COLLECTION MATHÉMATIQUE

ΣΥΝΑΓΩΓΉ ΜΑΘΗΜΑΤΙΚΉ

ABOUT

THE SYMMEDIAN's LINE

02. THE POINT

IS ON

A SYMMEDIAN

Jean - Louis AYME 1

A

B C

SaH

K

I

Ba

Résumé. L'article présente une compilation non exhaustive, appelée à être complété par de nouveaux exercices, concernant "des points remarquables sur une symédiane". Ce regroupement d'exercices autour d'un même thème permet à l'auteur de trouver quelques liens entre eux. Les figures sont toutes en position générale et tous les théorèmes cités peuvent tous être démontrés synthétiquement.

Abstract. he article presents a non-exhaustive compilation, to be supplemented by new exercises, concerning "remarkable points on a symedian". This grouping of exercises around the same theme allows the author to find some links between them.

The figures are all in general position and all cited theorems can all be shown synthetically.

1 St-Denis, Île de la Réunion (Océan indien, France), le 30/09/2019 ; jeanlouisayme@yahoo.fr

2

2

Resumen. El artículo presenta una compilación no exhaustiva, que se complementará

con nuevos ejercicios, relativos a "puntos notables sobre un symedian". Esta agrupación de ejercicios alrededor del mismo tema permite al autor encontrar algunos vínculos entre ellos.Las figuras están todos en posición general y todos los teoremas mencionados pueden todos ser demostrados sintéticamente.

Zusammenfassung. Der Artikel präsentiert eine nicht erschöpfende Zusammenstellung, die durch neue Übungen ergänzt werden soll, die "bemerkenswerte Punkte auf einem Symedian" betreffen. Diese Gruppierung von Übungen rund um das gleiche Thema ermöglicht es dem Autor, einige Verbindungen zwischen ihnen zu finden.

Die Figuren sind alle in einer allgemeinen Position und alle genannten Lehrsätze synthetisch nachgewiesen werden können.

Summary

A. Récapitulation 4

B. Les problèmes 7

C. Lexique Français-Anglais

3

3

A. RÉCAPITULATION

1. R is on Sa 2. I is on Sa

3. BY is parallel to CZ 4. P is on Sa.

5. Sa, Pb, DE and A'C' are concurrent 6. C* is on the C-symmedian of ABC

A

B C

Sa

Ba B'

C'

R L

A

B C

Sa H

K

I

Ba

A

B C

P

TcTb N

L

M

Z

Y

0

A

B C

B"

B'

C"

C'

P

Sa

A

B C

E

D A'

C'

0

Sa

Ta

Pb

A

B C

C*

4

4

7. A'' is on the A-symmedian of ABC 8. AD is the A-symmedian of ABC

A

B C

O

A'

A"

A

B C M

P

0

E F

N

D

0a

5

5

6

6

7

7

B. LES PROBLÈMES

8

8

PROBLÈME 1 2

A Maurice d'Ocagne's construction

(1880)

VISION

Figure :

A

B C

Sa

BaB'

C'

R

L

Features : ABC a triangle, B' the point on [AC[ so that AB' = AB, C' the point on [AB[ so that AC' = AC, L the point of intersection of B'C' and BC, R the midpoint of the segment B'C' and Sa the A-symmedian of ABC. Given : R is on Sa.

VISUALIZATION

2 d'Ocagne M., Journal de Mathématiques Élémentaires et Spéciales IV (1880) 539.

9

9

A

B C

Sa

BaB'

C'

R

L

0

0a

Ta

• Note 0 the circumcircle of ABC Ta the tangent to 0 at A. • Remarks : (1) Ba i.e. (AL) is the A-bissector of ABC

(2) CC' // BB'

(3) the trapez BB'CC' ic concyclic. • Note 0a this circle. • The circles 0 and 0a, the basic points B and C, the monians ABC' and ACB', lead to the Reim's theorem 1 ; consequently, Ta // C'B'. • According to 00. Problem 7, AR is the A-symmedian of AC'C. • Conclusion : the symmetric of AR wrt AL being the A-median of ABC, R is on Sa. Historical note : this construction is substancially equivalent to a more complicated one given by Constantin Harkema 3 of St. Petersburg

3 Harkema Const., Schlömilch’s Zeitschrift XVI (1871) 168

10

10

PROBLÈME 2 4

A Maurice d'Ocagne's construction

(1883)

VISION

Figure :

A

B C

SaH

K

I

Ba

Features : ABC a triangle, Ba the A-bissector of ABC, H, K the feet of the perpendiculars to Ba through B, C, I the point of intersection of the parallel to AB through H

and the parallel to AC through K and Sa the A-symmedian of ABC. Given : I is on Sa.

