Jakob Steiner1796 1863

– p.1/53

Jakob Steiner1796 1863

ne derange pas mes cercles ! – p.1/53

Jakob Steiner1796 1863

ne derange pas mes cercles ! – p.1/53

(Photos: Barbara Kummer)

born 1796in Utzenstorf (Bern)childhood: work at this farm,no school educationuntil age of 18

– p.2/53

1814 enters the Pestalozzi school in Yverdon ...

first as student ...... then as “instituteur”.

(Documents: Burgerbibliothek Bern)

With much energy:“Gefunden Samstag den 10. Christmonat 1814,3 + 3 + 4

St. daran gesucht, des Nachts um 1 Uhr gefunden.”

(Steiner, quoted from J.-P. Sydler,L’Enseig. Math.(1965), p. 241.)

Life-long admiration for Pestalozzis non-dogmatical teaching ...– p.3/53

Example. Th. Clausenpublished in Crelle Journal3, No. 17 thedetermination of the area ofT by complicated trigonometry.

(Steiner autogr. 1814)

kc

ka

kb

T

AB

C

D

F

HE

K L

(Steiner, Crelle J.3, No. 19, 1828)

Three pages later (Steiner’s answer):“Diese Aufgabe gehört zu einer Abtheilung

von Elementar-Aufgaben, die die Pestalozzische Schule wegen mancherlei

pädagogischer Vorzüge sehr in Betracht zog. Nach dieser Elementarübung wird die

obige Aufgabe ohne die Hülfe trigonometrischer Functionengelöset; auch kann sie

allgemeiner gestellt werden, ...”– p.4/53

Steiner’s solution: u, v, w ratios of side-lengths;

c1+w

wc1+w

a1+u

ua1+u

b1+v

vb1+v

C

BT

AB

C

D

F

GH

E

K L

DrawEG parall. toAFB;

by ThalesEG : EC = AF : AC

henceEG = c

(1+w)(1+v) ;

By Eucl. I.41 and Thales

C : B = EH : HB = EG : FB = 1(1+v)w ;

by Eucl. I.41 we haveC + B = 11+v

· A ;

can computeB = w

1+w+wv· A ;

cyclic permut. gives other triangles, subtract:

T = (1 − u1+u+uw − v

1+v+vu − w1+w+wv) · A

= (uvw−1)2

(1+u+uw)(1+v+vu)(1+w+wv) · A .

Observe: T = 0 ⇔ uvw = 1 (Ceva’s Theorem).Result later forgotten and rediscovered asDudeney–Steinhaustheoremor Routh’s theorem. – p.5/53

1821:→ Berlin, short career as teacher forlower mathsin Werder Gymnasium, then

“Privatlehrer” and starts writing a huge text on circle and sphere geometry ...

– p.6/53

1821:→ Berlin, short career as teacher forlower mathsin Werder Gymnasium, then

“Privatlehrer” and starts writing a huge text on circle and sphere geometry ...

... until he became a prolific contributor to the newly founded Crelle Journal.– p.6/53

Steiner’s“Programm”as “Privatlehrer” in Berlin:

(Steiner, Crelle J.1, p. 161, 1826)

• Apollonius (Pappus VII);• Malfatti;• Pappus IV.15;• conics.“Das Bestreben des Verfassers ... den ... gemeinschaftlichenZusammenhangzu finden”.⇒ General Theory of Circles. – p.7/53

Steiner’s “Book of Lemmas”:

1. Power of the pointE with respect to the circle:“In den Lehrbüchern der Geometrie findet man folgenden Satz bewiesen:”

α

α

ǫ

A

B

T

E(a)

α

α

δ

A

B

C

D

E(b)

t

r

r

d

r

A

B

T

E

O

(c)

EB · EA = ET 2 = ED · EC = d2 − r2 = t2 (Eucl. III.36).

“Dieses Product ... soll‘Potenz des Punkts in Bezug auf den Kreis’heissen.”– p.8/53

2. Line of equal powers for two circles:

A

B

E

C1

C2

r1r2

d1 d2

he1 e2

C1 C2D

E

“Die Bedeutung dergemeinschaftlichenPotenz zweier Kreise, wovon schon bei

Pappus und Vieta sich Spuren finden, lernte er durch ihre, vonihm gefundene

vielseitige Anwendbarkeit erkennen”.

