ISSN 0313 - 6825

FUNCTIONVolume 9 Part 4 " August 1985

----- ..---_. -- -- -_._-

A SCHOOL MATHEMATICS MAGAZINE

Published by Monash University

Function is a mathematics magazine addressed principallyto students in the upper forms of schools. Today mathematicsis used inmost of the sciences,physical, biolo9~cal andsoc i a 1, i n bus i ness management , in' e n9i nee r .in g . There are fewhum a n end e a v0 u r s, f rom we a the r p r ed i c t ion to s i tin9 0 f t r a f fie1 ights, that do not involve mathematics. F~nction containsasticles describing s6me of these uses of mathematics. It alsohas articles, for entertainment and instruction, about mathe­matics and its history. Each issue contains problems andsolutions are invited.

It is hoped that the student readers of F~nction will con­tribute material for publ ication. Articles, ideas, cartoons,comments, criticisms, advice are earnestly sought. Please sendto the editors your views about what can be done to makeFunction ~ore interesting for you.

EDITO RS : M• A. B. De a kin ( c hair man), J. J a f fa r, G. B. Pre s to n ,G.A. Watterson (all of Monash University); S.M.Brown( Swin bur ne Ins tit ute); K • McR. Evan s ( Sco t c h Col 1eg e) ;C. Fox (St Michael's Grammar Schoo,l); J.B. Henry( Vic tor i a Colle g e, Ru s den); P. E. K10 e den ( MurdochUniversity); J.M. Mack (Univ~rsity of Sydney).

BUSINESS MANAGER: . Joan Wi 11 iams(Tel. No.. (03) 541 0811Ext. 2591)

Articles, correspondence, problems (with or without solu­t ion s ) and other material for pub 1 i cat ion are i nv i ted. Addressthem to:

The Editors,Function,Department of Mathematics,Monash University,Clayton, Victoria, 3168.

Alternativelv correspondence may be addressed individuallyto any of the editors at the mathemati~s departments of theinstitutions shown above.

The magazine is published five times a year, appearing inFebruarv" Apri 1, June, August, October. Price for five issues(including postage): $8.o0~t~; sinqle issues $1.80. Paymentsshould be sent to the business manager at the above addr~ss:

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Reqistered for posting as. a periodical - "Category Btl

Mathematics finds, as we point out in our editorial state­ment opposite, many applications, some of them in fields thatdo not at first sight look "mathematical".

Professor Praeger's article on weaving (p.7) is a case inpoint, for the very simple question of how to tell if a certainweave, produced by a designer, will hang together or fall apartwhen it is actually woven, leads to some interesting and, inits details, quite difficult, mathematics.

perspective drawing is another such field." For themathematics this has produced, see Dr Stillwell's article onp.14.

Then mathematics may be applied to economics, and here thename of Vilfredo Pareto is much invoked, to industrial design(folding chairs) and to the calendar we use every day.

IJ

THE "FRONT COVER

This issue's front cover diagram shows a chair drawn froman unusual perspective. See pp.14 - 22.

CONTENTS

V~lfredo Pareto (1848-1923), a Distribution and a UsefulObservation. Neil S. Barnett 2

Mathematics and Weaving: I. Fabrics and how they hangtogether. Cheryl E. Praeger 7

The Calendar. S. Rowe 11

Susan's World First

Projective Geometry.

The Folding Chair

Perdix

Problem Section

J.C. Stillwell

13

14

23

27

28

VILFREDO PARETO (1848~1923)

A DISTRIBUTION AND A USEFUL

OBSERVATION

Neil S. Barnett,

Footscray Institute of Technology

Vilfredo Pareto was t~ained in the physical sciences and

practised as an engineer for twenty years before becoming in­

terested in the application of mathematics to economics. Al­

though born Italian he spent much of his working life in

Switzerland; "is first major work, 'Cours d'Economie Politique'

was based on a series of lectures that he had given at Lausanne.

Others of his works lnclude the'Manuale di, Economia Politica'

and 'Trait~ de sociologie g~n~rale'·.

One of the issues dealt with in the first of his works

concerned the so-called, 'law of income distribution'. He

concluded that income distribution shows a high degree.of

constancy for different times and countries. If the dist-ri­

bution is plotted on a logarithmic scale it appears as a line

with negative slope, the slope being remarkably consistent'

from population to population. This 'law'. has received much

criticism on various grounds but none-the-less seems a

reasonable approximation in many instances. It can be quan­

tified as follows:

Let the proportion of individuals of a population whoseB

income exceeds x be given by Px' then Px 8+1' hence

Log Px = LogB - (8 + l)Log x. (B and f3 a~e. ·population

constants B, S > 0). If the formula is applied only to those

above subsistence incomes it seems to be fairly generally

applicable; under such.circumstances 13 ~ 0·5.

There is a theoretical distribution that has evolved

from Pareto's observation, not s'urprisingly called the Pareto

distribution. A continuous random variable, X, is defined

(in fact replacing a + 1 above by a) s~ch that the proba­bility density function (p.d.f.) is

3

B

I a+lf(x) = ~

See the graph below.

where a > 0 and x ~ A

elsewhere.

f

A

Since f(x) is a p.d.f., then

f= 'J"" f(x)dx = Brx~:lJoox=A l x=A

Now, since a > 0, this gives

x

which means that. B

= 1

and

raA

af(x) = a+l' a > 0, x ~ A

x

and so the probability that x ~ X is

J;f(X)dX = (~)a for X > A.

