Jakob Steiner 1796 1863

– p.1/53 Jakob Steiner 1796 1863

ne derange pas mes cercles ! – p.1/53 Jakob Steiner 1796 1863

ne derange pas mes cercles ! – p.1/53 (Photos: Barbara Kummer) born 1796 in Utzenstorf (Bern) childhood: work at this farm, no school education

until age of 18 – p.2/53 1814 enters the Pestalozzi school in Yverdon ...

... then as “instituteur”. ﬁrst as student ... (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. Clausen published in Crelle Journal 3, No. 17 the determination of the area of by complicated trigonometry. TC kb E H D T K L ka (Steiner autogr. 1814) B (Steiner, Crelle J. 3, No.19, 1828) A kc F 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 Functionen gelöset; auch kann sie allgemeiner gestellt werden, ...” – p.4/53 Steiner’s solution: u, v, w ratios of side-lengths;

C Draw EG parall. to AF B; b 1+v ua by Thales EG : EC = AF : AC c 1+u hence EG = ; E C G (1+w)(1+v) H By Eucl. I.41 and Thales vb 1 B : = EH : HB = EG : F B = (1+v)w ; 1+v D C B a 1 T 1+u by Eucl. I.41 we have + = v ; K L C B 1+ ·A w B can compute = 1+w+wv ; wc B ·A A c 1+w cyclic permut. gives other triangles, subtract: 1+w F = (1 u v w ) T − 1+u+uw − 1+v+vu − 1+w+wv ·A (uvw 1)2 = − . (1+u+uw)(1+v+vu)(1+w+wv) ·A Observe: = 0 uvw = 1 (Ceva’s Theorem). T ⇔ Result later forgotten and rediscovered as Dudeney–Steinhaus

theorem or Routh’s theorem. – p.5/53 1821: Berlin, short career as teacher for lower maths in Werder Gymnasium, then → “Privatlehrer” and starts writing a huge text on circle and sphere geometry ...

– p.6/53 1821: Berlin, short career as teacher for lower maths in 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 ... gemeinschaftlichen Zusammenhang zu ﬁnden”. General Theory of Circles. ⇒ – p.7/53 Steiner’s “Book of Lemmas”: 1. Power of the point E with respect to the circle: “In den Lehrbüchern der Geometrie ﬁndet man folgenden Satz bewiesen:” E E (a) (b) (c) E ǫ B B α d t D B α δ r T T α r α O A A r C A

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:

C e e B 2 1 h 2

C1 d1 d2 D r2 A C1 r1 C2 E

“Die Bedeutung der gemeinschaftlichen Potenz zweier Kreise, wovon schon bei Pappus und Vieta sich Spuren ﬁnden, lernte er durch ihre, von ihm gefundene vielseitige Anwendbarkeit erkennen”.

– p.9/53 3. Point of equal powers for three circles:

E C3

C1 C2

“Wir wollen diesen Punct p(123) hinfort

‘Punct der gleichen Potenz der drei Kreise M1,M2,M3’ nennen.”

– p.10/53 4. Similarity centers of two circles:

Fermat 1679 Steiner 1826

2r 2 1 r 3a 3a C I C E 2 1 r 4a a 3 a 2a

“Die Franzosen nennen diese Punkte centre de similitude de deux cercles. Wir haben diese Benennung zuerst aus einer Abhandlung von Euler genommen, siehe

L. Euler: de centro similitudinis, 1777”. – p.11/53 5. Common power of two circles with respect to their similarity center.

D Y ′

Y C C X 2 X′ C1

E A′ A B B′ EC = θ EC 2 · 1 C3 EX′ EX = q · 2 EX EY = EX′ EY ′ = θq = EF F · · – p.12/53 1. “Libris antiqua propositio” (“von Pappus überliefert”):

Archimedes’ “Lemma 6”

5 centuries later ...

C r3 C r2 r 1 r2 r4 1 3 m2 r4 m2 r1 C2 h3 m1 h h h2 C2 h4 2 3 h4 r1 h1

m i i 1 h , , , ,... h , , , ,... ri = 0 2 4 6 ri = 1 3 5 7 Pappus IV.16 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.) (a) Tang. in B line of equal powers for C1 and C2, 2; A D (b) Sim. center A for m1 and m2 r2 C1 r b same power, on tang. 4,5; 1 m2 ⇒ (c) Thales: P1B : P2B = r1 : r2; m1 C h (d) AB2 = com.pow. m , m 5,1; h 2 1 2 1 (e) Ab2 = AX AY = AB2 5; C2 · (f) BAb Cm2b Dm1b P1 P2 B DbCB∼ aligned;∼ ⇒ h + r h r h h (g) Thales and (c): 1 1 = 2 − 2 or 2 = 1 + 2 . r1 r2 r2 r1 Followed by many generalizations.