– p.9/53

3. Point of equal powers for three circles:

C1 C2

C3E

“Wir wollen diesen Punctp(123) hinfort

‘Punct der gleichen Potenz der drei KreiseM1,M2,M3’ nennen.”

– p.10/53

4. Similarity centers of two circles:

Fermat 1679

Steiner 1826

2rr

r

23a

13a

43a

aa2a

C2 I C1 E

“Die Franzosen nennen diese Punktecentre de similitude de deux cercles. Wir

haben diese Benennung zuerst aus einer Abhandlung von Eulergenommen, siehe

L. Euler: de centro similitudinis, 1777”. – p.11/53

5. Common power of two circles with respect to theirsimilarity center .

C1

C2

C3

E

C

D

F

X ′X

Y

Y ′

A′ A B B′EC2 = θ · EC1

EX ′ · EX = q

EX · EY = EX ′ · EY ′ = θq = EF 2

– p.12/53

1. “Libris antiqua propositio” (“von Pappus überliefert”):

Archimedes’“Lemma 6”

5 centuries later ...

r1

m1

r2

h2

m2

r3

h3

r4

h4

C1

C2

hi

ri

= 0, 2, 4, 6, . . .

Pappus IV.16

r1m1

h1

r2

h2

m2

r3

h3

r4

h4

C1

C2

hi

ri

= 1, 3, 5, 7, . . .

Pappus IV.18

Steiner:“Den Pappischen Satz kannte ich nur ohne Beweis.”– p.13/53

Steiner’s proof of Pappus’ “antiqua” theorem:(Crelle J.1, 1826, p. 261-262.)

r1

h1

r2

h2

C1

C2

A

B

C

D

b

P1 P2

m1

m2

(a) Tang. inB line of equal powersfor C1 andC2, 2;

(b) Sim. centerA for m1 andm2

same power,⇒ on tang.4,5;(c) Thales:P1B : P2B = r1 : r2;(d) AB2 = com.pow.m1,m2 5,1;(e)Ab2 = AX · AY = AB2 5;(f) BAb ∼ Cm2b ∼ Dm1b

⇒ DbCB aligned;

(g) Thales and (c):h1 + r1

r1=

h2 − r2

r2or

h2

r2=

h1

r1+ 2 .

Followed by many generalizations.

– p.14/53

2. “Lösung derMalfatti schen Aufgabe...”(Crelle J.1, 1826, p. 178.)

Problem:Given triangleABC

find 3 circles such that ...

Problem appeared 1810/11 in Gergonne J;“les rédacteurs desAnnales” received letter from Italy that“M. Malfatti, géomètre italien très-distingué”had solved the problem, but “proof to long for this letter”.

Proof was also to long for Malfatti’s own publication(Mem. Soc. It. d. sc., Modena 1803,10, p.235).

– p.15/53

Steiner’s geometric solution of the Malfatti problem:I incenter ofABC;draw incircles ofAIB,BIC,CIA;D point of contact;DEF common tang. to other incircles;incircles ofADE andDBF⇒ two Malfatti circles.NO PROOF,NO EXPLANATION.

A

B

C

D

E

I

F

later reconstructions of Steiner’sproof long and without insight (Carrega 1981).Elegant trigonometric solution: K.Schellbach, Crelle 1853. – p.16/53

Immediately following was the“allgemeinere Aufgabe”(Crelle1, p.180):

Given 3 circles,find 3 “Malfatti” circles.

and “Aufgaben” for spheresand spherical triangles.

– p.17/53

3. Apollonius’ three circle problem.Viète (1595) challenged Adriaan van Roomen with theProblem: Given 3circles, find circleswhich touch all 3.

r1r2

r3

r

−r

Pappus (Intro. to Book VII): reports on (lost) books ofApolloniusOn Contacts(11 probl., 21 lemmas, 60 thms !,“propositiones ... esse plures ... sed nos unam posuimus”):Three objects given (“Punctis, et rectis lineis, et circulis”)describe a circle tangent to all three ( “datarum contingat”).