The mean and variance of Pareto's distribution can be shown

4

to be E(X) if ex > 1

and V(X) =(Ct - 2)(ex - 1)2

if ex > 2.

It is interesting to note that:

. (i) if 0 < ex ~ 1 then the distribution has neithera finite mean nor finite variance,

(ii) if 1 < a ~ 2 the distribution has a finite meanbut no .variance.

Although many of Pareto's ideas on social and economicmatters were controversial, his 'law' of income distributiondeveloped from the very simple, indisputable observation, atthe time, that just a few people held ,most of the nationalwealth. In this observation he seems to have struck upon aprinciple relevant to many cause-and-effect relationships.Put simply, it has generally been found that for many observedeffects, of all the contributing causes, only a few contributemost to the effect. This may sound vague and rather trite butit has proved a very useful concept in helping solve manyindustrial problems.

To be more specific, it is often the case that approxi­mately 80% of a company's profit is made on 20% of thedifferent items that it sells and apprOXimately 80% of productdefects will be the result vf approximately 20% of the totalfault types that can occur. This latter notion that mostproblems stem from a few causes can be used to give directionto the problem solving process. The 80%/20% 'guide is of coursenot rigid; it is merely an approximatiori, but in many instancesit is a remarkably good one.

An example of the application of Pareto's principle waspassed on to me recently by the Production Management Team atBradford University. As I understand it,'a student wasworking in conjunction with a company manufacturing tea-bagsthat was plagued with quality problems. The student, bystudying the manufacturing process and liaising with those in­volved in the manufacture~ was able to characterize differentquality problems into seven disjoint categories. These wereweight problems, bag problems, dirt problems, machine problems,bag formation problems, carton problems and paper problems.All quality problems arising in a two week production'runwere recorded in detail, and by type. The following data wereobtained:

5

Problem Frequency % CumulativeType Frequency %

Paper 55-95 56Machine 17-98 74Carton 11-22 85Dirt 6-85 92Bag Formation 4·93 97

, Weight 1-91 99'Bag 1·16 100

From this table the following cumulative frequency diagramwas drawn. The problem types are drawn in rank order (fromhighest contributor to overall problems to the lowest con­tributor) .

100il

>- 800c

60Q):Jc:rQ)

40..u..20

'- Q) c .. E .. 0»Q) C .. .c asQ. 0 C5.. Q

CIS :c .. 0 -i CD()

.. u..Q.

".(I

~0 Q~ CIS

CD

The cumulative frequency diagram pictures what is inunediatelyapparent from the table (as it is displayed in rank order),that 74% of the problems are caused by 29% of the primarycauses. It is immediately apparent then that tackling paperand machine problems will have the greatest effect on thetotal number of problems arising. In this particular caseboth paper and machine problems proved relatively easy tosolve. Prior to this simple'analysis management were un­aware that minor adjustments would have a "major impaGt onthe reduction of production quality problems.

It must of course be acknowledged, when tackling indus­trial problems of this nature, that a business exists tomake profit. Manufacturing problems cause financial loss soa company wants to know what are the most costly problems.The arralysi·s of problem costs in the above ci ted exampleshowed carton problems to represen~ the largest proportionnT n-rnh 1 ~m l"n~T: Tn 11 ntl1~n hu m~ l't.h ; np f~l1' i:~ _ 'Pan~r "faul tS:

6

were the third major contributor. Between them carton andmachine problems (29% of problem types) represented 79% ofthe total cost.

Thus, presented with two cumulative frequency diagrams,one relating problems to types and one relating cost toproble~ types, management had at its disposal information onwhere most profitably and effectively to concentrate its.effects on improvement.

This systematic search for the main components in an 80%/20~ rela~ionship is often called a Pareto analysis, in ack­nowledgement of Pareto's observation of wealth distributionlast century_ Perhaps national and international economieshave become sufficiently complex that in this realm Pareto'sprinciple is no longer a useful "one: None-the-less, in manyother areas his principle has ~proved to be a very useful ob­servation.

TWO MORE BREAKTHROUGHS

Two further long-standing conjectures 'have now beenresolved.

The first is the so~cal1ed Mertens conjecture which hasbeen disproved. This was first proposed by Stieltjes (whosediscoveries in integration were described in the previous issueof Function) in "1885, and reformulated by Mertens in 1897.This was a rather technical result in number "theory, whosetruth would have irntflied the truth of another , much more famousconjecture, known as the Riemann hypothesis."

It is now known that the Mertens conjecture is false how­ever. This was discovered by Andrew Odlyzko of Bell Labora­tories (USA) and Herman Te Riele of Amsterdam. They have notdiscover~d a specific counterexample, but have demonstratedthat one must exist.

The Riemann hypothesis itself is now believed, however, tobe true, despite this apparent set-back. The proof is due toa Japanese mathematician, Hideya Matsumoto, now resident inParis.

There is just one word of warning here, however. Thefull proof is not yet pUblished, and, till it is, there isstill some ground for scepticism.

For more on this,see New Scientist, 18 April 1985.

7

MATHEMATICS AND WEAVING:I. FABRICS AND HOW THEY HANG

TOGETHERt

Cheryl E. Praeger, University ofVVestern ~ustraJia

Although weav~ng is one of the ol~est activities of man­kind, and much has been written about it, it is only in thelast five years that the theoretical aspects of weaving havereceived much attention. In this paper I will give a mathe­matical description of a fabric and then discuss the problemof when a fabric hangs together.

