in turn we have that FHS DKR and finally LKHare straight

In class using Desarguersquos Theorem and its dual which is itsconverse I proved again the Quadrangle Theorem

Pappusrsquos proofs rely heavily on the proposition known inmodern times in some countries as Thalesrsquos Theorem []a straight line cutting two sides of a triangle cuts them pro-portionally if and only if it is parallel to the base This isldquoequationrdquo () on page with respect to Figure Euclidproves this theorem at the beginning of Book VI of the Ele-

ments using the theory of proportion developed in Book VThis theory relies on the so-called Archimedean Axiom thatof two line segments either can be multiplied so as to ex-ceed the other In fact one does not need this axiom but onecan develop a theory of proportion sufficient for establishing

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Friday

Addition and multiplication A non-commutative field Pappusrsquos Theorem intersecting case Pappusrsquos Theorem a second proof

Saturday

Cartesian coordinates Barycentric coordinates Projective coordinates Pappusrsquos Theorem third proof

Sunday

The projective plane over a field The Fano Plane Desarguesrsquos Theorem a proof Duality Quadrangle Theorem second proof

Bibliography

Contents

List of Figures

Pappusrsquos Hexagon Theorem The Quadrangle Theorem Desarguesrsquos Theorem

Degenerate case of Desarguesrsquos Theorem Points in involution An involution A five-line locus Solution of the five-line locus problem A diagram for Lemma I of Pappus Pappusrsquos Theorem in modern notation

The Quadrangle Theorem set up The Quadrangle Theorem ldquoCompleterdquo figures Desarguesrsquos Theorem Pappusrsquos Hexagon Theorem Triangles on the same base

Equal parallelograms Thalesrsquos Theorem Dates of some ancient writers and thinkers The angle in a semicircle Straight lines in the hyperbolic plane For a definition of proportion A circle of implications

Cases of Pappusrsquos Theorem A case of Pappusrsquos Theorem Lemma IV

Lemma from Lemma IV The Quadrangle Theorem

Thalesrsquos Theorem A proportion of lengths and areas A definition of proportion Lemma III Invariance of cross ratio in Pappus Lemma X Lemma III reconfigured Lemma X two halves of the proof

Addition of line segments Sum of two points with respect to a third Descartesrsquos definition of multiplication Commutativity of Descartesrsquos multiplication Lemma XI Lemma XI variant Lemma XII Steps of Lemma XII Pappusrsquos Theorem by projection

Line segments in the ratio of rectangles Alternation of ratio Associativity of addition Cevarsquos Theorem Pappusrsquos Hexagon Theorem

The Fano Plane Desarguesrsquos Theorem Proof of Desarguesrsquos Theorem The dual of Pappusrsquos Theorem The Quadrangle Theorem

List of Figures

Introduction

Overview

Of Pappusrsquos lemmas for Euclidrsquos Porisms students presentedsix VIII IV III X XI and XII in that order Lemmas VIIIXII and XIII are cases of what is now known as PappusrsquosTheorem while Lemmas III X and XI are needed to proveXII and XIII We skipped Lemma XIII in class its proof beingsimilar to that of XII Lemma IV is effectively what I shallcall the Quadrangle Theorem although Coxeter gives it noname [ p ] There is a related theorem calledDesarguesrsquos Involution Theorem by Field and Gray [ p ]Coxeter describes this as ldquothe theorem of the quadrangular setrdquo[ p ]

Two students volunteered to present the first two (VIII andIV) of Pappusrsquos lemmas above For the next two lemmas (IIIand X) volunteers were not forthcoming and so I picked twomore students When they had fulfilled their assignments onlytwo more students were still in class I asked them to preparethe last two lemmas (XI and XII) for the next day Class metat am a difficult time for many Nonetheless some absentstudents did return the next day

Every presenter of a lemma came more or less prepared forthe job though sometimes needing help from the audienceClass was mostly in Turkish except on the last day or twowhen only I was speaking with the remaining studentsrsquo per-mission I switched mostly to English

In the following week class was in the afternoon as one re-turning student had begged for it to be Most students in thesecond week were new With a couple of notable exceptionsthey did not prepare their presentations well Some of themleft the Village early earlier than I did without telling meand having accepted assignments for the day (Friday) whenthey would be gone I have doubts about how well even theremaining students understood Lobachevskirsquos non-Euclideanconception of parallelism they seemed to persist in their Eu-clidean notions

On the first day of that second week I reviewed the Eu-clidean geometry not requiring the Fifth Postulate that Loba-chevski summarizes in his Theorems ndash This is the geome-try of Propositions ndash of Book I of the Elements There isalso some solid geometry from Book XI though I did not gointo this I do not know how much the review of Euclidean ge-ometry meant to students who unlike those at Mimar Sinanhad not read Euclid in the first place I gave away the plotby describing the Poincareacute half-plane model for Lobachevskiangeometry but given the quality of later student presentationsI have doubts that the model made much sense I shall notsay more about that second week except that if that part ofthe course is repeated it should probably be coupled with areading of Euclid and then it would need a full week if nottwo

Details

One may refer to Pappusrsquos Theorem more precisely as Pap-pusrsquos Hexagon Theorem It is the theorem that if the ver-tices of a hexagon lie alternately on two straight lines thenthe intersection points of the three pairs of opposite sides of

Introduction

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

the hexagon lie on a straight line See for example Figure which is the same as Figure on page Pappusrsquos theo-rem remains true if some of the opposite sides of the hexagonare parallel provided one allows that parallel lines meet onthe ldquoline at infinityrdquo See Figure on page and Figure on page In any case the two straight lines holding thesix vertices between them can be considered as a degenerateconic section Pascal generalized to an arbitrary conic sectionthough without proof [ ]

The Quadrangle Theorem is that if a straight line intersectssix straight lines each of which passes through two of fourgiven points no three of the four being collinear then five ofthe intersection points determine the sixth regardless of thechoice of the four given points Thus in Figure which is thesame as Figure on page the points A B C D and Eon the same straight line determine the point L on that lineprovided it is understood that L is found by first picking apoint F not on the line AB then picking a point G on thestraight line AF letting GC and GE intersect FB and FD

Geometries

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

respectively at H and K and letting HK intersect the originalline AB

In class after we saw Pappusrsquos treatment of the Quadrangleand Hexagon Theorems I sketched a proof of all cases of thelatter by passing to a third dimension and projecting I gaveanother proof by introducing projective coordinates

Desarguesrsquos Theorem is that if the straight lines throughcorresponding vertices of two triangles intersect at one pointthen the intersection points of the corresponding sides of thetriangles lie on one straight line For example if the trianglesare ABC and DEF as in Figure which is the same as Fig-ure on page then HKL is straight I gave the proofattributed to Hessenberg from using three applicationsof Pappusrsquos Theorem as sketched in Figure on page Strictly this proof assumes that all points named in the dia-gram are distinct in the other case Cronheim gave in [] the argument sketched in Figure where by applying

Introduction

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

Pappusrsquos Theorem to hexagons GCELBA GAELBC andSRDCAF in turn we have that FHS DKR and finally LKHare straight

In class using Desarguersquos Theorem and its dual which is itsconverse I proved again the Quadrangle Theorem

Pappusrsquos proofs rely heavily on the proposition known inmodern times in some countries as Thalesrsquos Theorem []a straight line cutting two sides of a triangle cuts them pro-portionally if and only if it is parallel to the base This isldquoequationrdquo () on page with respect to Figure Euclidproves this theorem at the beginning of Book VI of the Ele-

ments using the theory of proportion developed in Book VThis theory relies on the so-called Archimedean Axiom thatof two line segments either can be multiplied so as to ex-ceed the other In fact one does not need this axiom but onecan develop a theory of proportion sufficient for establishing

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

List of Figures

Pappusrsquos Hexagon Theorem The Quadrangle Theorem Desarguesrsquos Theorem

Degenerate case of Desarguesrsquos Theorem Points in involution An involution A five-line locus Solution of the five-line locus problem A diagram for Lemma I of Pappus Pappusrsquos Theorem in modern notation

The Quadrangle Theorem set up The Quadrangle Theorem ldquoCompleterdquo figures Desarguesrsquos Theorem Pappusrsquos Hexagon Theorem Triangles on the same base

Equal parallelograms Thalesrsquos Theorem Dates of some ancient writers and thinkers The angle in a semicircle Straight lines in the hyperbolic plane For a definition of proportion A circle of implications

Cases of Pappusrsquos Theorem A case of Pappusrsquos Theorem Lemma IV

Lemma from Lemma IV The Quadrangle Theorem

Thalesrsquos Theorem A proportion of lengths and areas A definition of proportion Lemma III Invariance of cross ratio in Pappus Lemma X Lemma III reconfigured Lemma X two halves of the proof

Addition of line segments Sum of two points with respect to a third Descartesrsquos definition of multiplication Commutativity of Descartesrsquos multiplication Lemma XI Lemma XI variant Lemma XII Steps of Lemma XII Pappusrsquos Theorem by projection

Line segments in the ratio of rectangles Alternation of ratio Associativity of addition Cevarsquos Theorem Pappusrsquos Hexagon Theorem

The Fano Plane Desarguesrsquos Theorem Proof of Desarguesrsquos Theorem The dual of Pappusrsquos Theorem The Quadrangle Theorem

List of Figures

Introduction

Overview

Of Pappusrsquos lemmas for Euclidrsquos Porisms students presentedsix VIII IV III X XI and XII in that order Lemmas VIIIXII and XIII are cases of what is now known as PappusrsquosTheorem while Lemmas III X and XI are needed to proveXII and XIII We skipped Lemma XIII in class its proof beingsimilar to that of XII Lemma IV is effectively what I shallcall the Quadrangle Theorem although Coxeter gives it noname [ p ] There is a related theorem calledDesarguesrsquos Involution Theorem by Field and Gray [ p ]Coxeter describes this as ldquothe theorem of the quadrangular setrdquo[ p ]

Two students volunteered to present the first two (VIII andIV) of Pappusrsquos lemmas above For the next two lemmas (IIIand X) volunteers were not forthcoming and so I picked twomore students When they had fulfilled their assignments onlytwo more students were still in class I asked them to preparethe last two lemmas (XI and XII) for the next day Class metat am a difficult time for many Nonetheless some absentstudents did return the next day

Every presenter of a lemma came more or less prepared forthe job though sometimes needing help from the audienceClass was mostly in Turkish except on the last day or twowhen only I was speaking with the remaining studentsrsquo per-mission I switched mostly to English

In the following week class was in the afternoon as one re-turning student had begged for it to be Most students in thesecond week were new With a couple of notable exceptionsthey did not prepare their presentations well Some of themleft the Village early earlier than I did without telling meand having accepted assignments for the day (Friday) whenthey would be gone I have doubts about how well even theremaining students understood Lobachevskirsquos non-Euclideanconception of parallelism they seemed to persist in their Eu-clidean notions

On the first day of that second week I reviewed the Eu-clidean geometry not requiring the Fifth Postulate that Loba-chevski summarizes in his Theorems ndash This is the geome-try of Propositions ndash of Book I of the Elements There isalso some solid geometry from Book XI though I did not gointo this I do not know how much the review of Euclidean ge-ometry meant to students who unlike those at Mimar Sinanhad not read Euclid in the first place I gave away the plotby describing the Poincareacute half-plane model for Lobachevskiangeometry but given the quality of later student presentationsI have doubts that the model made much sense I shall notsay more about that second week except that if that part ofthe course is repeated it should probably be coupled with areading of Euclid and then it would need a full week if nottwo

Details

One may refer to Pappusrsquos Theorem more precisely as Pap-pusrsquos Hexagon Theorem It is the theorem that if the ver-tices of a hexagon lie alternately on two straight lines thenthe intersection points of the three pairs of opposite sides of

Introduction

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

the hexagon lie on a straight line See for example Figure which is the same as Figure on page Pappusrsquos theo-rem remains true if some of the opposite sides of the hexagonare parallel provided one allows that parallel lines meet onthe ldquoline at infinityrdquo See Figure on page and Figure on page In any case the two straight lines holding thesix vertices between them can be considered as a degenerateconic section Pascal generalized to an arbitrary conic sectionthough without proof [ ]

The Quadrangle Theorem is that if a straight line intersectssix straight lines each of which passes through two of fourgiven points no three of the four being collinear then five ofthe intersection points determine the sixth regardless of thechoice of the four given points Thus in Figure which is thesame as Figure on page the points A B C D and Eon the same straight line determine the point L on that lineprovided it is understood that L is found by first picking apoint F not on the line AB then picking a point G on thestraight line AF letting GC and GE intersect FB and FD

Geometries

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

respectively at H and K and letting HK intersect the originalline AB

In class after we saw Pappusrsquos treatment of the Quadrangleand Hexagon Theorems I sketched a proof of all cases of thelatter by passing to a third dimension and projecting I gaveanother proof by introducing projective coordinates