– p.18/53

Van Roomenanswered:easy !Find the center of the circle using 2 hyperbolas,(the sets of points with same distance from 2 circles).

Vi ete: Remember the requirements of Greek culture ...

– p.19/53

Three Types of Constructions(Pappus, p. 7):

• “plane” problems (επιπεδα): with ruler and compass;• “solid” problems (στερεα): using conics;• “grammical” problems (γραµµικα): using other curves.

(Ptolemy, Almagest, ed. 1496)

u · v = z

v = u2

u – p.20/53

Francois Viete(1600):“clarissime Apolloni Belga”

(Viète,Apollonius Gallus1600, Opera p. 325) (Viète,Apoll. Gallus1600, Opera p. 338)

(Viète,Apollonius Gallus1600, Opera p. 325)

the problem is planar, no hyperbolas required !!Viète solves Pappus’ 10 problems geometr. one after the other.– p.21/53

Surprise: Newton. (Principia, ed.1726, Lemma XVI).

F

P

van Roomen’s solution is planar too !!

• because both hyperbolas have a common focus;

• by using Pappus VII.238:PF =1

ei· dist. P -directrixi ;

gives additional linear equation.– p.22/53

Descartes.(1643): lettres to Elisabeth, princess of Bohemia.Mathematical problems are best remedy for cultivating the human mind.

Three circle problem of Apollonius is best math. problem.!

“Vostre Altesse” did not answer for a while ...

Desc.: Problem difficult; even an Angel finds solution only bymiracle!

z

y

d f

e

a

x

b

x

c

xG

EFA

B

C

D

(c)

Desc: slant lines in above pict. ... terrible calc. ... no result before three months;

Desc: intr. orthog. lines below. unknownsx, y, z. ... (birth orth. Cart. coord!)Pythagoras: 3 quadratic eqns; subtract ... linear equs;‘->“le Probleme est plan”.all these tedious calculations,“trop ennuyeux à Vostre Altesse”(-> Gergonne)– p.23/53

Gergonne’s solution(1813; fig. from Dörrie 1932)

(I,II,III =ext. sim. centers)

( 1, 2, 3 =poles of similarity axis)

(O = point ofcommon power.)

Proofby tediousanalytic calculations(see Descartes).

– p.24/53

Jakob Steiner: (> 300 pages)Allgemeine Theorie über dasBerühren und Schneiden der Kreise und Kugeln, from the years1825-1826, unpublished unti 1931,> 300 pages;discusses Apollonius 3-circle problem in detail (p.175):

“... ein Muster von Eleganz und Einfachheit” (R.Fueter, F.Gonseth 1930)– p.25/53

complete algorithm (drawn by Dr. F. Züllig, 1930)

Similar to Gergonne, but elegant geom. proof andall 8 sols.– p.26/53

1936, Frederick Soddy, Nobel in chemistry, publ. inNature:

Understand ?Go back again three centuries...– p.27/53

Descartes.(Nov. 1643): next letter to Elisabeth.Desc: try the simpler case where three given circles of radiusa, b, c touch each other.

Descartes’ drawing, 1901 edition of Œuvres, vol. 4, p. 48, BGE Ca1227/4

Here, says Descartes, “Vostre Altesse” would arrive at thisequation

a2b2c2 + a2b2x2 + a2c2x2 + b2c2x2 =

2(

abc2x2 + ab2cx2 + a2bcx2 + ab2c2x + a2bc2x + a2b2cx)

(without proof !) or, as we write it today1a2 + 1

b2 + 1c2 + 1

x2 = 2(

1ab + 1

ac + 1bc + 1

ax + 1bx + 1

cx

)

.– p.28/53

Steiner’s proof. (Crelle1, p. 273):

a b

c

pc

hc

A B

C

Pappus’ “ancient” thm:pc

x=

hc

c+ 2

orpc

hc

= x

(

1

c+

2

hc

)

;

“Known formula”pa

ha

+pb

hb

+pc

hc

= 1

(Euler) gives

1 = x

(

1

a+

1

b+

1

c+

2

ha

+2

hb

+2

hc

)

;

Insert Eucl. I.41:2

hc

=a + b

A and∼ for a, b;

Finally insert Heron’s formula forA:

1

x=

1

a+

1

b+

1

c+

b + c

A +c + a

A +a + b

A

=1

a+

1

b+

1

c+

2(a + b + c)√

(a + b + c)abc=

1

a+

1

b+

1

c± 2

√

1

ab+

1

bc+

1

ca.