– p.14/53 2. “Lösung der Malfattischen Aufgabe...” (Crelle J. 1, 1826, p.178.)

Problem: Given triangle ABC ﬁnd 3 circles such that ...

Problem appeared 1810/11 in Gergonne J; “les rédacteurs des Annales” 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 of ABC; F draw incircles of AIB,BIC,CIA; C D point of contact; E DEF common tang. to other incircles; incircles of ADE and DBF two Malfatti circles. ⇒ NO PROOF, NO EXPLANATION. I

A later reconstructions of Steiner’s D proof long and without insight (Carrega 1981). B

Elegant trigonometric solution: K.Schellbach, Crelle 1853. – p.16/53 Immediately following was the “allgemeinere Aufgabe” (Crelle 1, p.180): Given 3 circles, ﬁnd 3 “Malfatti” circles. and “Aufgaben” for spheres and spherical triangles.

– p.17/53 3. Apollonius’ three circle problem. Viète (1595) challenged Adriaan van Roomen with the Problem: Given 3 circles, ﬁnd circles which touch all 3.

r r2 r1

r r−3

Pappus (Intro.to Book VII): reports on (lost) books of Apollonius On 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 Roomen answered: easy ! Find the center of the circle using 2 hyperbolas, (the sets of points with same distance from 2 circles).

Viete:` Remember the requirements of Greek culture ...

– p.19/53 Three Types of Constructions (Pappus, p. 7):

“plane” problems (επ´ιπεδα): with ruler and compass; • “solid” problems (στερεα´): using conics; • “grammical” problems (γραµµικα´): using other curves. • v = u2

u v = z ·

(Ptolemy, Almagest, ed. 1496) u – p.20/53 Franc¸ois Viete` (1600): “clarissime Apolloni Belga”

(Viète, Apollonius Gallus 1600, Opera p. 325) (Viète, Apoll. Gallus 1600, Opera p. 338)

(Viète, Apollonius Gallus 1600, 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).

van Roomen’s solution is planar too !! because both hyperbolas have a common focus; • 1 by using Pappus VII.238: P F = dist. P -directrixi ; • ei · 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 by miracle! (c) B

x z e xD G x a y c

A C F E d f Desc: slant lines in above pict. ... terrible calc. ... no result before three months; Desc: intr. orthog. lines below. unknowns x,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 of common power.) Proof by tedious analytic calculations (see Descartes).

– p.24/53 Jakob Steiner:(> 300 pages) Allgemeine Theorie über das Berühren und Schneiden der Kreise und Kugeln, from the years 1825-1826, 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 and all 8 sols.– p.26/53 1936, Frederick Soddy, Nobel in chemistry, publ. in Nature:

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 radius a,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 this equation a2b2c2 + a2b2x2 + a2c2x2 + b2c2x2 = 2 abc2x2 + ab2cx2 + a2bcx2 + ab2c2x + a2bc2x + a2b2cx (without proof !) or, as we write it today 1 1 1 1 1 1 1 1 1 1 2 + 2 + 2 + 2 = 2 + + + + + . a b c x ab ac bc ax bx cx – p.28/53 Steiner’s proof. (Crelle 1, p.273):

pc hc Pappus’ “ancient” thm: = + 2 x c pc 1 2 c or = x + ; C hc c hc pa pb pc hc “Known formula” + + = 1 ha hb hc b (Euler) gives a pc 1 1 1 2 2 2 A B 1= x + + + + + ; a b c ha hb hc 2 a + b Insert Eucl. I.41: = and for a,b; hc ∼ A Finally insert Heron’s formula for : A 1 1 1 1 b + c c + a a + b = + + + + + x a b c A A A 1 1 1 2(a + b + c) 1 1 1 1 1 1 = + + + = + + 2 + + . a b c (a + b + c)abc a b c ± rab bc ca Formula becomes simplerp 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, then the √... is integer (use formula forwards and backwards), then all bends remain integer.