Desarguesrsquos Theorem is that if the straight lines throughcorresponding vertices of two triangles intersect at one pointthen the intersection points of the corresponding sides of thetriangles lie on one straight line For example if the trianglesare ABC and DEF as in Figure which is the same as Fig-ure on page then HKL is straight I gave the proofattributed to Hessenberg from using three applicationsof Pappusrsquos Theorem as sketched in Figure on page Strictly this proof assumes that all points named in the dia-gram are distinct in the other case Cronheim gave in [] the argument sketched in Figure where by applying

Introduction

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

Pappusrsquos Theorem to hexagons GCELBA GAELBC andSRDCAF in turn we have that FHS DKR and finally LKHare straight

In class using Desarguersquos Theorem and its dual which is itsconverse I proved again the Quadrangle Theorem

Pappusrsquos proofs rely heavily on the proposition known inmodern times in some countries as Thalesrsquos Theorem []a straight line cutting two sides of a triangle cuts them pro-portionally if and only if it is parallel to the base This isldquoequationrdquo () on page with respect to Figure Euclidproves this theorem at the beginning of Book VI of the Ele-

ments using the theory of proportion developed in Book VThis theory relies on the so-called Archimedean Axiom thatof two line segments either can be multiplied so as to ex-ceed the other In fact one does not need this axiom but onecan develop a theory of proportion sufficient for establishing

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Lemma from Lemma IV The Quadrangle Theorem

Thalesrsquos Theorem A proportion of lengths and areas A definition of proportion Lemma III Invariance of cross ratio in Pappus Lemma X Lemma III reconfigured Lemma X two halves of the proof

Addition of line segments Sum of two points with respect to a third Descartesrsquos definition of multiplication Commutativity of Descartesrsquos multiplication Lemma XI Lemma XI variant Lemma XII Steps of Lemma XII Pappusrsquos Theorem by projection

Line segments in the ratio of rectangles Alternation of ratio Associativity of addition Cevarsquos Theorem Pappusrsquos Hexagon Theorem

The Fano Plane Desarguesrsquos Theorem Proof of Desarguesrsquos Theorem The dual of Pappusrsquos Theorem The Quadrangle Theorem

List of Figures

Introduction

Overview

Of Pappusrsquos lemmas for Euclidrsquos Porisms students presentedsix VIII IV III X XI and XII in that order Lemmas VIIIXII and XIII are cases of what is now known as PappusrsquosTheorem while Lemmas III X and XI are needed to proveXII and XIII We skipped Lemma XIII in class its proof beingsimilar to that of XII Lemma IV is effectively what I shallcall the Quadrangle Theorem although Coxeter gives it noname [ p ] There is a related theorem calledDesarguesrsquos Involution Theorem by Field and Gray [ p ]Coxeter describes this as ldquothe theorem of the quadrangular setrdquo[ p ]

Two students volunteered to present the first two (VIII andIV) of Pappusrsquos lemmas above For the next two lemmas (IIIand X) volunteers were not forthcoming and so I picked twomore students When they had fulfilled their assignments onlytwo more students were still in class I asked them to preparethe last two lemmas (XI and XII) for the next day Class metat am a difficult time for many Nonetheless some absentstudents did return the next day

Every presenter of a lemma came more or less prepared forthe job though sometimes needing help from the audienceClass was mostly in Turkish except on the last day or twowhen only I was speaking with the remaining studentsrsquo per-mission I switched mostly to English

In the following week class was in the afternoon as one re-turning student had begged for it to be Most students in thesecond week were new With a couple of notable exceptionsthey did not prepare their presentations well Some of themleft the Village early earlier than I did without telling meand having accepted assignments for the day (Friday) whenthey would be gone I have doubts about how well even theremaining students understood Lobachevskirsquos non-Euclideanconception of parallelism they seemed to persist in their Eu-clidean notions

On the first day of that second week I reviewed the Eu-clidean geometry not requiring the Fifth Postulate that Loba-chevski summarizes in his Theorems ndash This is the geome-try of Propositions ndash of Book I of the Elements There isalso some solid geometry from Book XI though I did not gointo this I do not know how much the review of Euclidean ge-ometry meant to students who unlike those at Mimar Sinanhad not read Euclid in the first place I gave away the plotby describing the Poincareacute half-plane model for Lobachevskiangeometry but given the quality of later student presentationsI have doubts that the model made much sense I shall notsay more about that second week except that if that part ofthe course is repeated it should probably be coupled with areading of Euclid and then it would need a full week if nottwo

Details

One may refer to Pappusrsquos Theorem more precisely as Pap-pusrsquos Hexagon Theorem It is the theorem that if the ver-tices of a hexagon lie alternately on two straight lines thenthe intersection points of the three pairs of opposite sides of

Introduction

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

the hexagon lie on a straight line See for example Figure which is the same as Figure on page Pappusrsquos theo-rem remains true if some of the opposite sides of the hexagonare parallel provided one allows that parallel lines meet onthe ldquoline at infinityrdquo See Figure on page and Figure on page In any case the two straight lines holding thesix vertices between them can be considered as a degenerateconic section Pascal generalized to an arbitrary conic sectionthough without proof [ ]

The Quadrangle Theorem is that if a straight line intersectssix straight lines each of which passes through two of fourgiven points no three of the four being collinear then five ofthe intersection points determine the sixth regardless of thechoice of the four given points Thus in Figure which is thesame as Figure on page the points A B C D and Eon the same straight line determine the point L on that lineprovided it is understood that L is found by first picking apoint F not on the line AB then picking a point G on thestraight line AF letting GC and GE intersect FB and FD

Geometries

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

respectively at H and K and letting HK intersect the originalline AB

In class after we saw Pappusrsquos treatment of the Quadrangleand Hexagon Theorems I sketched a proof of all cases of thelatter by passing to a third dimension and projecting I gaveanother proof by introducing projective coordinates

Desarguesrsquos Theorem is that if the straight lines throughcorresponding vertices of two triangles intersect at one pointthen the intersection points of the corresponding sides of thetriangles lie on one straight line For example if the trianglesare ABC and DEF as in Figure which is the same as Fig-ure on page then HKL is straight I gave the proofattributed to Hessenberg from using three applicationsof Pappusrsquos Theorem as sketched in Figure on page Strictly this proof assumes that all points named in the dia-gram are distinct in the other case Cronheim gave in [] the argument sketched in Figure where by applying

Introduction

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

Pappusrsquos Theorem to hexagons GCELBA GAELBC andSRDCAF in turn we have that FHS DKR and finally LKHare straight

In class using Desarguersquos Theorem and its dual which is itsconverse I proved again the Quadrangle Theorem

Pappusrsquos proofs rely heavily on the proposition known inmodern times in some countries as Thalesrsquos Theorem []a straight line cutting two sides of a triangle cuts them pro-portionally if and only if it is parallel to the base This isldquoequationrdquo () on page with respect to Figure Euclidproves this theorem at the beginning of Book VI of the Ele-

ments using the theory of proportion developed in Book VThis theory relies on the so-called Archimedean Axiom thatof two line segments either can be multiplied so as to ex-ceed the other In fact one does not need this axiom but onecan develop a theory of proportion sufficient for establishing

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Introduction

Overview

Of Pappusrsquos lemmas for Euclidrsquos Porisms students presentedsix VIII IV III X XI and XII in that order Lemmas VIIIXII and XIII are cases of what is now known as PappusrsquosTheorem while Lemmas III X and XI are needed to proveXII and XIII We skipped Lemma XIII in class its proof beingsimilar to that of XII Lemma IV is effectively what I shallcall the Quadrangle Theorem although Coxeter gives it noname [ p ] There is a related theorem calledDesarguesrsquos Involution Theorem by Field and Gray [ p ]Coxeter describes this as ldquothe theorem of the quadrangular setrdquo[ p ]

Two students volunteered to present the first two (VIII andIV) of Pappusrsquos lemmas above For the next two lemmas (IIIand X) volunteers were not forthcoming and so I picked twomore students When they had fulfilled their assignments onlytwo more students were still in class I asked them to preparethe last two lemmas (XI and XII) for the next day Class metat am a difficult time for many Nonetheless some absentstudents did return the next day

Every presenter of a lemma came more or less prepared forthe job though sometimes needing help from the audienceClass was mostly in Turkish except on the last day or twowhen only I was speaking with the remaining studentsrsquo per-mission I switched mostly to English

In the following week class was in the afternoon as one re-turning student had begged for it to be Most students in thesecond week were new With a couple of notable exceptionsthey did not prepare their presentations well Some of themleft the Village early earlier than I did without telling meand having accepted assignments for the day (Friday) whenthey would be gone I have doubts about how well even theremaining students understood Lobachevskirsquos non-Euclideanconception of parallelism they seemed to persist in their Eu-clidean notions

On the first day of that second week I reviewed the Eu-clidean geometry not requiring the Fifth Postulate that Loba-chevski summarizes in his Theorems ndash This is the geome-try of Propositions ndash of Book I of the Elements There isalso some solid geometry from Book XI though I did not gointo this I do not know how much the review of Euclidean ge-ometry meant to students who unlike those at Mimar Sinanhad not read Euclid in the first place I gave away the plotby describing the Poincareacute half-plane model for Lobachevskiangeometry but given the quality of later student presentationsI have doubts that the model made much sense I shall notsay more about that second week except that if that part ofthe course is repeated it should probably be coupled with areading of Euclid and then it would need a full week if nottwo

Details

One may refer to Pappusrsquos Theorem more precisely as Pap-pusrsquos Hexagon Theorem It is the theorem that if the ver-tices of a hexagon lie alternately on two straight lines thenthe intersection points of the three pairs of opposite sides of

Introduction

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

the hexagon lie on a straight line See for example Figure which is the same as Figure on page Pappusrsquos theo-rem remains true if some of the opposite sides of the hexagonare parallel provided one allows that parallel lines meet onthe ldquoline at infinityrdquo See Figure on page and Figure on page In any case the two straight lines holding thesix vertices between them can be considered as a degenerateconic section Pascal generalized to an arbitrary conic sectionthough without proof [ ]

The Quadrangle Theorem is that if a straight line intersectssix straight lines each of which passes through two of fourgiven points no three of the four being collinear then five ofthe intersection points determine the sixth regardless of thechoice of the four given points Thus in Figure which is thesame as Figure on page the points A B C D and Eon the same straight line determine the point L on that lineprovided it is understood that L is found by first picking apoint F not on the line AB then picking a point G on thestraight line AF letting GC and GE intersect FB and FD

Geometries

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

respectively at H and K and letting HK intersect the originalline AB

In class after we saw Pappusrsquos treatment of the Quadrangleand Hexagon Theorems I sketched a proof of all cases of thelatter by passing to a third dimension and projecting I gaveanother proof by introducing projective coordinates

Desarguesrsquos Theorem is that if the straight lines throughcorresponding vertices of two triangles intersect at one pointthen the intersection points of the corresponding sides of thetriangles lie on one straight line For example if the trianglesare ABC and DEF as in Figure which is the same as Fig-ure on page then HKL is straight I gave the proofattributed to Hessenberg from using three applicationsof Pappusrsquos Theorem as sketched in Figure on page Strictly this proof assumes that all points named in the dia-gram are distinct in the other case Cronheim gave in [] the argument sketched in Figure where by applying

Introduction

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

Pappusrsquos Theorem to hexagons GCELBA GAELBC andSRDCAF in turn we have that FHS DKR and finally LKHare straight

In class using Desarguersquos Theorem and its dual which is itsconverse I proved again the Quadrangle Theorem

Pappusrsquos proofs rely heavily on the proposition known inmodern times in some countries as Thalesrsquos Theorem []a straight line cutting two sides of a triangle cuts them pro-portionally if and only if it is parallel to the base This isldquoequationrdquo () on page with respect to Figure Euclidproves this theorem at the beginning of Book VI of the Ele-

ments using the theory of proportion developed in Book VThis theory relies on the so-called Archimedean Axiom thatof two line segments either can be multiplied so as to ex-ceed the other In fact one does not need this axiom but onecan develop a theory of proportion sufficient for establishing

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

In the following week class was in the afternoon as one re-turning student had begged for it to be Most students in thesecond week were new With a couple of notable exceptionsthey did not prepare their presentations well Some of themleft the Village early earlier than I did without telling meand having accepted assignments for the day (Friday) whenthey would be gone I have doubts about how well even theremaining students understood Lobachevskirsquos non-Euclideanconception of parallelism they seemed to persist in their Eu-clidean notions

On the first day of that second week I reviewed the Eu-clidean geometry not requiring the Fifth Postulate that Loba-chevski summarizes in his Theorems ndash This is the geome-try of Propositions ndash of Book I of the Elements There isalso some solid geometry from Book XI though I did not gointo this I do not know how much the review of Euclidean ge-ometry meant to students who unlike those at Mimar Sinanhad not read Euclid in the first place I gave away the plotby describing the Poincareacute half-plane model for Lobachevskiangeometry but given the quality of later student presentationsI have doubts that the model made much sense I shall notsay more about that second week except that if that part ofthe course is repeated it should probably be coupled with areading of Euclid and then it would need a full week if nottwo