Formula becomes simpler if radii are replaced by inverses (the “bends” of Soddy).– p.29/53

“Apollonian packings” (Lagarias, Mallows and Wilks 2002):Nice discovery: if four consecutive “bends” are integers, thenthe√... is integer (use formula forwards and backwards), thenall bends remain integer.Examples.

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919

922

823

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55

139

154

207

234306

247

274

331

235391

442

291

426

498

427

562

583

310511

547

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462

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223454

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487619

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583

715814298

559

610

522699

834

658

835

919

478

847862

702

987

742

343694

715

727934

783

990

418799

835

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858

523

967982

867907

195

447

474

571682

850

643 754

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495867

954711

943

630

846

267 582

618

691847

787

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522939

810

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483

738

783

370

847

859

895

955

442

982

922

538

199

490514

687

795751

859

694

918

814

511

811

703

234

567

594

778

907

850

979

759

894

522

867

738

514 802

826

55

139

154

207

234306

247

274

331

235391

442

291

426

498

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562

583

310511

547

366 546

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642663

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522699

834

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478

847862

702

987

742

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727934

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990

418799

835

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858

523

967982

867907

195

447

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571

682

850

643 754

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495867

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943

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267 582

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738

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370

847

859

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955

442

982

922

538

199

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795751

859

694

918

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511

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234

567

594

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907

850

979

759

894

522

867

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514 802

826

439

474

579

439

474

579

70

187

202

294

327 426

334

367

451

355

586

658

427

631

730

619

823

850

430706

763

502751

835

654

903

930

358727

778

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910

463

874

946

823

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478967

583

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315726

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931

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723

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262

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934

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742

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307799

826

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307

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522

567 738

562

607

763

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795

766

870

778

943

898

771

987

946

682

927

922

547

787

862

426

615

826

403

483

106

307

322

522

567 738

562

607

763

691

795

766

870

778

943

898

771

987

946

682

927

922

547

787

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426

615

826

403

483

898

978

898

978

127

379

394

663

714 930

703

754

955

907

982

883

687

523

763

487

586

127

379

394

663

714 930

703

754

955

907

982

883

687

523

763

487

586

150

459

474

822

879

862

919

843

630

927

579

699

150

459

474

822

879

862

919

843

630

927

579

699

175

547

562

999

747

679

822

175

547

562

999

747

679

822

202

643

658

874

787

955

202

643

658

874

787

955

231

747762

903

231

747

762

903

262

859874

262

859

874

295

979994

295

979994

330

330

367

367

406

406

447

447

490

490

535

535

582

582

631

631

682

682

735

735

790

790

847

847

906

906

967

967

-2– p.30/53

Towards Steiner’s Porism.What happens with the “Shoemaker’s knife” if the two circlesare no longer tangent ? Then the line of common power is nolonger tangent:

C1

C2

Steiner 1826, Crelle Journal1, p.288

Long search for an interesting theorem also in this case ...

– p.31/53

Steiner’s Porism(Crelle1, 1826, p.254, Art.22)“... nachstehende merkwürdige Folgerungen ziehen.”

Γ1

Γ2

Γ3

Γ4

Γ5

Γ6Γ7

Γ8

Γ9

C1

C2

Γ1

Γ2

Γ3

Γ4

Γ5

Γ6Γ7

Γ8

Γ9

C1

C2

If the space between two fixed circlesC1 andC2 is filled withcirclesΓ1,Γ2, ... all touching each other and the circles, and iffor somen Γn+1 = Γ1, then the same holds forany initialposition ofΓ1.Exists simple proof with inversion map;known to Steiner ???