1. D~8cription of a fabric.

As you probably know a roll of woven fabric has somethreads, called warp threads, running along the length ofthe fabric and other threads, called weft threads runningacross the fabric. However I want to give a precise mathe­matical description of a fabric. This is most easily donediagrammatically. The real fabric (Figure lea)) is idealizedby considering a warp thread as a vertical strand (a strandis the set of points in the plane lying strictly between twoinfinite parallel lines) and a weft thread as a hprizontalstrand; these horizontal and vertical strands are wovenunder and over each other to form an interlacement pattern(Figure l(b». The intersection of a vertical strand and ahorizontal strand, called a square, is coloured black if thevertical strand is on top and white if the vertical strand isunderneath (Figure l(c). This diagram of black and whitesquares is called by weavers a design, diagram,draft, draw­down diagram or point diagram and is the way a weaver wouldusually describe a weaving pattern. However we (and theweaver) will not be considering a point diagram covering the

t This article is based on the Professor Praeger's HannaNeumann memorial lecture delivered to the Fifth InternationalConference on Mathematics Education, August 1984.

8

whole plane. All fabric diagrams which we shall consider

will be periodia, that .is they will consist of a finite

block of squ~res which is simp~y repeated as we move across

and/or down the fabric. A block of squares of smallest size

with this property is called a fundamental bloak for the

fabric. There may be more than one fundamenta.l block but

they al~ have the same "size"; see. Figur~ 2(a) for the funda­

mental blocks for ·the "plain weave ll fabr,ic represented. in

Figure 1. In practise the point diagram is taken as either a

fundamental block of the fabric or some union of fundamental

blocks. To be able to use some mathematics in investigating

problems in weaving we take one further step: we replace all

black squares by a 1 and all white squares by a 0 so that the

point diagram becomes a binary' matrix D, that is a matrix with

entries 0 or 1.

"2. Hanging together.

Now a moment's thought should convince" you that not all

binary matrices will represent ·a· fabric which hangs together,

that is a fabric such that no proper subset of warp and weft

threads can be completely lifted off the rest of the fabric.

For example a whole column of zeros or ones means that the

corresponding warp thread is· not woven into the fa.bric. How­

ever there are many examples of a fabricts not hanging to~

gether which are much less "obvious than these trivial examples

(where an all-zero 'or an all-one row or column is present).

I am told that in the past the usual way in which a weaver

decided whether or not a fabric would hang together was to do

a trivial weave. However in 1980 the British mathematician

C.J.C. Clapham produced a mathematical algorithm to answer

this question.

To see the problem more clearly, let D = (d~.) be an'l,J'

m x n .binary matrix, that is the row i column j entry dij

is 0 or 1. Suppose that the fabric corresponding to D does

not hang together - say a set T of vertical strands and a set

S of horizontal strands lifts off the rest of the fabric. We

can regard T as a subset of {1,2, ... ,n} and S as a subset

of {l,2, .. o,m}. Now the strands of Sand T lift off the

rest of the fabric if and only if

(i) every ~ertical strand not in T goes under every

horizontal s~rand in S, and

(ii) every horizontal strand not in S goes under every

vertical strand in T.

These two conditions are equivalent to

JET.

j ¥ T,

i , 5, and

i E 5, and

for all

for all

1

o(i)' d •.l"J

(ii)' d."1,J

Thus the fabric will not hang together if and only if

there are subsets T,S of . {1,2, ..·. ,n} .and {l,2', ... ,m}

resnectively (with at least one a proper non~empty subset)

and

(b)

9

(e)

Figure 1

(a)

(~~) (~~)(b)

Figure 2

10

such .. that (i)' and (ii)' are true. In other words the fabricwill not hang together if and only if it is possible to re­arrange the rows and columns of D in such a way that the re­sulting matrix has the form

where 0 is an all-ze~o matrix, J is an all-one matrix (withat least one of 0 and J non-vacuous) and X and Yare not re­stricted. However to run through all possible sets.S,T tosee whether they have this property would take far too longwhenm and n are large. Clapham introduced an alternativemethod'~ whose details are somewhat technical for Function,but interested readers can find it in the Bulletin of theLondon Mathematical So.ciety, Vol.12 (1980), pp.161-164.Essentially it consists of the evaluation of n numbers~ Ofthese one is equal to zero necessarily, but if any of .theothers is zero, then the fabric does not hang together.Otherwise it does. This algorithm is easily implemented ona computer.

The only theoretical advance I know of on Clapham's workis due to T.e. Enns and was published in the technical journalGeometrica Dedicata, Vol.15 (1984), pp.259-260. He has givenan efficient algorithm for determining whether or not a fabricwoven with any number of sets of parallel threads (not justone set of warp threads and one set of weft threads) will hangtogether.

For more on the mathematics of weaving, see the articleby B. Grunbaum and J.G.' Shepherd "Satins and twills: anintroduction to the geometry of fabrics" in Mathematics Maga­zine, Vol.53 (l980}, pp.139~161. In a sequel to the presentarticle, I shall discuss the math~matics involved in theactual setting up of the 100m.

BEWARE OF MATHEMATICIANS

Canon XXXVI of the Council of Laodicea, held some time inthe period 343-381 AD, reads:

They. who are of the priesthood, or of the clergy, shallnot be magicians, enc~anters, mathematicians, or astrologers;nor shall they make what areocalled amulets, which are chainsfor their own souls. And those who wear such, we command tobe cast out of the Church.

This looks as if several eminent mathematicians who werealso in holy orders, were in breach of Canon Law. However, anote explains the apparent anomaly.

"Mathematicians"!' are they who hold the opinion that thebelestial bodies rule the .universe, and that all earthly thingsare ruled by their influence.