Details

One may refer to Pappusrsquos Theorem more precisely as Pap-pusrsquos Hexagon Theorem It is the theorem that if the ver-tices of a hexagon lie alternately on two straight lines thenthe intersection points of the three pairs of opposite sides of

Introduction

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

the hexagon lie on a straight line See for example Figure which is the same as Figure on page Pappusrsquos theo-rem remains true if some of the opposite sides of the hexagonare parallel provided one allows that parallel lines meet onthe ldquoline at infinityrdquo See Figure on page and Figure on page In any case the two straight lines holding thesix vertices between them can be considered as a degenerateconic section Pascal generalized to an arbitrary conic sectionthough without proof [ ]

The Quadrangle Theorem is that if a straight line intersectssix straight lines each of which passes through two of fourgiven points no three of the four being collinear then five ofthe intersection points determine the sixth regardless of thechoice of the four given points Thus in Figure which is thesame as Figure on page the points A B C D and Eon the same straight line determine the point L on that lineprovided it is understood that L is found by first picking apoint F not on the line AB then picking a point G on thestraight line AF letting GC and GE intersect FB and FD

Geometries

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

respectively at H and K and letting HK intersect the originalline AB

In class after we saw Pappusrsquos treatment of the Quadrangleand Hexagon Theorems I sketched a proof of all cases of thelatter by passing to a third dimension and projecting I gaveanother proof by introducing projective coordinates

Desarguesrsquos Theorem is that if the straight lines throughcorresponding vertices of two triangles intersect at one pointthen the intersection points of the corresponding sides of thetriangles lie on one straight line For example if the trianglesare ABC and DEF as in Figure which is the same as Fig-ure on page then HKL is straight I gave the proofattributed to Hessenberg from using three applicationsof Pappusrsquos Theorem as sketched in Figure on page Strictly this proof assumes that all points named in the dia-gram are distinct in the other case Cronheim gave in [] the argument sketched in Figure where by applying

Introduction

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

Pappusrsquos Theorem to hexagons GCELBA GAELBC andSRDCAF in turn we have that FHS DKR and finally LKHare straight

In class using Desarguersquos Theorem and its dual which is itsconverse I proved again the Quadrangle Theorem

Pappusrsquos proofs rely heavily on the proposition known inmodern times in some countries as Thalesrsquos Theorem []a straight line cutting two sides of a triangle cuts them pro-portionally if and only if it is parallel to the base This isldquoequationrdquo () on page with respect to Figure Euclidproves this theorem at the beginning of Book VI of the Ele-

ments using the theory of proportion developed in Book VThis theory relies on the so-called Archimedean Axiom thatof two line segments either can be multiplied so as to ex-ceed the other In fact one does not need this axiom but onecan develop a theory of proportion sufficient for establishing

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

the hexagon lie on a straight line See for example Figure which is the same as Figure on page Pappusrsquos theo-rem remains true if some of the opposite sides of the hexagonare parallel provided one allows that parallel lines meet onthe ldquoline at infinityrdquo See Figure on page and Figure on page In any case the two straight lines holding thesix vertices between them can be considered as a degenerateconic section Pascal generalized to an arbitrary conic sectionthough without proof [ ]

The Quadrangle Theorem is that if a straight line intersectssix straight lines each of which passes through two of fourgiven points no three of the four being collinear then five ofthe intersection points determine the sixth regardless of thechoice of the four given points Thus in Figure which is thesame as Figure on page the points A B C D and Eon the same straight line determine the point L on that lineprovided it is understood that L is found by first picking apoint F not on the line AB then picking a point G on thestraight line AF letting GC and GE intersect FB and FD

Geometries

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

respectively at H and K and letting HK intersect the originalline AB

In class after we saw Pappusrsquos treatment of the Quadrangleand Hexagon Theorems I sketched a proof of all cases of thelatter by passing to a third dimension and projecting I gaveanother proof by introducing projective coordinates

Desarguesrsquos Theorem is that if the straight lines throughcorresponding vertices of two triangles intersect at one pointthen the intersection points of the corresponding sides of thetriangles lie on one straight line For example if the trianglesare ABC and DEF as in Figure which is the same as Fig-ure on page then HKL is straight I gave the proofattributed to Hessenberg from using three applicationsof Pappusrsquos Theorem as sketched in Figure on page Strictly this proof assumes that all points named in the dia-gram are distinct in the other case Cronheim gave in [] the argument sketched in Figure where by applying

Introduction

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

Pappusrsquos Theorem to hexagons GCELBA GAELBC andSRDCAF in turn we have that FHS DKR and finally LKHare straight

In class using Desarguersquos Theorem and its dual which is itsconverse I proved again the Quadrangle Theorem

Pappusrsquos proofs rely heavily on the proposition known inmodern times in some countries as Thalesrsquos Theorem []a straight line cutting two sides of a triangle cuts them pro-portionally if and only if it is parallel to the base This isldquoequationrdquo () on page with respect to Figure Euclidproves this theorem at the beginning of Book VI of the Ele-

ments using the theory of proportion developed in Book VThis theory relies on the so-called Archimedean Axiom thatof two line segments either can be multiplied so as to ex-ceed the other In fact one does not need this axiom but onecan develop a theory of proportion sufficient for establishing

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

respectively at H and K and letting HK intersect the originalline AB

In class after we saw Pappusrsquos treatment of the Quadrangleand Hexagon Theorems I sketched a proof of all cases of thelatter by passing to a third dimension and projecting I gaveanother proof by introducing projective coordinates

Desarguesrsquos Theorem is that if the straight lines throughcorresponding vertices of two triangles intersect at one pointthen the intersection points of the corresponding sides of thetriangles lie on one straight line For example if the trianglesare ABC and DEF as in Figure which is the same as Fig-ure on page then HKL is straight I gave the proofattributed to Hessenberg from using three applicationsof Pappusrsquos Theorem as sketched in Figure on page Strictly this proof assumes that all points named in the dia-gram are distinct in the other case Cronheim gave in [] the argument sketched in Figure where by applying

Introduction

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

Pappusrsquos Theorem to hexagons GCELBA GAELBC andSRDCAF in turn we have that FHS DKR and finally LKHare straight

In class using Desarguersquos Theorem and its dual which is itsconverse I proved again the Quadrangle Theorem

Pappusrsquos proofs rely heavily on the proposition known inmodern times in some countries as Thalesrsquos Theorem []a straight line cutting two sides of a triangle cuts them pro-portionally if and only if it is parallel to the base This isldquoequationrdquo () on page with respect to Figure Euclidproves this theorem at the beginning of Book VI of the Ele-

ments using the theory of proportion developed in Book VThis theory relies on the so-called Archimedean Axiom thatof two line segments either can be multiplied so as to ex-ceed the other In fact one does not need this axiom but onecan develop a theory of proportion sufficient for establishing

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

Pappusrsquos Theorem to hexagons GCELBA GAELBC andSRDCAF in turn we have that FHS DKR and finally LKHare straight

In class using Desarguersquos Theorem and its dual which is itsconverse I proved again the Quadrangle Theorem

Pappusrsquos proofs rely heavily on the proposition known inmodern times in some countries as Thalesrsquos Theorem []a straight line cutting two sides of a triangle cuts them pro-portionally if and only if it is parallel to the base This isldquoequationrdquo () on page with respect to Figure Euclidproves this theorem at the beginning of Book VI of the Ele-

ments using the theory of proportion developed in Book VThis theory relies on the so-called Archimedean Axiom thatof two line segments either can be multiplied so as to ex-ceed the other In fact one does not need this axiom but onecan develop a theory of proportion sufficient for establishing

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A

B

C

D

E

F

G

H

K

L

S

R

Figure Degenerate case of Desarguesrsquos Theorem

Introduction

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Thalesrsquos Theorem on the basis of Book I of the Elements Onecan take Thalesrsquos Theorem as a definition of proportion ex-cept that one should fix one of the base angles of the triangleBut then one can prove as I did in class that the base an-gles do not matter A neat way of proving this is DesarguesrsquosTheorem applied to Figure on page In short Thalesproves Pappus which proves Desargues which proves Thales

In modern axiomatic projective plane geometry the theo-rems of Pappus and Desargues are not equivalent In class weproved not exactly their equivalence with Thalesrsquos Theorembut simply their truth in the geometry of Book I of EuclidrsquosElements They are true in the projective plane over a com-mutative field but the ancient proofs use not just the algebraof fields but the geometry of areas In Lemma VIII for exam-ple which is the case of the Hexagon Theorem when two pairsof opposite sides are parallel Pappusrsquos proof relies on addingand subtracting triangles that are equal because they are onthe same base and between the same parallels

Following first Descartes [] and then Hilbert [] we canobtain a field from Euclidrsquos geometry This may not be thebest way to think about that geometry

Origins

The origin of this course is my interest in the origins of mathe-matics This interest goes back at least to a tenth-grade geom-etry class in ndash where we students were taught to writeproofs in the two-column statementndashreason format I did notmuch care for our textbook which for an example of congru-ence used a photo of a machine stamping out foil trays forTV dinners [ p ] In ldquoCommensurability and Symmetryrdquo

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

[] I mention how the WeeksndashAdkins text confuses equalitywith sameness A geometrical equation like AB = CD meansnot that the segments AB and CD are the same but thattheir lengths are the same Length is an abstraction from asegment as ratio is an abstraction from two segments This iswhy Euclid uses ldquoequalrdquo to describe two equal segments butldquosamerdquo to describe the ratios of segments in a proportion Onecan maintain the distinction symbolically by writing a propor-tion as A B C D rather than as AB = CD I noticedthe distinction many years after high school but even in tenthgrade I thought we should read Euclid I went on to read himalong with Homer and Aeschylus and Plato and others at StJohnrsquos College []

In the first course I taught at the Nesin MathematicsVillage was an opportunity to review something I had read atSt Johnrsquos Called ldquoConic Sections agrave la Apollonius of Pergardquomy course reviewed the propositions of Book I of the Conics

[ ] that pertained to the parabola I shall say more aboutthis later for now I shall note that while the course was greatfor me I donrsquot think it meant much for the students who justsat and watched me at the board One has to engage withthe mathematics for oneself especially when it is somethingso unusual for today as Apollonius A good way to do this isto have to go to the board and present the mathematics as atSt Johnrsquos

In at METU in Ankara I taught the course called His-tory of Mathematical Concepts in the manner of St JohnrsquosWe studied Euclid Apollonius and (briefly) Archimedes inthe first semester Al-Khwarizmı Thabit ibn Qurra OmarKhayyaacutem Cardano Viegravete Descartes and Newton in the sec-ond []

Introduction

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

At METU I loved the content of the course called Funda-mentals of Mathematics which was required of all first-yearstudents I even wrote a text for the course a rigorous textthat might overwhelm students but whose contents I thoughtat least teachers should know In the end I didnrsquot thinkit was right to try to teach equivalence-relations and proofsto beginning students independently from a course of realmathematics Moving to Mimar Sinan in I was able(in collaboration) to develop a course in which first-year stu-dents read and presented the proofs that taught mathematicsto practically all mathematicians until about a century agoAmong other things students would learn the non-trivial (be-cause non-identical) equivalence-relation of congruence I didnot actually recognize this opportunity until I had seen theway students tended to confuse equality of line-segments withsameness

Our first-semester Euclid course is followed by an analyticgeometry course Pondering the transition from the one courseto the other led to some of the ideas about ratio and proportionthat are worked out in the present course My study of Pap-pusrsquos Theorem arose in this context and I was disappointedto find that the Wikipedia article called ldquoPappusrsquos HexagonTheoremrdquo did not provide a precise reference to its namesakeI rectified this condition on May when I added tothe article a section called ldquoOriginsrdquo giving Pappusrsquos proof

In order to track down that proof I had relied on Heath whoin A History of Greek Mathematics summarizes most of Pap-pusrsquos lemmas for Euclidrsquos lost Porisms [ p ndash] In thissummary Heath may give the serial numbers of the lemmas assuch these are the numbers given above as Roman numeralsHeath always gives the numbers of the lemmas as propositionswithin Book VII of Pappusrsquos Collection according to the enu-

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

meration of Hultsch [] Apparently this enumeration wasmade originally in the th century by Commandino in hisLatin translation [ pp ndash ]

According to Heath Pappusrsquos Lemmas XII XIII XV andXVII for the Porisms or Propositions and of Book VII establish the Hexagon Theorem The latter twopropositions can be considered as converses of the former twowhich consider the hexagon lying respectively between paralleland intersecting straight lines

In Mathematical Thought from Ancient to Modern Times

Kline cites only Proposition as giving Pappusrsquos Theorem[ p ] This proposition Lemma XIII follows from Lem-mas III and X as XII follows from XI and X For PappusrsquosTheorem in the most general sense one should cite also Propo-sition Lemma VIII which as we have noted is the casewhere two pairs of opposite sides of the hexagon are paral-lel the conclusion is then that the third pair are also parallelHeathrsquos summary does not seem to mention this lemma at allThe omission must be a simple oversight