– p.32/53

Excursion: Stereographic projection.

S2

N

R2

Proj. from pole to equator plane:• maps circles to circles (Ptolemy)

(rediscovered Miquel 1838)

• preserves angles (Halley 1696)

N

A

B

O

A′

α1

α2

α3

α4

γ

γ′

A

B

E

F

N

A′

– p.33/53

1 x3

x4

P1 P2P3 P4O

S

DN

Inversion Map P3 7→ P4:composition stereogr. mapS → P3 → D

with stereogr. mapN → D → P4

⇒ preserves circles and angles !⇒ OP3 · OP4 = R2, P1P2P3P4 harm.

P3

P3

P4

P4

O

– p.34/53

Steiner Porism become trivialityand Kissing Circles after inversion !

A′

D′

DB′C ′CA B

O

OO

– p.35/53

Another Excursion:J.-V. Poncelet1812 (Russia) – 1822 (publ):

Given triangleOPQ

given “unit” pointU ,find central projectiontransformingOPQU

to orthonormal grid↓A′

B′

A

B

V

O

U

PQ R

O′

U ′

– p.36/53

is possible: placeQRP horizontally and determineC by Eucl. III.20:

P ′

R′

Q′

C

P

R

Q

O

U

O′

U ′

45◦45◦

Q R P

CD

– p.37/53

transforms circles to conics and vice-versa:

C

P

Q

O

O′

– p.38/53

Pappus VII.143 and Desargues’ theorem becomes triviality:

A

BC

A′ B′ C ′

NM L

a

b111

1c

d

A′ B′

C ′

AB

C

NM L

AB

C

A′

B′

C ′

N ML

F

M

L

NA

B

C

F

A′

B′

C ′

– p.39/53

... as well as Pascal’s theorem(“hexagramma mysticum”⇒ Eucl. III.21) :

P1

P2P3

P4

P5

P6

K LM

P1

P2

P3

P4

P5

P6

γ

γ

– p.40/53

The Cross Ratio:Remember the def. ofharmonic mean:

(Pappus, Book III, p. 12 of ed. 1660)

q e dh p

x1 x2x3 x4

P1 P2P3 P4

h = harm. m. ofp, q ⇔ q

e=

p

d⇔ − x3 − x1

x3 − x2=

x4 − x1

x4 − x2.

or if thecross-ratio

XR (P1, P2, P3, P4) =x3 − x1

x3 − x2:x4 − x1

x4 − x2= −1 .

The four pointsP1, P2, P3, P4 are then calledharmonic.– p.41/53

J.Steiner,“Systematische Entwickelung ...”(240 pages, 1832):Theorem (Pappus VII.129). The cross-ratio of four points

XR (P1, P2, P3, P4) =P1P3

P2P3:P1P4

P2P4=

x3 − x1

x3 − x2:x4 − x1

x4 − x2

is invariant under projective transformations.Steiner’s Proof:

sine rule:

P1P3

P2P3=

sin a1a3

sin a2a3· sin β

sinα,

P1P4

P2P4=

sin a1a4

sin a2a4· sin β

sinα.

⇒ XR =sin a1a3

sin a2a3:sin a1a4

sin a2a4.

a1 a2 a3 a4

C

P1 P2 P3 P4

P ′

1

P ′

2 P ′

3

P ′

4

α β

XR depends only on angles atC, ⇒ XR (a1, a2, a3, a4).– p.42/53

Example: The Butterfly. 4 pointsU,A,B′, U ′ on circle;(Eucl. III.2: α = α, β = β, γ = γ and Thm. Steiner)⇒XR (UB,AB,B′B,U ′B) = XR (UA′, AA′, B′A′, U ′A′)

⇒ XR (U,F,C, U ′) = XR (U,C, F ′, U ′) or1

m+

1

q=

1

p+

1

n.

q pn m

α

α

β

β

γ

γ

A

B

A′

B′

CF F ′U U ′

same is truefor any conic(after projection)“Steiner’s Theorem”

– p.43/53

The Triangle. Steiner’s view of the Euler line.