11

THE· CALENDARt

S. Rowe, Student,Swinburne. Institute of Technology

Historically~ all measurements of time have depended onastronomical observations - the day is measured from the ro­tation of the Earth, the week approximates the changingphases of the moon; the month is measured from the· revolutionof the"moon around the Earth and the year is ,measured fromthe revolution of th~ Ear~h around the Sun.

Many ancient civilizations, particularly the Babylonian,based their calendars on the cycles of the moon, and thelunar measurement of years has been preserved in the modernJewish, Chinese and Mos~em calendars. Against this, theEgyptians based their calendars on the Sun (which also figuredprominently in their religion). The Egyptian civilizationdepended upon the .seasonal rising of the Nile; which wasclosely associated with the solar cycle. Ancient peoplesdetermined the solar year by observing the rising of.a brightstar after it had been invisible because of its proximity tothe Sun. A conunon star used for this purpose was Sirius. Byaveraging many such observations, the solar year was found to.be very close to 365 days.

In ancient Rome, months were based on lunar cycles. Thepontifices watched for the first appearance of the thincrescent moon after the new moon so that they could declarethe beginni'ngof the newmonth. This first day , shouted fromthe steps of the Capitol, was termed Kalendae, which meansthe calling. Our word calendar is derived from this term.

Unfortunately for our measurement of time, the lunarcycle is not a whole number of days, nor is the time Earthtakes to complete an orbit of the Sun relative to ~he stars.The Moon's cycle is 29.53059 days, while the Earth's orbitaround the Sun takes 365.242196 days. So 12 months are shortof a year, and 13 months would give us a year that is toolong. And our seven day week (which is based on religion),alth9ugh close to the lunar phases, is not a factor- of thelunar period, the month ·or the year.

When the Romans adopted the Egyptian solar year at thetime of Julius Caesar, their own lunar-solar calendar wasvery much in error. Introduced to Rome by an astronomer,the Egyptian calendar was ordered into official Romap use byJulius Caesar in 45 Be. It w~s called the ~ulian calendar, .and was based on a sola~ year of 365.25 days. The year wasdivided into months, of which eleven contained 30 or 31 days

:This article first appeared in the Bulletin of the Lions Clubof Belgrave, March, 1985.

12

and the twelth had 28 days only. The first month was Marchand the last was _February. July is named after Julius Caesarand August after Augustus Caesar L both months being allocat-edthe full 31 days, as befitted a Caesar. The seventh monthwas named September, the eighth October, the ninth November andthe tenth December after the Latin septem, octo, novem anddecem for seven, eight, nine and ten respectively. '

The Julian calendar lost approximately one-quarter dayeach year. This loss was corrected by adding an extra day tothe twelth month (February) every fourth year, which was theleap year. Nevertheless, this calendar gradually became outof step with the seasonal position tif the Sun t~lative to thestars. The year of the Julian calendar was actually 11minutes 4 seconds longer than -the time it takes the apparentSun to revolve to precisely the same position. By 1500 theerror amounted to approximately 11 days. Christian religiousfestivities based on ~aster assumed a fixed vernal equinox ofMarch 21, and as a consequence they were becoming graduallyout of step with the seasons. Accordingly, Pope Gregory XIIIentrusted a reformation of the calendar to a German Jes~it

whose latinized name is Clavius. Clavius used a scheme de­vised by a Neapolitan astronomer in which centuries would notbe leap years unless perfectly divisible by 400'. To correctthe calendar, Pope Gregory ordered that October 15, 1582,­should follow October 4. Despite prote~ts from angry mobs,who thought that ten days of their lives were being stolen,·the correction was made and the new calendar ~as called theGregorian calendar. The new calendar also moved the beginningof the year from March 25 to January 1.

The Gregorian calendar was adopted by most of the RomanCatholic countries and by Denmark and the Netherlan-ds in1582. But it was nearly two centuries before it was generallyaccepted. During that time, a traveller could leave Englandin February 1679, for example, and find that it was February1680 in some parts of Europe and Scotland. The day of themonth was also different between England and some parts ofEurope.

Finally, other countries began to accept the new calendar.The Protestants in Germany and Switzerland adopted it in 1700,Britain and the American colonies in 1752 (omitting the elevendays between September 2 and 14), Prussia began to use it in1778, Ireland in 1782, Russia in- 1902. Following the FrenchRevolution, a new calendar was adopted in France, the firstday of the year being September 22, 1792. This calendar wasused until December 1805, when France accepted the Gregoriancalendar again.

Other calendars are still in use, however, particularlyin regard to religious events. The Jewish calendar uses alunar cycle and a sola~ cycle. The months are lunar ,months,but they are about 11 days short of a solar year. A thir­teenth mont~ periodically has to be intercalated to maintainsome synchronism with the solar cycle. The Moslem calendarignores the solar cycle completely and is tied to lunar cycleswith alternate months of 30 a-nd 29 days. The year begins at

13

different seasons over a 32.5 year cycle.

Prior tp World War II there was an attempt among somebusiness in Europe to introduce a 13 month calendar in. whichall months would ~ave four weeks. This business calendarwould have allowed more meaningful financial comparisons,but it did not receive wide acceptance. .

SUSAN'S WORLD FIRSTWe have all heard of the pr~s'tigious .Rh9d~.~. ·sc1}olar$hips.Eac~ year, 'one is awarded from ea~h of- the Australian statesand from other parts of the former British Empire (including,. quaintly, the U.S.A.). But how many of us have heard of theRhodes post-doctoral fellowship? Yes, fellowship, not fellow­ships. There is only one, and this is of~ered world.wide eachyear.