In the catalogue of my home department at Mimar Sinanthere is an elective course called Geometries meeting twohours a week The course had last been taught in the fallof when I offered it in the fall of For use in thecourse I translated the first of Pappusrsquos lemmas for Eu-

Searching for Commandinorsquos name in Jonesrsquos book reveals an in-teresting tidbit on page Book III of the Collection is addressed toan otherwise-unknown teacher of mathematics called Pandrosion Shemust be a woman since she is given the feminine form of the adjec-tive κράτιστος -η -ον (ldquomost excellentrdquo) but ldquoin Commandinorsquos Latintranslation her name vanishes leaving the absurdity of the polite epithetκρατίστη being treated as a name lsquoCratistersquo while for no good reasonHultsch alters the text to make the name masculinerdquo

Introduction

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

clidrsquos Porisms I had not thought there was an English versionbut at the end of my work I found Jonesrsquos This helped meto parse a few confusing words What I found first on Library

Genesis was the first volume of Jonesrsquos work [] ProfessorJones himself supplied me with Volume II the one with thecommentary and diagrams []

Book VII of Pappusrsquos Collection is an account of the so-called Treasury of Analysis (ἀναλυόμενος τόπος) This Trea-

sury consisted of works by Euclid Apollonius Aristaeus andEratosthenes most of them now lost As a reminder of thewealth of knowledge that is no longer ours I ultimately wroteout on the back of my Pappus translation a table of thecontents of the Treasury Pappusrsquos list of the contents is in-cluded by Thomas in his Selections Illustrating the History

of Greek Mathematics in the Loeb series [ pp ndash]Thomasrsquos anthology includes more selections from PappusrsquosCollection but none involving the Hexagon Theorem He doesprovide Proposition of Book VII that is Lemma IV forthe Porisms of Euclid the lemma that I am calling the Quad-rangle Theorem

Additions

Here are some further developments of topics of the course

Involutions

Lemma IV is that if the points ABCDEL in Figure above(or in Figure on page ) satisfy the proportion

AL middot BC AB middot CL AL middotDE AD middot EL ()

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

then HKL is straight Chasles [ p ] rewrites the condi-tion in the form

BC middot AD middot EL = AB middotDE middot CL

We can also write

BC middotAD AB middotDE CL EL

this makes it easier to see that when ABCDE are giventhen some unique L exists so as to satisfy () As L variesthe ratio CL EL takes on all possible values but unity IfBC middotAD = AB middotDE that is

AB BC AD DE

then HK AE by Lemma I Otherwise by Lemma IV HKmust pass through the L that satisfies () Thus the converseof the lemma holds as well We might speculate whether thisconverse was one of Euclidrsquos original porisms Chasles seemsto think it was

Thomas observes [ pp ndash] that

[the converse of Lemma IV] is one of the ways of expressingthe proposition enunciated by Desargues The three pairs

of opposite sides of a complete quadrilateral are cut by any

transversal in three pairs of conjugate points of an involution

Following Coxeter I would call the ldquocomplete quadrilateralrdquohere a complete quadrangle (see Figure page ) Desarguesproves the Involution Theorem in his Rough Draft of an

Essay on the results of taking plane sections of a cone [ p] One way to interpret the theorem is to observe that inFigure or if the points BCDE are conceived of as fixed

Introduction

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

b b b b b bbc bc

A B C D F G H

Figure Points in involution

then A determines L Moreover this operation transposes Aand L and so it is an involution of the straight line BE

Desargues proceeds towards the Involution Theorem by firstobserving that if seven points ABCDFGH are arranged asin Figure on a straight line so that

AB middot AH = AC middot AG = AD middot AF ()

then without reference to A we have [ p ]

DG middot FG CD middot CF BG middotGH BC middot CH ()

Indeed by second equation in ()

AG AF AD AC ()

so by addition or subtraction of the terms of the first ratio toor from the terms of the second

AG AF DG CF ()

Similarly after alternating () we obtain

AF AC AG AD

AF AC FG CD

Composing the latter with () yields

AG AC DG middot FG CD middot CF ()

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Replacing D with B and F with H in the second equation of() yields the first equation Hence we can do the same in ()obtaining

AG AC BG middotGH BC middot CH

Eliminating the common ratio from the last two proportionsyields () For meeting this condition the three pairs BH CGand DF of points are said to be in involution by Desarguesrsquosdefinition

Now suppose these pairs consist of those points where atransversal cuts the pairs of opposite sides of a complete quad-rangle as in Thomasrsquos description For example the pointsBCDFGH could be respectively the points ABCDEL inFigure or or ΑΒΓ∆ΕΖ in Figure on page Thenthe converse of Pappusrsquos Lemma IV expressed in () gives usnow

BH middot CD BC middotDH BH middot FG BF middotGH ()

This proportion is equivalent to () However the best way toshow this may not be obvious One approach is to introducethe cross-ratio though apparently Desargues does not do this[ p ] Despite the misgivings expressed above (page )we turn to modern notation for ratios If ABCD are pointson a straight line we let

AB middot CD

AD middot CB= (ABCD) ()

by definition Note the pattern of repeated letters on the leftWe could use a different pattern we just have to be consistentWe take line segments to be directed so that CB = minusBCThen () is equivalent to

(BHDC) = (BHGF ) ()

Introduction

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

while since () is in modern notation

DG middot FG

CD middot CF=

BG middotGH

BC middot CH

we obtain from this

DG middot BC

CD middot BG=

CF middotGH

FG middot CH

that is

(DGBC) = (CFGH) ()

But () obtained from Pappus must still hold if we permutethe pairs BH CG and DF Sending each pair to the next(and the last to the first) we obtain from () the equivalentequation

(CGBD) = (CGFH) ()

We shall have that this is equivalent to () once we under-stand how cross-ratios are affected by permutations of theirentries

The cross-ratios that can be formed from ABCD are per-muted transitively by a group of order 24 Then there are atmost 6 different cross-ratios since

(CDAB) = (ABCD)

(BADC) = (ABCD)

Moreover from () we can read off

(ADCB) =1

(ABCD)

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

while

(ACBD) =AC middot BD

AD middotBC=

(AB +BC) middot (BC + CD)

AD middot BC

=AC middot BC +BC middot CD

AD middotBCminus (ABCD)

= 1minus (ABCD)

The involutions x 7rarr 1x and x 7rarr 1 minus x of the set of ratiosgenerate a group of order 6 (Here we either exclude the ratios0 and 1 or allow them along with infin) Now we have accountedfor all permutations of points We have

(CGBD) = (CGFH) lArrrArr (CBGD) = (CFGH)

lArrrArr (BCDG) = (CFGH)

lArrrArr (DGBC) = (CFGH)

that is () and () are equivalent as desiredThe involution of the straight line BE in Figure or

that transposes A and L will transpose B and E and alsoC and D Following Coxeter [ p ] we obtainthis transposition as in Figure as a composition of threeprojections of one straight line onto another In Coxeterrsquos no-

tation ABELG=and

FBNM means AGF EGN and LGM are

all straight (and implicitly ABEL and FBNM are straight)It follows from Lemma IV that this transformation is an invo-lution and is uniquely determined by the pairs AL and BE

Locus problems

Thomasrsquos anthology [ ndash] includes the account by Pap-pus of five- and six-line locus problems that Descartes quotes

Introduction

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

AB

Eb

b

b

F

G

L

b

b

b

AB

E

F

G

L

M

N

b

b

b

b

b

b

bb

ABELG=and

FBNM

BE

F

G

L

M

N

P

b

b

b

b

b

bb

b

FBNML=and

PENG

AB

E

F

G

LN

P

b

b

b

b

b

bb

b

PENGF=and

LEBA

Figure An involution

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4

Figure A five-line locus

in the Geometry [ pp ndash] Pappus suggests no solution tosuch problems but later in the Geometry Descartes solves aspecial case of the five-line problem where four of the straightlinesmdashsay ℓ0 ℓ1 ℓ2 and ℓ3mdashare parallel to one another eacha distance a from the previous while the fifth linemdashℓ4mdashis per-pendicular to them What is the locus of points such that theproduct of their distances to ℓ0 ℓ1 and ℓ3 is equal to the prod-uct of a with the distances to ℓ2 and ℓ4 One can write downan equation for the locus and Descartes does Effectively let-ting the x- and y-axes be ℓ2 and ℓ4 the positive direction ofthe latter being from ℓ3 towards ℓ0 Descartes obtains

y3 minus 2ay2 minus a2y + 2a3 = axy

This may allow us to plot points on the desired locus as inFigure but we could already do that The equation is thusnot a solution to the locus problem since it does not tell uswhat the locus is But Descartes shows that the locus is traced

Introduction

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

ℓ0 ℓ1 ℓ2 ℓ3

ℓ4 bc

bc

Figure Solution of the five-line locus problem

by the intersection of a moving parabola with a straight linepassing through one fixed point and one point that moves withthe parabola See Figure The parabola has axis slidingalong ℓ2 and a is its latus rectum The straight line passesthrough the intersection of ℓ0 and ℓ4 and through the pointon the axis of the parabola whose distance from the vertex isa

Descartesrsquos solution is apparently one that Pappus wouldrecognize as such Thus Descartesrsquos algebraic methods wouldseem to represent an advance and not just a different way ofdoing mathematics

See my article ldquoAbscissas and Ordinatesrdquo [] for more than you everimagined wanting to know about the term latus rectum

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Ancients and moderns

Modern mathematics is literal in the sense of relying on theletters that might appear in a diagram rather than the dia-gram itself The diagram is an integral part of ancient proofsPappus may put points of a diagram in a list without givingall of their relations For example the enunciation of LemmaI of our text is

῎Εστω καταγραφὴ Let the diagram beἡ ΑΒΓ∆ΕΖΗ ΑΒΓ∆ΕΖΗκαὶ ἔστω and let it be thatὡς ἡ ΑΖ πρὸς τὴν ΖΗ as ΑΖ is to ΖΗοὕτως ἡ Α∆ πρὸς τὴν ∆Γ so is Α∆ to ∆Γκαὶ ἐπεζεύχθω ἡ ΘΚ and let ΘΚ have been joinedὅτι [I say] thatπαράλληλός ἐστιν ἡ ΘΚ τῇ ΑΓ ΘΚ is parallel to ΑΓ

Reconstruct the diagram from that One needs to know firstof all that ΑΖΗ∆Γ are on a straight line ΑΒΕ are also on astraight line and one needs to know that Λ is to be on thisline when ΖΛ is constructed parallel to Β∆ We can infer thelocations of Θ and Κ from Pappusrsquos assertion that as ΕΒ isto ΒΛ so are ΕΚ to ΚΖ and ΕΘ to ΘΗ but still we mustunderstand that each ratio involves collinear points In hisdiagrams Hultsch gives five possible configurations meetingthese conditions my translation gives a sixth as in Figure where ΒΕΚΘ is emboldened to indicate that the lemma isa version of the Quadrangle Theorem where one of the sixstraight lines through the vertices of the quadrangle is parallelto a straight line cutting the others

Netz observes that Greek mathematics never simply declareswhat letters in a diagram stand for [ pp ndash]

Introduction

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Α

Β

Γ∆

Ε

Ζ Η

Θ Κ

Λ

Figure A diagram for Lemma I of Pappus

Nowhere in Greek mathematics do we find a moment of spec-ification per se a moment whose purpose is to make sure thatthe attribution of letters in the diagram is fixed

It may well be Descartes he says who first fixes such attribu-tions I note an early passage in La Geacuteomeacutetrie [ p ]

Mais souvent on nrsquoa pas besoin de tracer ainsi ces lignessur le papier et il suffit de les deacutesigner par quelques lettreschacune par une seule Comme pour ajouter la ligne BD agraveGH je nomme lrsquoune a et lrsquoautre b et eacutecris a + b et a minus b

pour soustraire b de a et ab pour les multiplier lrsquoune parlrsquoautre

We have already seen how Descartesrsquos abbreviations make so-lutions to ancient unsolved problems possible

Nonetheless it it possible to take cleverness with notationtoo far The third proof of Pappusrsquos Theorem given in thecourse is taken from Coxeter [ p ] and it may appealto modern Cartesian sensibilities but it gives less sense ofwhy the theorem is true than the second proof which involves

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A2

B1

C2

A1

B2

C1

C3

A3B3

Figure Pappusrsquos Theorem in modern notation

passing to a third dimension and projecting starting from thediagram of Lemma VIII Coxeterrsquos diagram for Pappusrsquos The-orem is labelled as in Figure it allows us to observe oncefor all that AiBjCk is straight whenever i j k = 123If one is going to be this clever it seems to me one might aswell go all the way and let the indices be 0 1 and 2 so thatone can use for the set of them the notation 3