687843 927 763 Examples. 965 845 615 886 788 680 726 423547 483 629 533 367 582 848 488 968 931 315 847 404 454 267 691 365 632 619 223 293 922 342 810 183 810 536 260 912 487 762 147 427 741 741 246 618 939 893 200 368 855 980 588 522 754 115 522775 173 699 490 319 296 677 860 271 166 453 87 943 559 378 909 581 125 834 514 742 989 951 787 644 814 495 867 869 715 104 548 787598435298187 298 336 336 812 282 63 610102 954 176 711 413 919 379 704 984 583 835 990 990 151 861 861 549 322 523 574 967 341 658 466 571 380 237 883 202 768557 128 708 68 768 682 943 682 773 43 279 507 415 512 358 850 934 405 893 702 702 308 727 742 447 828 573 624 235 382 367 987 987 53 989 778 655 391 511 895 774 54 91 221 138 138 138 903 389728 643306754 474 908 869 156 682195 862 847 677 823 451846 930262 933 456 456 442 178 339 426 910 283 426 480 480 485 783 27 370 523 773 630 630 478 478 958 747 984 984 984 874 874 317 723 173 581 159 247718 463291946 675 498 67 927 667 703 536 284692 562 454738 188 357 238 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543 187558 452 771 596 540 718 567 93 917 93 944 22427 605 605 643 869 583210 357 238 536 692 562 90 454738 927 703 284 675 667 188 463 946 498 581 291 718 723 247 984 984 984 874 874 56 958 317 159 478 478 747 173 773 630 630 20 485 910 783 523 480 480 370 933 426 426 677 456 456 823 442 283 451846 930 339 67 862 847 908 869 682 643 262 728 178 754 474 389 306 903 774 511 195 391 895 989 778 655 221 156 987138 138987 138 828 573 624 382 742 893 727 367 447 235 702 702 29 308 934 405 15 91 512 358 415 850 773 507 768 708 768 682 943 682 557 12 883 279 380 658 466 54 571 237 202 967 341 835 523 861 861 549 322 990 990 574 984 704128 919 583 711 610 954 379 413 812 151 548 787598 10 435 715 27 869 53 282 336 336 867 298 814 298 644 176989 951187 787 495 742 909 581 68 102834 514 559 453 943378 677 860 43 699271 980 588 522 754 490 522775 296104 893 855 319 741 741 618 63 166939 125368 427 912 810487246 87 762 810 173 922 536 200632 619 115 260 147 342691 848 454 847 365 293968 931 223 183582 488629 533404 726 267 788965 845680 315423547 615483367 886 -3 687843 927 763 -2 – p.30/53 Towards Steiner’s Porism. What happens with the “Shoemaker’s knife” if the two circles are no longer tangent ? Then the line of common power is no longer tangent:

Steiner 1826, Crelle Journal 1, p.288

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

– p.31/53 Steiner’s Porism (Crelle 1, 1826, p.254, Art.22) “... nachstehende merkwürdige Folgerungen ziehen.”

C1 Γ Γ1 9 Γ1 Γ 9 Γ2 Γ2 Γ C2 C2 Γ3 8 Γ3 Γ4 Γ4 Γ8 Γ5 Γ5 Γ7 Γ6 Γ6 Γ7 C1

If the space between two ﬁxed circles C1 and C2 is ﬁlled with circles Γ1, Γ2, ... all touching each other and the circles, and if for some n Γn+1 =Γ1, then the same holds for any initial position of Γ1. Exists simple proof with inversion map; known to Steiner ??? – p.32/53 Excursion: Stereographic projection. N α1 S2 R2 N O α2 A α3 α4 A′ N B

A Proj. from pole to equator plane: γ maps circles to circles (Ptolemy) E • (rediscovered Miquel 1838) B γ′ A′ preserves angles (Halley 1696) F •

– p.33/53 Inversion Map P P : N 3 7→ 4 D composition stereogr. map S P3 D with stereogr. map N D →P → → → 4 x4 preserves circles and angles ! x3 ⇒ 2 P1 1 O P3 P2 P OP3 OP4 = R , P1P2P3P4 harm. 4 ⇒ · P4

O P3 P4

– p.34/53 SteinerPorism becometriviality and Kissing Circles after inversion !

D ′ ′ A B C ′ B A O C D′

O O

– p.35/53 Another Excursion: V J.-V. Poncelet 1812 (Russia) – 1822 (publ): B Given triangle OP Q A given “unit” point U, ′ ﬁnd central projection B transforming OPQU A′ to orthonormal grid ↓

R Q P

U U ′ O O′ – p.36/53 is possible: place QRP horizontally and determine C by Eucl. III.20: Q Q R R P C ◦ 45 45◦ D P C

Q′ ′ O U ′ R O′

P ′

– p.37/53 transforms circles to conics and vice-versa: Q

O O′

– p.38/53 Pappus VII.143 and Desargues’ theorem becomes triviality: A′ B′ C′ ′ ′ B A C′ N N M L M L d a 1 1 C 1 1 b B c C B A L ′ N M A C C′ A′ L B′ ′ M C A

A ′ B B N C A F B F – p.39/53 ... as well as Pascal’s theorem (“hexagramma mysticum” Eucl.III.21) : ⇒ KM L P2

P P3 2 γ P1 P3 P1

P4 P4 P6 γ P6 P P5 5

– p.40/53 The Cross Ratio: Remember the def. of harmonic mean :