The holder for 1986 will "be Susan Scott, the firstAustralian ever to win .the award. .

Dr Scott graduated from Monash University in 1979 and con­tinued her studies at post-graduate level at the University ofAdelaide, where she has been conducting resear~~ ·in. mathematicalphysics.

The $40,000 award will cover her fares, accommodat"ion andresearch expenses and will take her to Oxford for two yearsfrom January 1986.

Two ot the- areas she explores in her'work are the originof the uhiverse and the nature of black holes. "I .am fascin­ated by phys~cs,tf Dr Scott said, "1 just want to know as muchabout physics as ~ can. 1f

Dr Scott also looks at whether there are other universesbeyond our own.

"It's quite possible there ·are other universes," she said.f1If there are, we have no w~y of telling.we are unique." It's possible

Dr Scott has been enthusiastic about science since she wasa teen-ager and is concerned aQout the small number of womeninvolved in mathematical research.

She is keen to pursue her research abroad for a few yearsbut eventually wants to teach so she can pass on her enthusiasmfor science to future generations~ .

Based on· an article by Jo McKenna, The HeFaZd 22/7/85.

14 '

PROJECTIVE GEOMETRY

J.C.. Stillwell, Monash UniversityPepspective.

Perspective may be simply described as the realisticrepresentation of spatial scenes on a plane. This of courSehas been a concern of painters since ancient times, and someRoman artists seem to have achieved correct perspective by thefirst century Be. However, this may 'have been a stroke of

Figure 1

15

individual genius rather than the success of a theory,because the vast majority of ancient paintings show incorr­ect perspective. If indeed there was a.classical theory ofperspective, it was well and truly lost during the dark ages.Medieval artists made some charming attempts at perspective,but always got it wrong, and errors persisted well into the15th century (Figure 1).

The discovery of a method for correct perspective isusually attributed to the Florentine painter-architectBrunelleschi (1377-1446), around 1420. The first publishedmethod appears in the treatise On Painting, Alberti (1436).

-The latter method, which came to be known as Alberti's veil,was to set up a piece of transparent cloth, stretched OJ) aframe·, in front of the scene to be painted. Then, viewingthe Scene with one eye, in a fixed position, one could tracethe scene directly onto the veil. Figure 2 shows this method,with a peephole to maintain a fixed eye position, as depictedbr Durer (1525).

Figure 2

16

Alberti's veil was fine for painting actual scenes, butto paint an imaginary sc~ne in perspective some theory wasrequired. The basic principles used by Renaissance artistswere:

(1) a straight line in perspective remains straight(ii) parallel lines either remain parallel or converge

to a single point (their vanishing point).

These princiPles suffice to solve a problem artists frequentlyencountered ~ the perspective depiction of a square-tiledfloor. Alberti (1436) solved the special case of thisproblem in which one set of floor lines is horizontal, i.e.parallel to the horizon. His method, which became known asthe aostruzione legittima, is indicated in simplified form inFigure 3. .

Figure 3

The non-horizontal floor lines are determined'by spacl.ng themequally along the base line (imagined to' touch the, floor) andletting them converge to a vanishin~ point on thehori~on.

The horizont~l floor lines are then determined by choosingone of.them arbitrarily, thus determintng one tile in thefloor, and then producing the diagonal of this tile to thehorizon. The intersections of this diagonal with the noo­horizontal lines are the points·through which the horizontallines pass. This is c~rtainly true on the actual floor(Figure 4), hence it remains true in the perspective view.

Ii

)1

17

i/V

//

//

Figure 4

The same principles in fact suffice to generate a pers­pective view of a tiled floor given an arbitrarily situated,tile.

Anamoraphosis.

It is clear from the Alberti veil construction that aperspective view will not look absolutely correct exc~pt whenseen from the viewpoint used by the artist. Experience shows,however, that distortion ~s not noticeable except from extremeviewing positions. Following the mastery of perspective bythe Italian artists, an inter~sting variation developed, inwhich the picture looks right only from one, extreme, view­point. The first known example of this style, known asanamorphosis, is an undated drawing by Leonardo da Vinci fromthe Codex Atlanticus (compiled between 1483 and 1518).Figure 5 shows part of this drawing, a childts face which looksright when viewed with the eye near the right hand edge of thepage.

18

Figure 5

The idea was taken up by German artists around 1530.The most famous example occurs in Holbein-' s painting The twoambassadors (1533). A mysterious streak across the bottomof the picture becomes a skull "when viewed from near thepicture's edge. The art of anamorphosis reach~d its tech­nically most advanced form in France in the early 17th century.It seems no coincidence that this was also the time and placeof the birth of projective geometry. In fact the key figuresin the two fields, Niceron and Desargues, were well aware ofeach others' work.

Niceron (1613 - ~646) was a student of Mersenne and, likehim, a monk in the order of Minims. He executed some extra­ordinary anamorphic wall paintings, up to 55 metres long, andalso explained the theory in a book La perspective curieuse,pUblished in 1638. The cover illustration is one of hisillustrations : anamorphosis of a chair.

The anamorphosis, viewed normally, shows a chair likenone ever seen, yet from a sui tably extreme point one s"ees anordinary chair in perspective. This example encapsulates animportant mathematical fact :a perspective view of a pers­pective view is not in general a perspective view. Iterationof perspective views gives what we now call a projective view,and Niceron's chair shows that proj~ctivity is a broader con­cept than perspectivity. As a consequence, projectivegeometry, which studies the properties which are invar.tantunder projection, is bro~derthan the theory of perspective.Perspective itself did not develop into a mathematical theory,descriptive geometry, until the end of the 18th century_

19

Desargues' projective geometry.