I do not know whether Pappus (or Euclid) recognized a singletheorem lying behind Lemmas VIII XII and XIII He mayhave recognized a similarity between the lemmas but not asatisfactory way to prove the lemmas once for all He mayhave thought it important to treat each case individually

There is a point of view that blurs the distinctions betweenthe three conic sections but it not necessarily the best point ofview When Omar Khayyam used conic sections to solve whatwe should call cubic equations he had to consider several casesdepending on what we should call the signs of the coefficientsof the equations [ pp ndash] For example for the casewhere ldquoa cube and sides are equal to squares and numbersrdquowe can write the problem as the equation

x3 + b2x = cx2 + b2d

Introduction

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

which we manipulate into

x2

b2=

dminus x

xminus c

We can define y in terms of a solution so that

x

b=

y

xminus c=

dminus x

y

Thus we solve the original equation by finding the intersectionof the two conics given by

x2 minus cx = by y2 + (xminus c) middot (xminus d) = 0

These are normally a parabola and a circle however if wehave allowed negative coefficients then we may have had tolet b be imaginary This matters if we want to construct realsolutions

In Rule Four of the posthumously published Rules for theDirection of the Mind [ p ] Descartes writes of amethod that is

so useful that without it the pursuit of learning would Ithink be more harmful than profitable Hence I can readilybelieve that the great minds of the past were to some ex-tent aware of it guided to it even by nature alone Thisis our experience in the simplest of sciences arithmetic andgeometry we are well aware that the geometers of antiquityemployed a sort of analysis which they went on to apply tothe solution of every problem though they begrudged reveal-ing it to posterity At the present time a sort of arithmeticcalled ldquoalgebrardquo is flourishing and this is achieving for num-bers what the ancients did for figures

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

We have already observed that Book VII of Pappusrsquos Collec-

tion concerned the Treasury of Analysis The term ldquotreasuryrdquois a modern flourish but our word ldquoanalysisrdquo is a translitera-tion of the Greek of writers like Pappus It means freeing upor dissolving As Pappus describes it mathematical analysisis assuming what you are trying to find so that you can workbackwards to see how to get there We do this today by givingwhat we want to find a name x

Today we think of conic sections as having axes one forthe parabola and to each for the ellipse and hyperbola Thenotion comes from Apollonius but for him an axis is justa special case of a diameter A diameter of a conic sectionbisects the chords of the section that are parallel to a cer-tain straight line This straight line is the tangent drawn at apoint where the diameter meets the section Apollonius showsthat every straight line through the center of an ellipse or hy-perbola is a diameter in this sense and every straight lineparallel to the axis is a diameter of a parabola As far as I cantell it is difficult to show this by the methodrsquos of Descartesrsquosanalytic geometry Like Euclidrsquos and Pappusrsquos proofs Apollo-niusrsquos proofs rely on areas There are areas of scalene trianglesand non-rectangular parallelograms Earlier I discussed a lo-cus problem in terms of distances from several given straightlines For Pappus what is involved is not distances as suchbut the lengths of segments drawn to the given lines at givenangles which are not necessarily right angles The apparentlygreater generality is trivial This is why Descartes can solvea five-line problem using algebra But if one is going to provethat a straight line parallel to the axis of a parabola is a di-ameter one cannot just treat all angles as right Apolloniusmay have had a secret weapon in coming up with his proposi-tions about conic sections but I donrsquot think it was Cartesian

Introduction

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

analysis Such at least was my impression when going throughApollonius for my course in Şirince in

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Monday September

Statement of the Quadrangle Theorem

Suppose five points A through E fall on a straight line as inFigure a and F is a random point not on the straight lineJoin FA FB and FD Now let G be a random point on FAas in Figure b and join GC and GE Supposing these twostraight lines cross FB and FD at H and K respectively joinHK as in Figure If this straight line crosses the originalstraight line AB at L then L depends only on the original fivepoints not on F or G Let us call this result the Quadrangle

Theorem It is about how the straight line AB crosses the sixstraight lines that pass through pairs of the four points F GH and K Any such collection of four points no three of whichare collinear together with the six straight lines that they

AB

DCE

b

b

bb

b

Fbc

(a) Point F is chosen

AB

DCE

b

b

bb

b

F

G

bc

bc

(b) Point G is chosen

Figure The Quadrangle Theorem set up

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A

B

DC

E

b

b

bb

b

F

G

HK

L

bc

bc

bc bc bc

Figure The Quadrangle Theorem

determine as in Figure a is called a complete quadrangle

(tam doumlrtgen) Similarly any collection of four straight linesno three passing through the same point together with the sixpoints at the intersections of pairs of these six straight lines asin Figure b is a complete quadrilateral (tam doumlrtkenar)

As stated the Quadrangle Theorem is a consequence ofLemma IV of our text of Pappus Lemmas I II V VI andVII treat other cases such as when HK in Figure is parallelto AE

Statement of Desarguesrsquos Theorem

There is an alternative proof of the Quadrangle Theorem basedon Desarguesrsquos Theorem Desargues was a contemporary ofDescartes and the theorem named for him is that if thestraight lines through corresponding vertices of two triangles

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

bc

bc

bc

bc

(a) A complete quadrangle

bc

bc

bc

bc

bc

bc

(b) A complete quadrilateral

Figure ldquoCompleterdquo figures

intersect at a common point as AD BE and CF meet atH in Figure then the intersection points of correspondingsides of the triangles lie in a single straight line If two pairsof corresponding sides are parallel this means the third pairmust be parallel as well

Statement of Pappusrsquos Theorem

We shall prove Desarguesrsquos Theorem by means of Pappusrsquos

Hexagon Theorem This is the theorem that if the verticesof a hexagon lie alternately on two straight lines as do thevertices of ABCDEF in Figure then the points of inter-section as G H and K of the three pairs of opposite sides ofthe hexagon also lie on a straight line Again there are othercases as when two of the pairs of opposite sides are parallelhere the third pair must be parallel as well

Pappusrsquos Theorem is Lemmas VIII XII and XIII in our textIn the proofs of XII and XIII Lemmas III X and XI are usedLemma VIII needs only the theorem of Euclid (namely I

Monday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A

B

C

D

E

F

G

HK

Figure Pappusrsquos Hexagon Theorem

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A B

C D

Figure Triangles on the same base

and of the Elements []) that when two triangles have thesame base as in Figure then the triangles are equal to oneanother just in case the straight line joining the apices of thetriangles is parallel to their common base

Equality and sameness

Equality of triangles means not congruence of the trianglesbut sameness of their areas In fact Euclidrsquos proof that paral-

lelograms on the same base and in the same parallels are equalis by cutting the parallelograms into congruent pieces as inFigure where parallelograms ABDC and ABFE are equalbecause when CH = DG and FL = EG while CK and FMare both equal to DE then triangles CHK and MLF arecongruent to one another (and to DGE) and the pentagonsAGDKH and BLMEG are congruent to one another Beinghalf of equal parallelograms triangles ABC and ABE are nowequal to one another

Monday

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A B

C D E F

GH

K

L

M

Figure Equal parallelograms

Statement of Thalesrsquos Theorem

After Lemma VIII Pappusrsquos proofs of other cases of the Hexa-gon Theorem use what is known as Thalesrsquos Theorem Thisis that if the straight line DE cuts the sides of triangle ABCas in Figure then

DE BC lArrrArr AD DB AE EC ()

A

B C

D E

F

Figure Thalesrsquos Theorem

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

We obtain the alternative formulation

DE BC lArrrArr AD AB DE BC

by drawing DF parallel to AC and using a rule like

a b c d lArrrArr a+ b b c+ d d

The proportion AD DB AE EC is an expression not ofthe equality of the ratios AD DB and AE EC but of theirsameness An equation of bounded straight lines like CH =DG earlier says that the lengths of the straight lines are thesame But the length of a line cannot be drawn in a diagramit is an abstraction from the line itself Equality of boundedstraight lines is an equivalence relation and for the sake ofhaving a formal definition we can understand the length ofCH as the corresponding equivalence class consisting of all ofthe straight lines that are equal to CH Similarly the ratioof two straight lines is abstract and cannot be drawn in adiagram We shall discuss proportions later today and moreformally on Wednesday

Thales himself

There is little evidence that Thales knew in full generality thetheorem named for him I have learned this while preparing forthe Thales Meeting to be held on Saturday September inThalesrsquos home town of Miletus Thales supposedly measuredthe heights of the Pyramids by considering their shadows buthe may have done this just when his own shadow was as longas a person is tall since in this case the height of the pyra-mid would be the same as the length of its own shadow (asmeasured from the center of the base of course)

Monday

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

bce Thales Euclid ce Diogenes Laeumlrtius Pappus Proclus

Figure Dates of some ancient writers and thinkers

Thales may have recognized that two triangles are congruentif they have two angles equal respectively to two angles andthe common sides equal (This is the so-called Angle-Side-Angle or ASA Theorem) According to Proclus who wrote acommentary on the Book I of Euclidrsquos Elements [] Thalesalso knew the following three theorems found in that book

) the diameter of a circle divides the circle into two equalparts

) vertical angles formed by intersecting straight lines areequal to one another

) the base angles of an isosceles triangle are equal to oneanother

According to Diogenes Laeumlrtius [ indash] (who wrote biogra-phies of philosophers starting with Thales) Thales also knewthat

) the angle inscribed in a semicircle is rightDates of activity for the persons we have named are roughly asin Figure All four of the listed theorems can be understoodto be true by symmetry For example the equation

angABC = angCBA

basically establishes the equality of vertical angles Also sup-pose we complete the diagram of an angle inscribed in a semi-circle as in Figure Here the quadrilateral BCDE has four

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

AB

C

D

E

Figure The angle in a semicircle

equal angles If it follows that those angles must be right thenthe theorem of the semicircle is proved

Those four equal angles are right in Euclidean geometryHere by Euclidrsquos fifth postulate if the angles at DCB andCBE are together less then two right angles then CD andBE must intersect when extended In that case for the samereason they intersect when extended in the other directionbut this would be absurd

Non-Euclidean planes

In Lobachevskirsquos geometry Euclidrsquos fifth postulate is deniedIn this case if AB is perpendicular to AC as in Figure there will be an angle ABD less than a right angle such thatBD never intersects AC no matter how far extended thoughstraight lines through B making with BA an angle less than

Monday

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A

B

C

DE

Figure Straight lines in the hyperbolic plane

ABD will meet AC If EBA = ABD then BE too will nevermeet AC though BE and BD are different straight lines

Lobachevskirsquos geometry may be understood as taking placein the hyperbolic plane We can understand Pappusrsquos ge-ometry as taking place in the projective plane Here theseveral cases of (for example) the Hexagon Theorem can begiven a single expression because formerly parallel straightlines are now allowed to intersect ldquoat infinityrdquo We obtain theprojective plane from the Euclidean plane by adding

) for each family of parallel straight lines a new pointcommon to all of them namely their point at infinity

and) a new straight line the straight line at infinity which

consists precisely of all of the points at infinity

In the projective plane

) any two distinct points lie on a single straight line as inEuclidean geometry but now also

) any two distinct straight lines intersect at a single point(which could be at infinity)

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A B

C

D

E

F

G

Figure For a definition of proportion

Proportions

We said that Pappusrsquos proof of the Hexagon Theorem usedThalesrsquos Theorem How can we prove Thalesrsquos Theorem Wefirst have to define proportions One approach is to say thatby definition in Figure where angles ABC and ADE areright we have

AB BC AD DE

But then if BF = BC and DG = DE and BF is parallelto DG we want to show that AFG is a straight line Thisis a consequence of Desarguesrsquos Theorem Indeed in place ofthe assumption that DG = DE assume that AFG is straightSince BC DE and BF DG it follows by Desargues thatCF EG In that case since in triangle BCF the angles at Cand F are equal to one another the same is true of the anglesat E and G in triangle DEG (This uses the consequence ofEuclidrsquos fifth postulate that parallel straight lines make equalangles with the same straight line) As a result DE = DGThus if this equation is true by construction then AFG mustbe straight

Monday

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Desargues

Pappus

4ltclubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

clubsclubsclubsclubsclubsclubsclubsclubsclubsclubs

Thales

ck

Figure A circle of implications

In the end we shall have the implications in Figure atleast with the assumption of a certain part of Book I of EuclidrsquosElements

In fact the theorems of Desargues Pappus and Thales are allsimply true in the full geometry of Book I of Euclidrsquos ElementsWe shall show this for Thalesrsquos Theorem on Wednesday

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Tuesday September

The parallel case of Pappusrsquos Theorem

We can understand Pappusrsquos Theorem (or the HexagonTheorem) as having the six cases depicted in Figure Ineach case the vertices of hexagon ABCDEF lie alternatelyon two straight lines but these may be parallel or not andof the pairs (ABDE) and (BCEF ) of opposite sides of thehexagon two one or none may be parallel

Pappusrsquos Lemma VIII is the case where the two straightlines intersect and the two pairs of opposite sides of the hexa-gon are parallel In particular letting the hexagon be ΒΓΗΕ∆Ζin Figure a we suppose