Rr

r

2

3a 1

3a

4

3a

aa2a

O G N H

A

B

C

H

A′

B′

C′

H ′=O

G

A

B

C

H

G

A′

B′

C′

O

O′=N

DE

F

K

L

M

ABC → A′B′C ′ = medial red., side bisectors→ altitudes,circumcircleR → nine point circler = R

2 ,G inner,H outer sim. center⇒ aligned,GO = 2HG,NO = HN .– p.44/53

The Simson Line.(F.-J. Servois 1813):Géométrie pratique:“le théorème suivant, qui est, je crois, de Simson” (it was not!!):

P on circumcircle;D,E, F orth, proj. to three sides⇒ aligned.

RA

B

C

P

F

D

E

(a)

δδ

αR

A

B

C

P

F

D

E

(b)

Proof(Servois):• draw circles with diam.PA,PB,PC.;• Thales circles:D,E, F lie on them;• Eucl. III.22: δ = δ (opposite toα);• grey angles atP equal; Eucl. III.21: grey angles atD equal.– p.45/53

Steiner Deltoid. (Crelle-Borchardt 1857, “... schon beimgeradlinigen Dreieck

...”) A

B

C

P P ′

F

F ′D

E

H

O

N

Q Q′

R

R′

G

rot. on circumcirc., rot. on nine-point circ, ang.′ 1 ang. ′

– p.46/53

whathappens?

τ

t

Q

N P

Q

R

A

B

C

– p.47/53

either created by pointP on rolling circler = 13

t

t

t

t/2

2t

S A

Q

C

D

PU

V

or by diameterV U of rolling circle r = 23

S

S

– p.48/53

Our last challenge of Steiner for his readers.(Crelle 1846)

r

r

B

A

O

C

D

r radius of oscul. circ. of ellipse,

r “nach aussen verlängert...”,

- - - - - circle of radius√

a2 + b2;

show thatABCD are harmonic!

Steiner: no proof, no hint! – p.49/53

Our last challenge of Steiner for his readers.(Crelle 1846)

r

r

B

A

O

C

D

r radius of oscul. circ. of ellipse,

r “nach aussen verlängert...”,

- - - - - circle of radius√

a2 + b2;

show thatABCD are harmonic!

Steiner: no proof, no hint!

r

r

r′ ρ

r′

h

B

A

O

D

C ′

D′

S

A′

F

Proof. D′OBC ′A′ diam. of Monge circle,

coord.F = (c3(a − b2

a), s3(b − a2

b))

andB = (ca, sb) known since Newton,

have to showρ(ρ + r′) = a2 + b2.– p.49/53

... and a very last ...Crelle (1846b):

Given pointD on ellipse (other than a vertex). Then exist threepointsA,B,C on the same ellipse, whose osculatory circlepasses throughD.

Moreover,A,B,C,D are cocyclic.

A

B

C

D

– p.50/53

... and a very last ...Crelle (1846b):

Given pointD on ellipse (other than a vertex). Then exist threepointsA,B,C on the same ellipse, whose osculatory circlepasses throughD.

Moreover,A,B,C,D are cocyclic.

A

B

C

D

Hint for solution:

3αα

120◦

120◦

A

B

C

D

– p.50/53

Little autobiographic note of the speaker:When playing around, at the age of 17, with a stone attached toa rope, he discovered that

angleBON = 3 angleNOA

α3α

O

A

B

N

which he verified analytically.He now knows that this is nearly the same as above result ofSteiner :-)

– p.51/53

Postscriptum (“Schlussbemerkung”):

“In den Abhandlungen ... [Steiners] ...findet sich eine solche Ueberfülle

ohne Beweis ausgesprochener Sätze, dass ...auf eine eingehende Prüfung

ihrer Richtigkeit verzichtet werden musste.”

W[eierstrass].

– p.52/53

Towards the end of his life,Steiner (unmarried professor

in Berlin) became rich.In his last will,

he gave one third of his fortuneto the village of Utzenstorf,

so thatthesechildrenwould get a better school education

than he had.

– p.53/53