The mathematical setting in which one can understandAlberti's veil is the family of lines ("light ra"ysH)" througha point (the Heyen), together with a plane V (the "veil"),as in Figure 6.

Figure 6.

In this setting, the problems of perspective and"anamorph­osis were not very difficult, but the concepts were interestingand"a challenge to traditional geometric thought. Contraryto Euclid, one had

(i) points at infinity ("vanishing points ft) where parallels

met,and

(ii) transformations which changed lengths and angles(projections).

The first to construct a mathematical" theory incorpor­ating these ideas was Desargues (1591-1661), although the ideaof points at infinity had already been used" by Kepler in 1604.

20

Desargues' book Brouil1on project d'une atteinte auxevenemens des re,contres du Cone avec un Plan (1639)(Schematic sketch,of what happens when a cone meets a plane)suffered an extreme case of' delayed recognition, being com­pletely lost, for 200 years. Fortunately, his two most 'important theorems, the so.;..called Des,argues' theorem and theinvariance of the cross-ratio, were published in a book onperspective by Bosse (1648).

Kepler and Desargues both postulated one point atinfinity on each line, closing the line to a "circle ofinfinite radius". All lines in a family of parallels sharethe same point at infinity. Non-parallel lines, having afinite point in common, do not have the same point atinfinity. Thus any two distinct lines have exactly one pointin common - a simpler axiom than Euclid's. Strangely enough,the line at infinity was only introduced into the theory byPoncelet (1822), even though it is the most obvious line inperspective drawing, the,h9rizon. Desargues made'extensiveuse of projections in the Brouillon.project; he was thefirst to use- them to prove theorems about conic sections.

Desargues' theorem is a property of triang"les in pers­pective illustrated by Figure 7.

Figure 7.

21

The theorem states that the points X,Y,Z at the inter­sections of corresponding sides lie in a line. This isobvious if the triangles are in space - the line is theintersection of the planes containing them. . The theorem inthe plane is subtly but fundamentally different, and requiresa separate proof, as Desargues realised. In fact, Desargues'theorem was shown to play a key role in the 'foundations ofprojective geometry by Hilbert (1899).

The invariance of the cross ratio answers a naturalquestion first raised by Alberti : since length and angle arenot preserved by projection, what is? No property of threepoints on a line can be invariant because it is possible to'project any three points on a line to any three others. . Atleast four points are therefore n~eded, and the cross ratio isin fact a projective invariant of four points. The crossratio (ABCD) of points A, B, C, D on a line (in that order)

is ~~ / g~. Its invariance is most simply seen by re­

expressing it in terms of angles using Figure 8.

o

A B c

Figure 8.

Let 0 be any point outside the line and consider theareas of the triangles OCA, OCB, ODA, ODB. First use baseson the given line and height h, then recompute using OAand OB as base~, and heights expressed in terms of thesines of angles at 0

22

-i'h ·CA area OCA tOA -oe si.n !COA

*h ·CB area OCB lfOB ·OC sin !eOE

~h ·DA area aDA ~ OA ·OD sin !DOA

~h ·DB area ODB :J.jOB ·OD sin {DOB

SUbstituting the values of CA, CB, DA, DB from these equationswe find the cross ratio in terms of angles at 0

CA -/ DA sin/COA sin !DOACB DB sin/COB / sin /DOB

Any four points A', B' ~ C', D' in perspective with A,B,C,Dfrom a point 0 have the same angles (Figure 9), hence theywill have the same' cross ratio.

o

A 8 c

Figure 9

o

But then so will any four points AJf, B n

, C't., D" projectivelyrelated to A,B,C,D since a projectivi.ty is, by definition,the product of a sequence of perspectives.

23

THE FOLDING CHAIRt

In the market-place in Groningen there's a doughnutstand. The other day I let my doughnuts get cold, immersedas I was in the construction of the folding chairs. Look atthe photograph onp.24 and the drawing on p.27. (Incidentally,a small simplication has been made in the drawing but onlypettifoggers would take note of that.)

There are six turning points and no sliding mechanisms,as in many other types of folding chairs. The basic form isa quadrilateral ABeD which is more or less a parallelogram..Of course ~ otherwise you could -never fold the chair flat.

The striking feature is the linking piece AQ. What isits function? Why isn't the back leg simply connected to A?Th~ jam had already melted when I understood that then thechair 'could not be folded up. For then the triangle'APDwould be a rigid figure, and hence the quadrilateral ABeDwould be rigid too.

But how do you then get the le.ngth of AQ? And why doesthe chair remain .standing exactly in the position drawn, evenwhen you ~it down on it? .

The answer is reached from the picture on p. 24 and the o,therphotographS-. Dur-ing·J,J~J;l;:f.-9:t4:i.~ng,tIle ch:air _.~ngle.l)_beYHm~s

smaller and you can imagine how the weight of the "sitter" onthe chair will try to make that angle ·as small-as possible;his or her weight pushes the parallelogram ABOD into itslowest position. Now take the' back~rest AD as a fixed line.Then DP and AQ become rotating rays with D,A respective'ly ascentral points.