ΒΓ ∆Ε ΗΕ ΖΒ

We prove ΓΗ ∆Ζ using several equations of triangles

) ∆ΒΕ = ∆ΓΕ [Elements I since ΒΓ ∆Ε]

) ΑΒΕ = Γ∆Α [add ∆ΑΕ]

) ΒΖΕ = ΒΖΗ [Elements I since ΒΖ ΕΗ]

) ΑΒΕ = ΑΗΖ [subtract ΑΒΖ]

) ΑΓ∆ = ΑΗΖ [steps and ]

) Γ∆Η = ΓΖΗ [add ΑΓΗ]

) ΓΗ ∆Ζ [Elements I]

In fact if Pappusrsquos figure were more like those in Figure itmight be as in Figure b and then the proof would have to

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A B

C

DE

F

Lemma VIII

A B

C

DE

F

A B

C

DE

F

A B

C

DE

F

Lemma XII

A B

C

D

E

F

Lemma XIII

A B

C

D

E

F

Figure Cases of Pappusrsquos Theorem

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Α

Β Γ

∆ Ε

Ζ

Η

(a) Lemma VIII itself

Α

ΗΕ

∆

ΖΒ

Γ

(b) Another version of Lemma VIII

Figure A case of Pappusrsquos Theorem

Tuesday

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Α

ΒΓ

∆Ε

ΖΗΘ

Κ

Λ

ΜΝ

Ξ

Figure Lemma IV

be adjusted by for example subtracting ΒΖΕ and ΒΖΗ from

ΑΒΖ in step

Pappusrsquos proof of the Quadrangle Theorem

Lemma IV is that if as in Figure the given points Α ΒΓ ∆ Ε and Ζ along a straight line satisfy the proportion

ΑΖ middot ΒΓ ΑΒ middot ΓΖ ΑΖ middot ∆Ε Α∆ middot ΕΖ ()

then Θ Η and Ζ are in a straight line To prove this byalternation of () we obtain

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΑΒ middot ΓΖ Α∆ middot ΕΖ ()

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Considering first the left-hand member we have by simplifi-cation

ΑΖ middot ΒΓ ΑΖ middot ∆Ε ΒΓ ∆Ε

and we write the latter ratio as a composition of three ratios

ΒΓ ∆Ε ΒΓ ΚΝ amp ΚΝ ΚΜ amp ΚΜ ∆Ε

Now we analyze the right-hand member of () as a composite

ΑΒ middot ΓΖ Α∆ middot ΕΖ ΒΑ Α∆ amp ΓΖ ΖΕ

Assuming ΚΜ is drawn parallel to ΑΖ by Thalesrsquos Theoremwe have

ΝΚ ΚΜ ΒΑ Α∆

Eliminating this common ratio from either member of ()and reversing the order of the new members we obtain

ΓΖ ΖΕ ΒΓ ΚΝ amp ΚΜ ∆Ε

and therefore by Thalesrsquos Theorem applied to each ratio inthe compound

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΗ ΗΕ ()

Pappus says now that ΘΗΖ is indeed straight Although heprovides a reminder he may expect his readers to know assome students today know from high school the generalizationof Thalesrsquos Theorem known as Menelausrsquos Theorem whosediagram is in Figure Here ΕΞ ΘΓ and ΘΗ is extended

Menelausrsquos Sphaerica survives in Arabic translation [ p ] butwe also have Menelausrsquos Theorem in Ptolemy where I read it as a studentat St Johnrsquos College just before Toomerrsquos translation [] came out

Tuesday

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Γ

Ε

Ζ

ΗΘ

Κ

Ξ

ΓΖ ΖΕ ΓΘ ΘΚ amp ΚΗ ΗΕ lArrrArr ΘΗΖ is straight

Figure Lemma from Lemma IV

to Ξ Then from () we have

ΓΖ ΖΕ ΘΓ ΚΘ amp ΚΘ ΕΞ

ΘΓ ΕΞ

we used the translation that Taliaferro had made for the College [ Ip ] Thomas also puts Menelausrsquos Theorem in his anthology [ pp ff] In the commentary for their translation of Desargues Field andGray remark that Pappusrsquos Lemma IV is proved by ldquochasing ratios muchin the fashion Desargues was later to use In this case collinearity couldhave been established by appealing to the converse of Menelausrsquo theorembut when Pappus reached that point he missed that trick and continuedto chase ratios until the conclusion was establishedmdashin effect proving theconverse of Menelausrsquo theorem without saying sordquo [ pp ndash] I wouldadd that the similarity of ldquofashionrdquo in Pappus and Desargues is probablydue to the latterrsquos having studied the former Moreover Pappus seemsnot to have ldquomissed the trickrdquo since he asserts the desired collinearity at apoint when it can be recognized only by somebody who knows MenelausrsquosTheorem

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

ΑΒ Γ ∆ Ε

Ζ

Λ

Κ

Θ

Η

Τ

Σ

Ρ Π

Figure The Quadrangle Theorem

By Thalesrsquos Theorem again the points Θ Ξ and Ζ must becollinear and therefore the same is true for Θ Η and Ζ Thiscompletes the proof of Lemma VIII

The steps of the proof are reversible Thus if we are giventhe complete quadrangle ΗΘΚΛ of Figure and the pointsΑ Β Γ ∆ Ε and Ζ where its sides cross a given straight linethe proportion () must be satisfied Therefore if five sides ofanother complete quadrangle as ΠΡΣΤ in Figure shouldpass through the points Α Β Γ ∆ and Ε then the sixth sidemust pass through Ζ This is the Quadrangle Theorem

Tuesday

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Wednesday September

Euclidrsquos proof of Thalesrsquos Theorem

Suppose in Figure DE BC Then DEB = DEC so

AD DB ADE DBE

ADE ECD AE EC

Conversely if AD DB AE EC then

ADE DBC AD DB

AE EC ADE BCE

and so BCD = BCE which gives us DE BC This isEuclidrsquos proof of Thalesrsquos Theorem

In any proportion any of the four terms be it a straightline a plane figure or a solid can be replaced by a term of

A

B C

D E

Figure Thalesrsquos Theorem

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

a b

c d

Figure A proportion of lengths and areas

the same kind Beyond this the foregoing proof relies on twoproperties of proportions

If a and b are bounded straight lines and c and d areplane figures then the proportion

a b c d

is true if and only if there is a rectangle as in Figure

Sameness of ratio is indeed an equivalence relation Inparticular it is transitive

These properties follow from Euclidrsquos definition of proportionfound in Book V of the Elements But this definition relies onthe Archimedean Axiom that for any two magnitudes thathave a ratio some multiple of either of them must exceed theother

A simple definition of proportion

We can avoid using the Archimedean Axiom by first of allletting Property above be a definition Moreover if a b c

Wednesday

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

and d are all straight lines we define

a b c d

if for some plane figures e and f

a b e f c d e f

But we must check the following Suppose this condition ismet and also g and h are plane figures such that

a b g h

Does it follow that c d g h It does as is shown in Figurea which can be completed as in Figure b Here the lineacross the big parallelogram (which may not be a rectangle)really is straight so that the the smaller parallelograms areequal in pairs as shown

Cross ratio in Pappus

In Lemma III the straight lines ΘΕ and Θ∆ cut the straightlines ΑΒ ΓΑ and ∆Α as in Figure The diagram is com-pleted by making

ΚΛ ΖΓΑ ΛΜ ∆Α

We first haveΕΖ ΖΑ ΕΘ ΘΛ

ΑΖ ΖΗ ΘΛ ΘΜ()

so ex aequali

ΕΖ ΖΗ ΕΘ ΘΜ

ΘΕ middot ΗΖ = ΕΖ middotΘΜ

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

a b

c

d

e f

f

g h

(a)

a b

c

d

ef

f

g h

g

h

(b)

Figure A definition of proportion

Wednesday

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

b

Α

ΒΓ

∆

Ε ΖΗ

Θ

Κ

Λ

Μ

Figure Lemma III

Being equal these areas have the same ratio to ΕΖ middotΘΗ andso

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΕΖ middotΘΜ ΕΖ middotΘΗ

ΘΜ ΘΗ

ΛΘ ΘΚ ()

The ratio ΛΘ ΘΚ is independent of the choice of the straightline through Θ that cuts the three straight lines that passthrough Α In particular we can conclude immediately

ΘΕ middot ΗΖ ΕΖ middotΘΗ ΘΒ middot ∆Γ ΒΓ middotΘ∆

which is the theoremThe ratio ΘΕ middot ΗΖ ΕΖ middot ΘΗ can be called the cross ratio

(ccedilapraz oran) of the points Θ Ε Ζ and Η By the lemmaif four straight lines in a plane intersect at a point then thecross ratio of the four points where some other straight linecrosses the lines is always the same We can see this usingFigure

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Figure Invariance of cross ratio in Pappus

b

Α

Β

Γ

∆ Ε

Ζ

Η

ΘΚΛ

Μ Ν

Figure Lemma X

Wednesday

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

b

Α

Η ΖΕ

Θ

Λ

Κ

Μ

Figure Lemma III reconfigured

Lemma X is a converse to Lemma III The diagram is asin Figure where the hypothesis is

Θ∆ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΕ middot ΖΗ ()

We make ΚΛ parallel to ΓΑ and then extend ΑΒ and Α∆to meet ΚΛ at two points which Pappus writes as Κ and Λthough today we might say this is the wrong order We ensureΛΜ Α∆ and ΚΝ ΑΒ with ΕΘ extended to Μ and ∆Θto Ν Now we repeat part of the proof of in Lemma III atleast in the configuration of Figure with two of the threeconcurrent straight lines interchanged That is using only thepart of the diagram shown in Figure a we have

∆Θ ΘΝ ∆Γ ΓΒ

∆Θ middot ΓΒ = ∆Γ middotΘΝ

∆Θ middot ΒΓ ∆Γ middot ΒΘ Γ∆ middotΘΝ ∆Γ middot ΒΘ

ΘΝ ΘΒ

ΚΘ ΘΛ

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

b

Α

Β

Γ

∆

Θ

Κ

Λ

Ν

(a)

b

Α

Ε

Ζ

Η

Θ Κ

Λ

Μ

(b)

Figure Lemma X two halves of the proof

Now we move to the part of the diagram shown in Figure bwhere we have proportions corresponding to the last two

ΚΘ ΘΛ ΗΘ ΘΜ

ΘΗ middot ΖΕ ΘΜ middot ΖΕ

In sum we have shown

∆Θ middot ΒΓ ∆Γ middot ΒΘ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

But the left member already appears in () Hence the rightmembers of the two proportions are the same that is

ΘΗ middot ΖΕ ΘΕ middot ΖΗ ΘΗ middot ΖΕ ΘΜ middot ΖΕ

ΘΕ middot ΖΗ = ΘΜ middot ΖΕ

ΘΜ ΘΕ ΗΖ ΖΕ

Thus what we did in Figure a we have done in reverse inFigure b It remains to draw the conclusion ΑΖ ΚΛ cor-responding to the hypothesis ΑΓ ΚΛ Pappus argues as

Wednesday

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

follows (the bracketed proportions replaced with a referenceto addition and alternation)

[

ΘΜ+ΘΕ ΘΕ ΗΖ+ ΖΕ ΖΕ

ΜΕ ΘΕ ΗΕ ΖΕ

]

ΜΕ ΕΗ ΘΕ ΕΖ

ΛΕ ΕΑ ΘΕ ΕΖ

and so ΑΖ ΚΛ which means ΓΑΖ must be straight

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Friday September

Addition and multiplication

In Euclid a bounded straight line (sınırlanmış doğru ccedilizgi)is called more simply a straight line (doğru ccedilizgi) and moresimply still a ldquostraightrdquo (doğru) In English this is usuallycalled a line segment (doğru parccedilası) although for Euclidand sometimes in English too a line (ccedilizgi) may be curvedFor example a circle is a certain kind of line For us thoughhenceforth lines will always be straight and unbounded

It is clear how to add the lengths of two line segments placethe segments end to end in one straight line Thus in Figure

AB +BC = AC

It is implicit in Euclid and Pappus and we are making itexplicit now that AB for example can be reversed and placedon itself so that A is on B and B is on A briefly AB = BAThen

BC + AB = CB +BA = CA = AC

so addition is commutativeWe can consider addition in another way We select a point

A on a Euclidean unbounded straight line If B and C are

b b b

A B C

Figure Addition of line segments

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

b b b b

b b b b

C A D B

A C B D

Figure Sum of two points with respect to a third

also points on the line distinct from one another and from Athen there are two points D on the line such that

BD = AC

but for only one of them do we also have

CD = AB

See Figure We may refer to this D as the sum of B and

C with respect to A writing

D = B +A C = C +A B

If B and C are the same point then B +A C is the uniquepoint D different from A such that AB = BD Finally if oneof B and C is the same as A then B +A C is the same as theother Thus defined the operation +A makes the points of theline into an abelian group with neutral element A