VISUALIZATION

4 d'Ocagne M., Sur un élément du triangle rectiligne ; symédiane, Nouvelles Annales de Mathématiques, 3ème série, II (1883) 464,

exercice 7 ; http://www.numdam.org/journals/NAM/ Isogonal conjugate and symmedian, AoPS du 10/12/2013 ; http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=566565 Interesting symmedian problem, AoPS du 20/07/2016 ; http://www.artofproblemsolving.com/community/c6t48f6h1280041_interesting_symmedian_problem

11

11

A

B C

H

K

I

A'

1

2 3

4 5

6 Ba

• Note A' the midpoint of the segment BC. • According to "An unlikely concurrence" 5, A'H // AC.

• According to "The little Pappus theorem" 6 applied to the hexagon ABHA'KCA, A'K // AB. • Partial conclusion : the quadrilateral HIKA' is a parallelogram.

A

B C

H

K

I

A'L

Ba

• The diagonals IA' and HK of HIKA' intersect in their midpoint. • Note L this midpoint. • According the passing axiom IIIb applied to the band limited by BH and CK, A'L // BH ; by hypothesis, BH⊥ AL ; according to the axiom IVa of the perpendiculars, A'L⊥ AL ; consequently, A'L is the mediator of the segment HK. • Partial conclusion : HIKA' is a rhomb.

5 Ayme J.-L., An unlikely concurrence, revisited and generalized, G.G.G. vol. 4, p. 3-5 ; https://jl.ayme.pagesperso-orange.fr/ 6 Ayme J.-L., Une rêverie de Pappus, G.G.G. vol. 6, p. 2-5 ; https://jl.ayme.pagesperso-orange.fr/

12

12

A

B C

H

K

I

A'L

Ba

• Remark : (1) AL is the mediator of the segment A'I. (2) AA' is the A-median of ABC (3) AI is the symmetric of AA' wrt Ba. • Conclusion : A is on Sa. Commentaire : Ross Honsberger 7 a présenté ce problème dans son livre Episodes in Euclidean Geometry en le

mettant dans la catégorie "difficile".

7 Honsberger Ross, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, New Mathematical Library (1995)

exercice 7.6, p. 76.

13

13

PROBLÈME 3 8

A converse Maurice d'Ocagne's construction

VISION

Figure :

A

B C

P

TcTb

N

L

M

Z

Y

0

Features : ABC a triangle, Tb, Tc the tangents to 0 wrt B, C, P the point of intersection of Tb and Tc, LMN the medial triangle of ABC and Y, Z the point of intersection of AP wrt LN, LM. Given : BY is parallel to CZ.

VISUALIZATION

8 Parallels segments M, M-20, Mathlinks du 16/12/2009 ; http://www.mathlinks.ro/Forum/viewtopic.php?t=318910

14

14

A

B C

P

TcTb

N

L

M

Z

Y

0

Ta

• Note Ta the tangent to 0 at A. • According to 00. Problem 3. AP is the A-symmedian of ABC. • According to ''Another unlikely concurrence'' 9, BY // Ta and Ta // CZ ; the relation ''is parallel to'' being transitive, BY // CZ. • Conclusion : BY is parallel to CZ.

9 Ayme J.-L., Another unlikely concurrence, G.G.G. vol. 10 ; https://jl.ayme.pagesperso-orange.fr/ Crux Mathematicorum, (Canada) 8 (2003) 511-513 ; http://math.ca/crux/ ;

15

15

PROBLÈME 4 10

The accidental trase of Ernst Wilhem Grebe

(1847)

VISION Figure :

A

B C

B"

B'

C"

C'

P

Sa

Features : ABC a triangle, AC'C"B, CB'B"A two external squares, Sa the A-symmedian of ABC and P the point of intersection of B'B" and C'C". Given : P is on Sa.

VISUALIZATION

10 Grebe E. W., Das geradlinige Dreieck in Bezug auf die Quadrate der Perpendikel, die man von einem Punkte seiner Ebene auf

seine Seiten Fallen kann, Grünerts Archiv 9 (1847) 250-259 Nouvelles Correspondance (1876) 268 ; solution (1877) 400, (1880) 214, 365 Symmedian, AoPS du 02/01/2017 ; http://www.artofproblemsolving.com/community/c6t48f6h1362892_symmedian

16

16

A

B C

B"

B'

C"

C'

P

O B*

C*0

• According to "Triangle inscriptible in a half circle", the quadrilateral AB"PC' is cyclic. • Note 0 this circle O the center of 0 i.e. the midpoint of AP et B*, C* the second points of intersection of AC, AB with 0 resp.. • According to "Triangle inscriptible in a half circle", the triangles B*AP and C*PA are B*, C*-right angled resp..