So the question is! in Which posi-tion, is angle Dminimal? To see that we look at triangle ADP, of which VAand VP are the hinging sides. AngleD'is minimal if theQpposita side';4P (which you'll just have to draw· in yourmind) is as small as possible. Tha"t occurs (and now look attriangle AQP)' if angle Q is minimal. Well, angle Q is ofcourse minimally zero, that is, if A is onFQ. So the chair.remains in that posit-ion if you sit down on it!

t This 'article is a translatio~ from the Dutch byA.-M. Vandenberg. It first'appeared in the journal PythagorasVol.24, Part 4 and is reproduced under an exchange agreement.

24

This doesn't happen, of course, with the real chair,because A runs into the back leg PQ. There's a reason forthat, too: if A could pass the line PQ (wh~ch could be quitefeasible'technically') the chair ,would "spring" a Ii ttle be- 'cause you can move through that·minimum. Something like amarble rolling to and fro in a little hollow.

After all, the lengths of DA, DP, AQ and PQ can't bechosen independently of each other. Look at the photographof· the folded chair, and you'll see that VA + DP = AQ + PQ.

25

26

27

PERDIXAustralians are good at competitive sport and the sport of

mathematical olympiad contests is proving no exception.42 countries sent teams to take part in the 1985 InternationalMathematical Olympiad in Finland. The Australian team wasplaced 11th. Team members were:

Shane Booth, Shepparton, VictoriaJohn Graham, Sydney, N.S.W.Alisdair Grant, Melbourne, VictoriaAndrew Hassell, Perth, W.A.David Hogan, Sydney, N.S.W.Catherine P1ayout, Sydney, N.S.W.

Andrew Hassell was placed 7th out of the 209 competitorsand was awarded a gold medal (14 competitors' were awarded goldmedals). John Graham and A1isdair Grant were awarded silvermedals (35 silver medals were awarded). Shane Booth wasawarded a bronze medal (52 bronze medals were awarded).

Congratulations to the Australian team~

Continued on p.30.

28

PROBLEM SECTIONProblems 9.1.1, 9.1.3 are still outstanding, and we leave

them that way for this issue, as they are interesting and wewould like to give readers a further chance to do them.

We have solutions to 9.2.1, 9.2.2 which we print below, andhave also received correspondence on problems posed in Volume 9,Part 3; thi~ will be reported on in the next issue.

SOLUTION TO PROBLEM 9.2.1The problem, from Hall and Knight's Highep Algebpa, read as

follows.

"There are three Dutchmen of my acquaintance to see me,being lately married; they brought t.heir wives with them. Themen's names were Hendriek, Claas, and Cornelius; the women'sGeertruij, Catriin, and Anna; but I forgot the.name of eachmants wife. They told me they had been at market to bUy hegs;each per~on bought as many hogs as they give shillings for onehog; Hendriek bought 23 hogs more than Catriin; and Claasbought 11 more than Geertruij; likewise ,. each man laid out 3guineas more than his wife. I desire to know the name ofeach man f s wife." (Note -: . 1 gUinea =' 21 shillings.)

David Shaw of Geelong West Technical School, who submittedthe problem, also sent us his solution.

Let h =·number of hogs bought by each man and ~ =number of hogs bought by each wife. Then "each person boughtas many hogs as they gave shillings for one hog" and "eachman laid out 3 guineasmor~ than his wife" leads to theequation

The factor pairs of 63 are (63.,1), (21,3), (9.,7).

(h +w)(h - w) = 63h+w=63

h·- ~ = 1

gives h 32 w = 31.

(21, 3) gives h 12 w = 9,

(9, 7) gives h 8 'W = 1.

"Hendrick bought 23 hogs more than Catriin tr indicates thatHendrick bought 32 hogs and Catriin 9.

"Claas bought 11 more than Geertruijtl indicates that C1eesbought 12 and Geertruij 1.

29

So, Catriin is Claas t wifeGeertruij is Cornelius' wife andAnna is Hendrick's wife.

SOLUTION TO PROBLEM 9,2.2We asked for a proof that no (positive) integers exist

such that

4:x; 4y 2z

Devon Cook of Urrbrae Agricultural High School (Netterby,S.A.) solved this problem. He writes:

I will use the method of infinite descent, fir~t used Ibelieve, .by Fermat. The equation can be written as{:x;2)2 _ (y2)2 = z'2 , the Pythagorean identity which has thewell known solutions

x 2 n 2 + m2 ,y2 2nm, Z = n2 _ m2

It remains to show,that since the two variables re 2 and y2are perfect squares, there can be no solution. We have

2y = 2nm and thus there must be two integers p~q such that

P 2 , 2m = q2 h t b .n = as we may c oose' n , moe co-prJ.me.

But n2 = x 2- m2 which is itself Pythagorean and thus

n = 82

- 1;2 and m =281; (or vice versa, when the analysis'.is similar).

Thus 2m = 48t = q2 and following on again, there must be

integers x' and y' such that 8 = x,2 ,t y,2 There-fore

2,p where p < z .

Thus there is a set of integers which satisfy the equation butusing a value of z which is smaller than the original, viz: p.Since this process c'an be repeated ad in.finitum, there can beno such integers." -The problem is solved!

There does still.remain the possibility

x 2 = n2 + m2 ,y2 n2 _ m2 , z 2nm •

This, however, can be dealt with along similar liries and weleave the details to the reader.

The following fun problem came from Garnet A. Greenbury ofBrisbane.

PROBLEM 9.4.1A sequence of integers has been divided by the same number

giving these remainders: 2, 4, 8, 5.

30

(a) What number comes next?

(b) What is. the original sequence of numbers?

(c) What is the divisor?

(d) Write down the next five remainders following the 5.Call them a, b, a, d, e and show that a + 1,· b + 2, C + 4)d + 8, e + 5 are all equal. Can you explain this?