Can we define a multiplication that makes this group intoa field We have already treated the product of two line seg-ments as a rectangle having sides equal to those two segmentsBut now we should like the product of line segments to be an-other segment

If a and b are points on the real number line as studied incalculus then a+ b and a middot b are also points on that line Thedirected segment from a to a+ b is equal to that from 0 to bBut how is a middot b defined

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

1 b

a

a middot b

Figure Descartesrsquos definition of multiplication

Descartes showed the way [] The position of 1 on the realline must be known More generally on our Euclidean straightline with point A chosen let us select another point B Wecan then denote by 1 the length of AB If a and b are twolengths then by the proportion

1 a b a middot b

we can define the length a middot b To be more precise since thislength depends on the segment AB we might write the lengthas

a middotAB b

We can construct a line segment having this length by meansof Thalesrsquos Theorem as in Figure

Why is the multiplication middotAB commutative We do knowfrom Thalesrsquos Theorem that if A B C and D are now arbi-trary line segments and A B C D then A middot D = B middot Cwhere the products are rectangles as before Since then B middotC =C middot B we obtain A C B D Therefore

1 b a a middot b

Friday

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

1 1

ab

ab ba

Figure Commutativity of Descartesrsquos multiplication

and so by interchanging a and b

1 a b b middot a

b middot a = a middot b

For an alternative proof of commutativity of multiplication wecan apply the case of Pappusrsquos Theorem established in LemmaVIII as in Figure

We can do Euclidean geometry in the set R2 of ordered pairsof real numbers We can do some geometry in K2 for any fieldK We can even allow K to be a non-commutative field (alsoknown as a division ring) but then Pappusrsquos Theorem willnot be true in this geometry We shall see this in more detailtomorrow

A non-commutative field

Meanwhile an example of a non-commutative field is H thefield or rather skew field of quaternions discovered by

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Hamilton We can obtain this field from the field C of com-plex numbers roughly as C is obtained from R Indeed by onedefinition

C =

(

x yminusy x

)

x isin R amp y isin R

We can write x for(

x 00 x

)

and i for(

0 1minus1 0

)

so

(

x yminusy x

)

= x+ yi

Now C is a commutative field and it has the automorphismz 7rarr z given by

x+ yi = xminus yi

We define

H =

(

z wminusw z

)

z isin C amp w isin C

One shows that this is an additive abelian group and is closedunder multiplication and nonzero elements have multiplica-tive inverses it is a division ring We can write z for

(

z 00 z

)

and j for(

0 1minus1 0

)

so

(

z wminusw z

)

= z + wj

Then

ij =

(

i 00 minusi

)(

0 1minus1 0

)

=

(

0 i

i 0

)

ji =

(

0 1minus1 0

)(

i 00 minusi

)

=

(

0 minusi

minusi 0

)

Friday

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

b

Α

Β Γ

∆

Ε

Ζ Η

Θ

Figure Lemma XI

Intersecting case of Pappusrsquos Theorem

Pappusrsquos Lemma XI is that in Figure

∆Ε middot ΖΗ ΕΖ middot Η∆ ΓΒ ΒΕ

Since ΓΘ is drawn parallel to ΑΕ we have the two proportions

ΓΑ ΑΗ ΓΘ ΖΗ

ΓΑ ΑΗ Ε∆ ∆Η

and therefore

Ε∆ ∆Η ΘΓ ΖΗ

Ε∆ middot ΖΗ = ΓΘ middot ∆Η

Ε∆ middot ΖΗ ∆Η middot ΕΖ ΓΘ middot ∆Η ∆Η middot ΕΖ

ΓΘ ΕΖ

ΓΒ ΒΕ

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Α

ΒΓ

∆

b

Ε

Ζ

Η

Θ

Figure Lemma XI variant

as desired By Lemma III if ΕΓ meets Α∆ at a point Κ then

Ε∆ middot ΖΗ ΕΖ middot Η∆ ΕΚ middot ΓΒ ΕΒ middot ΓΚ

ΓΒ ΕΒ amp ΕΚ ΓΚ

In Lemma XI the point K has been sent off to infinity whichmeans ΕΚ ΓΚ has become the unit ratio

Pappus alludes to another case presumably as in Figure There is no change in the proof

Lemma XII is that in Figure the points Η Μ and Κare on a straight line The proof considers the parts of thediagram shown in Figure Pappusrsquos argument is

a) By Lemma XI

∆Ζ ΖΓ ΓΕ middot ΗΘ ΓΗ middotΘΕ ()

b) By Lemma XI again inversion and ()

ΓΖ Ζ∆ ∆Ε middot ΛΚ ∆Κ middot ΛΕ

∆Ζ ΖΓ ∆Κ middot ΛΕ ∆Ε middot ΛΚ

ΓΕ middot ΗΘ ΓΗ middotΘΕ ∆Κ middot ΛΕ ∆Ε middot ΚΛ

Friday

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Α Β

Γ ∆

Ε

Ζ

Η

Θ

Κ

Λ

Μ

Figure Lemma XII

Α

Γ ∆

Ε

Ζ

ΗΘ

(a)

Β

Γ ∆

Ε

Ζ

ΚΛ

(b)

Γ ∆

Ε

Η

Θ

Κ

Λ

Μb b

(c)

Figure Steps of Lemma XII

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

c) By Lemma X then ΗΜΚ is straightLemma XIII is similar with Lemma III used in place of

Lemma XI This gives us Pappusrsquos Theorem as proved by Pap-pus in the three cases labelled in Figure on page

A second proof of Pappusrsquos Theorem

Alternatively from Lemma VIII except for the easier casewhere everything that might be parallel is parallel all othercases of Pappusrsquos Theorem can be derived by projection

If a diagram is drawn on a transparent notebook cover andthe cover is raised at an angle to the first page and a shadow ofthe diagram is cast on that page all straight lines will remainstraight but some parallel lines will cease to be so and someintersecting lines will become parallel

Specifically we can choose one straight line in a diagramthat will become the line at infinity for a new diagram Callthis line in the original diagram A All straight lines parallelto A will remain parallel in the new diagram Also one lineparallel to these in the new diagram will represent the line atinfinity of the old diagram Call this line in the new diagramB Lines that intersected on A become parallel in the newdiagram lines that were parallel to one another but not to Awill intersect on B See for example Figure

Tomorrow we shall obtain a third proof of Pappusrsquos Theoremby coordinatizing the projective plane

Friday

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

new line at infin

A B

CF

E D

(a) Lemma VIII

A B

C

DE

F

G Hold line at infin

(b) A new case

Figure Pappusrsquos Theorem by projection

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Saturday September

Cartesian coordinates

We saw yesterday that by choosing a point A on an infinitestraight line we could make the points of the line into anabelian group with the addition denoted by +A By choos-ing another point B on the line we obtained a multiplicationmiddotAB of lengths This is not quite a multiplication of pointsHowever given points C and D on the line we can define

C middotAB D = E

where E is a point on the line such that if a and b are thelengths of AC and AD then a middotAB b is the length of AE andalso A lies between B and E just in case it lies between C andD Equipped with addition and multiplication so defined theEuclidean line becomes a field

We can obtain an isomorphic field by considering ratios ofline segments On Wednesday we defined ratios of line seg-ments and of rectangles By this definition any two line seg-ments have the ratio of two rectangles Conversely for any tworectangles we can find line segments that have their ratio Forexample in Figure

ABFE DFKH AB BC

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A

B

C

D

E

F

G

H

K

L

Figure Line segments in the ratio of rectangles

since DFKH = BCGF Ratios constitute a group the oper-ation being composition since

(a b amp b c) amp c d a c amp c d a d

a b amp (b c amp c d) a b amp b d a d

Composition is commutative since if

a b c d

then by Thalesrsquos Theorem

ad = bc a c b d

(see Figure ) so that

a b amp b c a c

b c amp a b b c amp c d b d a c

We have defined the product of two line segments in twodifferent ways

) as the rectangle bounded by them or

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

b

d

a c

bc

ad

Figure Alternation of ratio

) as the straight line that has the same ratio to one of thetwo segments that the other one has to some predeter-mined segment

In the former case commutativity is immediate in the lattercase it follows from the theorems of Thales and Pappus aswe showed In either case we can define sums of ratios ofline segments through the usual process of finding a commondenominator

(a b) + (c d) = ad+ bc bd

We get the same result no matter which multiplication ofsegments we use

If we allow negative lengths and a zero length then the ratiosof straight lines compose a field which we shall call K

We shall now work in the Euclidean plane the plane inwhich the propositions of Book I of the Elements are true Ifwe choose a neutral point A we obtain an additive abeliangroup as we did on a line Associativity can be checked asin Figure The group can also be understood as a vectorspace over the field K of ratios Here if s is such an ratio andB is a point of the plane different from A we define

s middotA B = C

Saturday

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A

B

CD

B +A C

C +A D

B +A C +A D

Figure Associativity of addition

where C is the point on AB such that

AC AB = s

Now suppose A B and C are vertices of a triangle thatis none of the three is on the straight line determined by theother two Then B and C compose a basis of the vector spacewhose neutral point is A In particular for any point D of theplane there is a unique ordered pair (s t) of ratios meaning(s t) isin K2 such that

D = s middotA B +A t middotA C

Here (s t) is the pair of Cartesian coordinates of D withrespect to ABC Conversely every element of K2 correspondsto a point of the plane in this way

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Barycentric coordinates

Again we have chosen triangle ABC and for some D in thesame plane we have

D = s middotA B +A t middotA C

This fails if we replace A with some other point However theequations

D = A+A s middotA (B minusA A) +A t middotA (C minusA A)

= (1minus sminus t) middotA A+A s middotA B +A t middotA C

are still correct if the subscripts A are replaced with any otherpoint Thus we may write simply

D = (1minus sminus t)A + sB + tC

Conversely if p q and r are three ratios whose sum is 1 thenthe linear combination

pA + qB + rC

is unambiguous Considering ABC as fixed we may write thesame point pA+ qB + rC as

(p q r)

But now we can allow

(p q r) = (tp tq tr)

if t 6= 0 Thus given arbitrary ratios p q and r whose sum isnot 0 we have

(p q r) =p

p+ q + rA +

q

p+ q + rB +

r

p+ q + rC

=q

p+ q + rmiddotA B +A

r

p+ q + rmiddotA C

Saturday

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A F B

D

C

E

Figure Cevarsquos Theorem

This point has the barycentric coordinates p q and r buteach must be considered together with the sum p+ q+ r Theidea is that the point is the center of gravity (the barycenter) ofthe system with weights p q and r at A B and C respectivelyIf we define

D =q

q + rB +

r

q + rC

then D is a point on BC and

(p q r) =p

p+ q + rA+

q + r

p+ q + rD

so (p q r) is a point on AD Similarly we can define E andF on AC and AB respectively so that (p q r) is on BE andCF Note that

BD DC r q

and so on so that we have Cevarsquos Theorem in Figure the lines AD BE and CF have a common point if and onlyif

BD DC amp CE EA amp AF FB 1

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

We can understand (p q r) as the equivalence class con-sisting of all ordered triples (px qx rx) where x is a nonzeroratio The notation makes sense for all ratios p q and r Weare going to show how to extend the Euclidean plane to theprojective plane by giving geometric meaning to (p q r)when p+ q + r = 0 but at least one of p q and r is not 0

Projective coordinates

For every straight line in the plane there are ratios a b and cwhere at least one of a and b is not 0 such that the straight lineconsists of the points such that if their Cartesian coordinatesare (s t) then

as + bt + c = 0

If the same point has barycentric coordinates (p q r) wherep+ q + r = 1 then (s t) = (q r) and so

0 = aq + br + c(p+ q + r)

= cp+ (a+ c)q + (b+ c)r

Thus the same line is given by

ax+ by + c = 0

in Cartesian coordinates and

cx+ (a+ c)y + (b+ c)z = 0

in barycentric coordinates The straight lines parallel to thisone are obtained by changing c alone Relabelling we nowhave that every straight line is given by an equation

ax+ by + cz = 0

Saturday

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

in barycentric coordinates where one of the coefficients a band c is different from the others (the coefficients are not allequal) We obtain parallel lines by adding the same ratio toeach coefficient

Thus if ap+ bq + cr = 0 and p+ q + r = 0 but (p q r) 6=(000) then (p q r) satisfies the equation of every straightline parallel to ax + by + cz = 0 and no other straight lineWe can understand (p q r) as the point at infinity of thestraight lines parallel to ax+ by+ cz = 0 The line at infinityis then defined by x+ y + z = 0

The projective plane now consists of points (p q r) where(p q r) 6= (000) The expression (p q r) consists ofprojective coordinates for the point

There is now a one-to-one correspondence between points ofthe projective plane and straight lines in that plane if thepoint is (p q r) in projective coordinates the straight lineis given by px+ qy + rz = 0

We may use the notation

P2(K) =

(x y z) (x y z) isin K3

(000)