A

B C

B"

B'

C"

C'

P

O B*

C*0

• The rectangles AB"PB* and AC*PC' being inscriptible in 0 and having the diagonal AOP in common, the diagonals B"B* and C'C* are two diametral lines of 0 ; consequently, the triangles PB*C* and AB"C' are perspective at O ;

17

17

A

B C

B"

B'

C"

C'

P

O B*

C*

0

• According to "The weak Desargues theorem" 11 applied to the triangles AB"C' and PB*C* which have two corresponding parallel sides, B"C' // B*C*.

A

B C

B"

B'

C"

C'

P

O B*

C*

0

I

• Note I the midpoint of BC. • According to Vecten "D'une médiane à une hauteur" 12, AI ⊥ B"C' ; we know that B"C' // B*C* ; according to the axiom IVa of the perpendiculars, AI ⊥ B*C*. • Conclusion : according to 07. Problem 1. A Maurice d'Ocagne exercise, P is on Sa. Historical note : the nature of this line which don't know the Doctor Ernst Wilhelm Grebe of Cassel

(Germany), has been discovered in 1876 by E. Hain 13 and proposed as exercise in 1883 by Maurice d'Ocagne 14.

11 Ayme J.-L., Une rêverie de Pappus, G.G.G. vol. 6, p. 6-10 ; https://jl.ayme.pagesperso-orange.fr/ 12 Ayme J.-L., La figure de Vecten. G.G.G. vol. 5, p.102 ; https://jl.ayme.pagesperso-orange.fr/ 13 Hain E., Ueber den Grebeschen Punkt, Archiv der Mathematik und Physik 58 (1876) 84-89 14 d'Ocagne M., Sur un élément du triangle rectiligne ; symédiane, Nouvelles Annales de Mathématiques, 3-ème série II (1883) 464,

exercice 2 ; http://www.numdam.org/journals/NAM/

18

18

PROBLÈME 5 15

Jean-Louis Ayme

An unlikely concurrence on a symmedian

(2003)

VISION

Figure :

A

B C

E

D A'

C'

0

Sa

Ta

Pb

Features : ABC a triangle, 0 the circumcircle of ABC, D, E the feet of the A, B-altitudes of ABC resp., A', C' the midpoints of BC, AB resp., Ta the tangent to 0 at A, Sa the A-symmedian of ABC

and Pb the parallel to Ta passing through B. Given : Sa, Pb, DE and A'C' are concurrent.

VISUALIZATION

15 Ayme J.-L., Crux Mathematicorum, (Canada) 8 (2003) 511-513 ; http://math.ca/crux/ ; Another unlikely concurrence, G.G.G. vol. 10 ; https://jl.ayme.pagesperso-orange.fr/

19

19

A

B C

M

E

D

N

T

S

A'

C'

L

0

Ta

Tb TcPb

1

Pn

• Note Ta, Tb, Tc the tangents to 0 at A, B, C resp., L, M the points of intersection of Pb with AC, DE resp., N the point of intersection of Tb and Tc, Pn the parallel to Ta passing through N and S, T the intersections of Pn with AC, AB resp..

• According to 00. Problems 3. and 5., Sa = AN.

• According to Boutin's theorem 16, B, C, S and T lie on a circle having ST for diameter and N for center we have, CT ⊥ ACS ; but since, ACS ⊥ BE ; we have, CT // BE.

• According to Carnot's theorem 17, Tc // DE.

• Since, Pb // Ta and Ta // Pn, we get : Pb // Pn.

• According to Desargues's theorem 18 applied to the homothetic triangles BME and TNC, M, N and A are collinear. • According to 00. Problem 6. Symmedian and antiparallel, M is the midpoint of BL ; consequently, A', M et C' are collinear.

• Conclusion : Sa = AN, Pb, DE and A'C' are concurrent. Remarks : (1) a more complete figure from the vertex A 16 Ayme J.-L., A propos du théorème de Boutin, G.G.G. vol. 1 ; https://jl.ayme.pagesperso-orange.fr/ 17 The sides of the orthic triangle are parallel to the tangents to the circumcircle at the vertices. 18 If two triangles have an axis of perspective (here the line at infinity), they have a center of perspective.

20

20

A

B C

E

D A'

C'

0

Sa

Ta

Pb

B'

F

Pc

• Note F the foot of the C-altitudes of ABC,

B' the midpoints of AC and Pc the parallel to Ta passing through C.

• Conclusion : mutatis mutandis, we would prove that Sa, Pc, EF and A'B' are concurrent. (2) The line CM

A

B C

M

E

D A'

C'

0

Sa

Ta

Pb

B'

F

Pc

• Conclusion : Ta, B'C' and CM are concurrent. 19

19 Ayme J.-L. ; Another unlikely concurrence, revisited and generalized, G.G.G. vol. 4, p. 20-22 ; https://jl.ayme.pagesperso-orange.fr/.