(Pappus, Book III, p. 12 of ed. 1660) P1 P3 P2 P4 x1 x3 x2 x4 q e d h p q p x x x x h = harm.m. of p, q = 3 − 1 = 4 − 1 . ⇔ e d ⇔ − x3 x2 x4 x2 or if the cross-ratio − − x x x x XR (P , P , P , P )= 3 − 1 : 4 − 1 = 1 . 1 2 3 4 x x x x − 3 − 2 4 − 2 The four points P , P , P , P are then called harmonic. 1 2 3 4 – p.41/53 J.Steiner, “Systematische Entwickelung ...” (240 pages, 1832): Theorem (Pappus VII.129). The cross-ratio of four points P P P P x x x x XR (P , P , P , P )= 1 3 : 1 4 = 3 − 1 : 4 − 1 1 2 3 4 P P P P x x x x 2 3 2 4 3 − 2 4 − 2 is invariant under projective transformations. Steiner’s Proof: C sine rule: P P sin a a sin β P ′ 1 3 = 1 3 , 4 P2P3 sin a2a3 · sin α ′ P2 ′ P1P4 sin a1a4 sin β ′ P3 = . P1 P2P4 sin a2a4 · sin α sin a a sin a a XR = 1 3 : 1 4 . P1 α P2 β P3 P4 ⇒ sin a2a3 sin a2a4 a1 a2 a3 a4 XR depends only on angles at C, XR (a , a , a , a ). ⇒ 1 2 3 4 – p.42/53 Example: The Butterﬂy. 4 points U, 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′) 1 1 1 1 XR (U,F,C,U ′)= XR (U, C, F ′,U ′) or + = + . ⇒ m q p n A

B′ same is true n q p m for any conic U U ′ (after projection) F C F ′ α “Steiner’s Theorem” β γ B γ α β

A′ – p.43/53 The Triangle. Steiner’s view of the Euler line.

R 2 a 1 a 3 3 r M O N G r 4 H 3 a

a a ′ A 2a C F C′ A B O B H′ =O O′ =N G G B′ ′ ′ A B H A′ H D K E C L C ABC A′B′C′ = medial red., side bisectors altitudes, circumcircle→ R nine point circle r = 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. ⇒ A A R R F α F B B D D C C δ E E δ (a) (b) P P Proof (Servois): draw circles with diam. PA,PB,PC.; Thales circles: •D,E,F lie on them; • Eucl. III.22: δ = δ (opposite to α); • grey angles at P equal; Eucl.III.21: grey angles at D equal.– p.45/53 • Steiner Deltoid. (Crelle-Borchardt 1857, “... schon beim geradlinigen Dreieck ...”) A

H R N R′ O F

′ B D F C ′ Q Q G E

P P ′ – p.46/53 rot. on circumcirc., rot. on nine-point circ, ang. ′ 1 ang. ′ what happens? A

Q Q τ C t R N P

– p.47/53 1 either created by point P on rolling circle r = 3

D S C V t Q 2t t t/2 t P S U A

or by diameter of rolling circle 2 VU r = 3 – p.48/53 Our last challenge of Steiner for his readers. (Crelle 1846) A

r C

B O r

D r radius of oscul. circ. of ellipse, r “nach aussen verlängert...”, ----- circle of radius √a2 + b2; show that ABCD are harmonic!

Steiner: no proof, no hint! – p.49/53 Our last challenge of Steiner for his readers. (Crelle 1846) A A

′ r r A C r′ C′ B B O r′ r O ρ S r h

F D′ D r radius of oscul. circ. of ellipse, D r “nach aussen verlängert...”, Proof. D′OBC′A′ diam. of Monge circle, b2 a2 ----- circle of radius √a2 + b2; coord. F =(c3(a ), s3(b )) − a − b show that ABCD are harmonic! and B =(ca, sb) known since Newton, ′ 2 2 Steiner: no proof, no hint! have to show ρ(ρ + r ) = a + b . – p.49/53 ... and a very last ... Crelle (1846b): Given point D on ellipse (other than a vertex). Then exist three points A, B, C on the same ellipse, whose osculatory circle passes through D. Moreover, A, B, C, D are cocyclic.

– p.50/53 ... and a very last ... Crelle (1846b): Given point D on ellipse (other than a vertex). Then exist three points A, B, C on the same ellipse, whose osculatory circle passes through D. Moreover, A, B, C, D are cocyclic. Hint for solution:

B B A A ◦ 120 α ◦ 3α 120

D D C C

– p.50/53 Little autobiographic note of the speaker: When playing around, at the age of 17, with a stone attached to a rope, he discovered that N A

α 3α angle BON = 3 angle NOA O B

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

– 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 fortune to the village of Utzenstorf, so that these children would get a better school education than he had.

– p.53/53