Any triangle ABC in the Euclidean plane determines a sys-tem of barycentric coordinates for the points of the planehence a system of projective coordinates for the projectiveplane However suppose a fourth point D in the Euclideanplane does not lie on any of the three sides of ABC Then Dhas barycentric coordinates (micro ν ρ) with microνρ 6= 0 Thereis now a bijection

(x y z) 7rarr

(

x

microy

νz

ρ

)

from the set of points of the projective plane to itself Thisbijection takes the line ax + by + cz = 0 to the line microax +

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

νby+ρcz = 0 while fixing A B and C which are (1 0 0)(0 1 0) and (0 0 1) respectively The points that usedto be on the line at infinity are sent to the line

microx+ νy + ρz = 0

while of course the points now at infinity are on x+y+z = 0The only former point at infinity that is still at infinity is(ν minus ρ ρminus micro microminus ν) In particular if E is a point that usedto be at infinity we can have chosen D so that E is no longerat infinity

In this way for any quadrangle ABCD in the projectiveplane there is a system of projective coordinates in whichthe vertices of the quadrangle are (1 0 0) (0 1 0)(0 0 1) and (1 1 1) respectively

A third proof of Pappusrsquos Theorem

To prove Pappusrsquos Hexagon Theorem in yet another way againwe suppose the vertices of the hexagon ABCDEF lie alter-nately on straight lines ACE and BDF and we letbull AB and DE intersect at Gbull BC and EF intersect at H bull CD and FA intersect at K

We may now suppose further

A = (1 1 1) B = (1 0 0)

E = (0 1 0) K = (0 0 1)

as in Figure The point of the choice is that no line of thediagram contains two of B E and K but these are on thelines through A If B E and K are all on a straight line we

Saturday

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

A

B

C D

E

F

G H

Kb

b

bb

b

b

bb

b

Figure Pappusrsquos Hexagon Theorem

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

may replace them with F C and H For the lines through Awe have

AB = y = z AE = x = z AK = x = y

and so for some p q and r

G = (p 1 1) C = (1 q 1) F = (1 1 r)

Consequently

BF = z = ry EG = x = pz KC = y = qx

Since these three lines have a common point namely D wemust have D = (1 q rq) = (pr 1 r) = (p qp 1) and so

1 = prq = qpr = rqp

But we have also

BC = y = qz EF = z = rx KG = x = py

In particular H being the intersection of the first two of theselines we have

H = (1 qr r)

and this lies on KG since pqr = 1 Figure shows thesituation when p = 2 and q = 2 so r = 14

Saturday

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Sunday September

The projective plane over a field

Given a field K (possibly a ldquoskewrdquo or non-commutative field)we have

P2(K) =

(

K3

(000)

)

sim

where(p q r) sim (pprime qprime rprime)

if and only if for some t in K 0

pprime = tp qprime = tq rprime = tr

We may understand this both as the set of points in the pro-jective plane and as the set of lines in the projective planeThe point (p q r) sits on the line (a b c) if and only ifap + bq + cr = 0 The line at infinity is (1 1 1) that isthe line given by the equation x+ y+ z = 0 So the points atinfinity satisfy this

The Fano Plane

For the simplest example we may let K be the two-elementfield F2 thus obtaining the Fano Plane The four points ofF22 in barycentric coordinates are

(1 0 0) (0 1 0) (0 0 1) (1 1 1)

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

(1 1 1)

(1 0 0) (0 1 0)

(0 0 1)

(0 1 1)(1 0 1)

(1 1 0)

Figure The Fano Plane

There are three points at infinity in P2(K)

(1 1 0) (1 0 1) (0 1 1)

If we call these A B C D E F and G respectively thenthe seven lines are as follows

x = 0 = BCG y = z = ADG

y = 0 = ACF x = z = BDF

z = 0 = ABE x = y = CDE

and the line at infinity EFG All of this can be depicted asin Figure

Sunday

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

bcA

bcB

bcC

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

L

Figure Desarguesrsquos Theorem

A proof of Desarguesrsquos Theorem

We now prove Desarguesrsquos Theorem in P2(K) for arbitrary KThe diagram is as in Figure repeated as Figure In K3if triangles ABC and DEF lie in two different planes theneach of H K and L lies in each of those planes and thereforethe three points lie in the intersection of the two planes whichis a straight line (If the planes are parallel they meet at theircommon line at infinity) If ABC and DEF lie in the sameplane the diagram can still be considered as the ldquoshadowrdquoor projection of the three-dimensional case So DesarguesrsquosTheorem holds in P2(K)

If K is not commutative then Pappusrsquos Theorem does nothold in P2(K) However with three applications of Pap-pusrsquos Theorem alone we can prove Desarguesrsquos Theorem assketched in Figure using in turn the hexagons ABGMDC

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Construction

bcA

bc

Bbc C

bc

bc

bc

D

E

F

bcG

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

HNP is straight

bcA

bcB

bcC

bc

D

bcG

bc H

bcM

bcN

bcP

KNQ is straight

bc C

bc

bc

bc

D

E

F

bcG bc K

bcM

bcN

bcQ

HKL is straight

bc C

bc

D

bc

bc

bc

H

K

LbcM

bcN

bc

bc

P

Q

Figure Proof of Desarguesrsquos Theorem

Sunday

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

bc

bc bc

bc

bc

bc

bc

bcbc

A

BC

DE

F

G

H

K

Figure The dual of Pappusrsquos Theorem

CDFEGM and CMDQNP Strictly we have now provedDesarguesrsquos Theorem on the assumption of three axioms

Two points determine a line Two lines determine a point Pappusrsquos Theorem is true

Duality

The dual of a statement about the projective plane is obtainedby interchanging points and lines Thus the dual of PappusrsquosTheorem is that if the sides of a hexagon alternately meettwo points then the straight lines met by pairs of oppositevertices meet a common point So in the hexagon ABCDEF let AB CD and EF intersect at G and let BC DE andFA intersect at H as in Figure If the diagonals AD and

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

AB C D E

F

G

H

KL

M

N

P Q

Figure The Quadrangle Theorem

BE meet at K then the diagonal CF also passes throughK For we can apply Pappusrsquos Theorem itself to the hexagonADGEBH since AGB and DEH are straight Since AD andEB intersect at K and DG and BH at C and GE and HAat F it follows that KCF is straight

It now follows from the three axioms above that the dualof Desarguesrsquos Theorem is true But the dual is precisely theconverse

A second proof of the Quadrangle Theorem

We now use the converse of Desarguesrsquos Theorem twice andthe original Theorem once to prove the Quadrangle TheoremWe shall show that in Figure the line PQ passes throughF

Sunday

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

by the converse of Desarguesrsquos Theorem applied to tri-angles GHL and MNQ sincebull GH and MN meet at Abull GL and MQ meet at D andbull HL and NQ meet at E

and ADE is straight it follows that GM HN and LQintersect at a common point R (not drawn)

Likewise in triangles GHK and MNP sincebull GH and MN meet at Abull GK and MP meet at B andbull HK and NP meet at C

and ABC is straight it follows that KP passes throughthe intersection point of GM and HN which is R

So now we know that HN KP and LQ intersect at RTherefore by Desarguesrsquos Theorem the respective sidesof triangles HKL and NPQ intersect along a straightline But HK and NP intersect at C and HL and NQintersect at E and KL intersects CE at F thereforePQ must also intersect CE at F

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Bibliography

[] Apollonius of Perga On conic sections Great Books of the WesternWorld no Encyclopaedia Britannica Inc Chicago LondonToronto

[] Apollonius of Perga Conics Books IndashIII Green Lion Press SantaFe NM revised edition Translated and with a note and an ap-pendix by R Catesby Taliaferro with a preface by Dana Densmoreand William H Donahue an introduction by Harvey Flaumenhaftand diagrams by Donahue edited by Densmore

[] M Chasles Les trois livres de porismes drsquoEuclide Mallet-BachelierParis Electronic source gallicabnffr Bibliothegraveque delrsquoEcole polytechnique

[] Frances Marguerite Clarke and David Eugene Smith ldquoEssay pourles Coniquesrdquo of Blaise Pascal Isis ()ndash March Trans-lation by Clarke Introductory Note by Smith wwwjstororg

stable224736

[] H S M Coxeter Introduction to Geometry John Wiley amp SonsNew York second edition First edition

[] Arno Cronheim A proof of Hessenbergrsquos theorem Proc Amer

Math Soc ndash

[] Reneacute Descartes The Geometry of Reneacute Descartes Dover Publica-tions Inc New York Translated from the French and Latinby David Eugene Smith and Marcia L Latham with a facsimile ofthe first edition of

[] Reneacute Descartes The Philosophical Writings of Descartes volume ICambridge University Press translated by John CottinghamRobert Stoothoff and Dugald Murdoch

[] Reneacute Descartes La Geacuteomeacutetrie Jacques Gabay Sceaux France Reprint of Hermann edition of

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

[] Diogenes Laertius Lives of Eminent Philosophers Number inthe Loeb Classical Library Harvard University Press and WilliamHeinemann Ltd Cambridge Massachusetts and London Volume I of two With an English translation by R D Hicks Firstpublished Available from archiveorg and wwwperseus

tuftsedu

[] Euclid The Thirteen Books of Euclidrsquos Elements Dover Publica-tions New York Translated from the text of Heiberg with in-troduction and commentary by Thomas L Heath In three volumesRepublication of the second edition of First edition

[] J V Field and J J Gray The geometrical work of Girard DesarguesSpringer-Verlag New York

[] Thomas Heath A History of Greek Mathematics Vol II From

Aristarchus to Diophantus Dover Publications Inc New York Corrected reprint of the original

[] David Hilbert The Foundations of Geometry Authorized transla-tion by E J Townsend Reprint edition The Open Court Publish-ing Co La Salle Ill Based on lectures ndash Translationcopyrighted Project Gutenberg edition released December (wwwgutenbergnet)

[] Victor J Katz editor The Mathematics of Egypt Mesopotamia

China India and Islam A Sourcebook Princeton University PressPrinceton and Oxford

[] Morris Kline Mathematical Thought from Ancient to Modern

Times Oxford University Press New York

[] Reviel Netz The Shaping of Deduction in Greek Mathematics vol-ume of Ideas in Context Cambridge University Press Cambridge A study in cognitive history

[] Pappus of Alexandria Pappus Alexandrini Collectionis Quae Su-

persunt volume II Weidmann Berlin E libris manu scriptisedidit Latina interpretatione et commentariis instruxit FridericusHultsch

[] Pappus of Alexandria Book of the Collection Part Introduc-

tion Text and Translation Springer Science+Business Media New

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

York Edited With Translation and Commentary by Alexan-der Jones

[] Pappus of Alexandria Book of the Collection Part Commen-

tary Index and Figures Springer Science+Business Media NewYork Edited With Translation and Commentary by Alexan-der Jones Pages numbered continuously with Part

[] Dimitris Patsopoulos and Tasos Patronis The theorem of ThalesA study of the naming of theorems in school geometry textbooksThe International Journal for the History of Mathematics Educa-

tion () wwwcomapcomhistoryjournalindexhtml

[] David Pierce History of mathematics Log of a course matmsgsuedutr~dpierceCoursesMath-history June EditedSeptember pp size A -pt type

[] David Pierce St Johnrsquos College The De Morgan Journal ()ndash Available from educationlmsacukwp-content

uploads201202st-johns-collegepdf accessed October

[] David Pierce Abscissas and ordinates J Humanist Math()ndash Available at scholarshipclaremontedu

jhmvol5iss114

[] David Pierce Commensurability and symmetry matmsgsuedu

tr~dpierceGeometry July Submitted for publication pp size A -pt type

[] Proclus A Commentary on the First Book of Euclidrsquos ElementsPrinceton Paperbacks Princeton University Press Princeton NJ Translated from the Greek and with an introduction and notesby Glenn R Morrow Reprint of the edition With a forewordby Ian Mueller

[] Ptolemy The Almagest In Robert Maynard Hutchins editorPtolemy Copernicus Kepler volume of Great Books of the West-

ern World pages ndash Encyclopaeligdia Britannica Chicago Translated by R Catesby Taliaferro

[] Ptolemy Ptolemyrsquos Almagest Princeton University Press Translated and annotated by G J Toomer With a foreward by

Bibliography

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries

Owen Gingerich Originally published by Gerald Duckworth Lon-don

[] Rene Taton Lrsquo laquo Essay pour les Coniques raquo de PascalRevue drsquohistoire de science et de leurs applications ()ndash wwwperseefrwebrevueshomeprescriptarticlerhs_0048-7996_1955_num_8_1_3488

[] Ivor Thomas editor Selections Illustrating the History of Greek

Mathematics Vol II From Aristarchus to Pappus Number inthe Loeb Classical Library Harvard University Press CambridgeMass With an English translation by the editor

[] Arthur W Weeks and Jackson B Adkins A Course in Geometry

Plane and Solid Ginn and Company Lexington Massachusetts

Geometries