21

21

PROBLÈME 6 20

Heinrich Dörrie

(1943)

VISION

Figure :

A

B C

C*

Features : ABC a triangle and C*AB a triangle C*-isoceles, outside ABC, such as <C*AB = <ACB. Given : C* is on the C-symmedian of ABC.

VISUALISATION

A

B C

C*

0

• Note 0 the circumcircle of ABC. 20 Dörrie H., Mathematische Miniaturen, F. Hirtin edition (1943) 46

22

22

• According to the Tangent-chordal angled theorem, (1) 0 is tangent to AC* at A

(2) 0 is tangent to BC* at B. • Conclusion : according to 00. Problems 3. and 5., CC* is the C-symmedian of ABC. Commentar : la solution de H. Dörrie a recours aux cercles d'Apollonius.

23

23

PROBLÈME 7 21

USAMO 1995

VISION

Figure :

A

B C

O

A'

A"

Features : ABC a triangle, O the center of the circumcircle of ABC, A' the midpoint of BC, and A'' the point on OA' so that the triangle AOA'' and A'OA are similar. Given : A'' is on the A-symmedian of ABC.

VISUALIZATION

21 Usamo 1995, Mathlinks du 31/12/2004 ; http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=22306 .

24

24

A

B C

O

A'

A"

H

A+

Ba

• Note H the orthocenter of ABC, A+ the foot of the A-symmedian of ABC

and Ba the A-inner bissector of ABC. • Remarks : (1) AA' is the A-median of ABC

(2) AH // OA' (3) H and O are two isogonal points wrt ABC.

• An angular chase : we have : <OA'A = <HAA' ; by symmetry wrt Ba, <HAA' = <A+AO ; by transitivity of the relation =, <OA'A = <A+AO ; AOA'' and A'OA being similar, A, A+ and A'' are collinear. • Conclusion : A'' is on the A-symmedian of ABC.

25

25

PROBLÈME 8 22

A construction of a symmedian

VISION

Figure :

A

B C M

P

0

E F

N

D

0a

Features : ABC a triangle, M the midpoint of the side BC, 0 the circumcircle of ABC, 0a the symmetric of 0 wrt BC, P the point of intersection of 0a with the segment AM, E, F the feet of BP, CP, N the midpoint of the segment EF and D the foot of the perpendicular to BC through N. Given : AD is the A-symmedian of ABC.

VISUALIZATION

22 Simedian, Mathlinks du 02/10/2011 ; http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=435433

26

26

A

B C M

P

0

E F N

D

0a

0'a

Tf Te

• Remarks : (1) AN is the A-median of ABC (2) EF // BC (3) N is on AM (4) the quadrilateral AFPE is cyclic. • Note 0'a this circle. • The circles 0 and 0'a, the basic point A, the monians BAF and CAE, the parallels BC and EF, lead to the Reim's theorem 7' ; consequently, 0 and 0'a are tangent at A. • Note Tf, Te the tangents to 0'a at F, E and D' the point of intersection of Tf and Te. • According to a particular case of MacLaurin- Pascal's theorem, BD'C is the pascal line of the degenerated hexagon AF Tf PE Te A. • ND' being the perpendicular bissector of EF, D' and D are identic. • Partial conclusion : AD is the A-symmedian of the triangle AFE. • Remark : the triangles AFE and ABC are homothetic. • Conclusion : AD is the A-symmedian of the triangle ABC.

27

27

C. LEXIQUE

FRANÇAIS - ANGLAIS A aligné collinear annexe annex axiome axiom appendice appendix adjoint associate a propos by the way btw acutangle acute angle axiome axiom B bissectrice bisector bande strip C centre incenter centre du cercle circonscrit circumcenter cercle circonscrit circumcircle cévienne cevian colinéaire collinear concourance concurrence coincide coincide confondu coincident côté side par conséquence consequently commentaire comment D d'après according to donc therefore droite line d'où hence distinct de different from E extérieur external F figure figure H hauteur altitude hypothèse hypothesis I intérieur internal identique identical i.e. namely incidence incidence L lemme lemma lisibilité legibility M mediane median médiatrice perpendicular bissector milieu midpoint

N Notons name nécessaire necessary note historique historic note O orthocentre orthocenter ou encore otherwise P parallèle parallel parallèles entre elles parallel to each other parallélogramme parallelogram pédal pedal perpendiculaire perpendicular pied foot point de vue point of view postulat postulate point point pour tout for any Q quadrilatère quadrilateral R remerciements thanks reconnaissance acknowledgement respectivement respectively rapport ratio répertorier to index S semblable similar sens clockwise in this order segment segment Sommaire summary symédiane symmedian suffisante sufficient sommet (s) vertex (vertice) T trapèze trapezium tel que such as théorème theorem triangle triangle triangle de contact contact triangle triangle rectangle right-angle triangle