(e) Show that the sequence formed by the first ten remaindersis the same as the pattern for the next ten remainders. Canyou explain this? .

Other problems occur in the Perdix section of each issueof Funation. Send solutions of those to Perdix. Solutionsto problems in this section should be sent to the Editor.

PERDIX (CONT.)

The competition took place on July 4 and July 5 withthree problems to solve. each' day in a session of 4t hours.The two problem papers are reproduced below.,

Problem I proved the easiest with 103 competitors gettingthe full 7 'points, the average number of points obtained being4.1. ,Problem 2 was a close runner up with 92 competitors getting7 points and with an average of 3.7. In increasing order ofdifficulty were problem 4 (average 2.4), problem 6 (average2.0), problem 5 (average 1.9), and problem 3 (average 0.8).Only 12 competitors got the full 7 points for problem 3(John Graham got 6),. while 153 scored 0 (includingAndrew Hassell, our gold medallist, who made up for this byscoring the full 7 points for each of the other problems).

Try the questions yourself. Do you have the same assess­ment of their ,difficulty as the competitors f scores'suggest?

FIRST DAY

Joutsa July 4, 1985

1. A circle has centre on the side AB of the cyclic quad­rilateral ABeD. The other three sides are tangent tothe circle. Prove that AD + Be = AB .

2. Let nand k be given relatively prime natural numbers,o < k < n. Each number in the set M = {I, 2, ... , n-l}is coloured either blue or white. It is given that

(i) for each i € M ,both i and n-i have the samecolour", and

(ii) for each i EM, i * k ,both i and Ii - klhave the same colour.

Prove that all numbers in M must have the same colour.

31

3. For any polynomial P(x) = a O + a l $ + ..• + akxk with

integer coefficients, the number of coefficients which areodd is denoted ~y w(P). For i = 0, 1, 2, ... let

'/;

Qi(x) =-(1 + x). Prove that if iI' i 2 , ... , in are

integers such that 0 ~ i 1 < i 2 < •.. < in ' then

W(Q. + Q. + ... + Qi ) ~ w(Q. ) ·~l ~2 . n ~l

Time allowed: 4! hoursEach problem is worth 7 points.

SECOND DAY

Joutsa July 5, 1985

4. Given a set M ~f 1985 distinct PQsitive integers, noneof which has a prime divisor greater than 26. Prove thatM contains at least one subset of four distinct elementswhose product is the fourth power of an integer.

5. A circle with centre 0 passes through the vertices Aand C of triangle ABC, and intersects the segments ABand Be again at distinct points K and N , respect­ively. The circumscribed circles of the triangles ABCand KBN intersect at exactly two distinct points BandM. Prove that angle OMB is a right angle.

6. For every real number xl ' construct the sequen~e

xl' x 2 ' ... by setting

x n +l = xn'(xn + ~)

for eachvalue of

n ~ 1. 'Prove that there exists exactly onexl for which 0 < x n < x n +l < 1 for every n .

Time allowed: 41 hoursEach problem is worth 7 points.

Send me your solutions. Perhaps you will find newmethods of soluti:on. For example, I found 3 -solutions toProblem 1 that are different. There may be better methods.

* * *I now give solutions to some of the problems I have set

earlier this year. Here are solutions to Probl~ms 7, 8, and9, provided by Hai Tan Tran, 15 Arthur Street,' Plympton Park,S.A., 5038.

PROBLEM 7. AB and CDE is' mid-point of segmentand AE meets BD at G.

32

Solution.

are two parallel straight lines,

CD. Line AC meets BE at FShow that FG is parallel to AB .

There are two variants of thefigure for which slightly di.ff­erent arguments are required.In each ·figure hI is distanceof F from CD, h 2 is the

distance of G from CD and his the distance between theparallel lines CD and AB.

A B In I1s CEF and ABF

h + hI AB0 E ChI CE

C- h AB·1- in the lowerh hI EC

figure )

A B

h + h 2 ABsimilarly, h2 ED

( h - h2 AB in the lower figure)h2 DE

Hence, since CE ED

h + hI h + h 2hI h2

( h - hI h - h2 )hI h2

whence, in both cases, hI = h 2

Thus FG is parallel to CD , and so to' AB .

PROBLEM 9. Let ABC and DBC be two triangles such that

AD is parallel to BC. , Let BD and AC meet at EDraw a line parallel to BC through E and let this linemeet AB at F and CD at G. Show that FE ,= EG.

SoZution. This problem asks for the proof of a resultconverse to that of problem 7.°

A

B

Similarly,

Hence

whence

oThere are again two variantsof the figure. We con­sider merely the one shown.

Let hI be the distance

from each of A and D toF7 and let h, be the dis­ta~ce between AD and BC.Then, since F~ is parallel'to BC, in 6s AFE and ABC

C hI FE

h+fii Be

hI EGh + hI BC

FE EGBC BC

FE EG

PROBLEM 8. Let ABC .be any tr_iangle and draw parallel lines

through the vertices A, B, and C to meet the opposite sidesin D, E, and F ,. respectively. Show that the area of 6DEFis twice the area of 6ABC.

SoZution. You will see how to use the result of problem 9 tosolve this result.' Alternatively we may proceed as follows.

o A 6BDE = 6BAE (1)

(6S with same base and ·equalheights] ;

similarly

F c

6BFE

and 6FDC

b.BCE

6FAC.

(2)

Subtracting AFBC from the latter gives ~DBF

Adding (1), (2) and (3) now gives the result.

6ABC (3)