199. Some properties of symmedian. Two extensions.

by Virgil Nicula, Dec 27, 2010, 9:23 AM

Let $ABC$ be a triangle with circumcircle $w$ . Denote $T\in BB\cap CC$ . Prove that the ray $[AT$ is the $A$-symmedian of $\triangle ABC$ .

Proof 1 (metrical). Denote $\{A,R\}=AT\cap w$ and $S\in AT\cap BC$ . Thus, $\left\{\begin{array}{ccc} \triangle ABR\ : & \frac {TR}{TA}=\left(\frac {BR}{BA}\right)^2\\\\ \triangle ACR\ : & \frac {TR}{TA}=\left(\frac {CR}{CA}\right)^2\end{array}\right\|\implies$

$\frac {RB}{RC}=\frac cb$ . Since $\frac {SB}{SC}=\frac {BA\cdot BR}{CA\cdot CR}$ obtain that $\frac {SB}{SC}=\left(\frac cb\right)^2$ , i.e. the ray $[AT$ is the $A$-symmedian of $\triangle ABC$ .

Remark. Since $\frac {RB}{RC}=\frac {AB}{AC}$ , i.e. $RB\cdot AC=RC\cdot AB$ obtain that the cyclical quadrilateral $ABRC$ is harmonically.

Proof 2 (synthetical). Denote $X\in AA\cap BB$ , $Y\in AA\cap CC$ , $U\in AB$ so that $TU\parallel AA$ and $V\in AC$ so that $TV\parallel AA$ . Prove easily that

$TU=TV=TB=TC$ , i.e. $BCVU$ is cyclically. Thus, $BC$ is antiparallel to $UV$ in $\triangle AUV$ . In conclusion, the ray $[AT$ is $A$-symmedian in $\triangle ABC$ .


Let $AS$ be the $A$-symmedian in the triangle $ABC$ , where $S\in (BC)$ . Denote the midpoints $M$ , $N$ of $[AC]$ , $[AB]$ respectively and

$D\in BM\cap AS$ , $E\in CN\cap AS$ . Prove that $ m\left(\widehat{ACD}\right)=m\left(\widehat{ABE}\right)=\phi$ , where $\tan\phi=\frac {4S}{a^2+b^2+c^2}$ and $\phi\le 30^{\circ}$ . See here.

Generalization 1. Let $AS$ be the $A$-cevian in $\triangle ABC$ , where $S\in (BC)$ . Denote the midpoints $M$ , $N$ of $[AC]$ , $[AB]$ respectively and $D\in BM\cap AS$ ,

$E\in CN\cap AS$ . Prove that $ m\left(\widehat{ACD}\right)=m\left(\widehat{ABE}\right)$ $\iff$ the cevian $AS$ is the $A$-symmedian and in this case $\tan\phi=\frac {4S}{a^2+b^2+c^2}$ , where $\phi\le 30^{\circ}$ .

Proof. Let $\begin{array}{c} G\in AB\cap CD\\\ F\in AC\cap BE\end{array}$ . Since $ D$ , $ E$ belong to the medians $ BM$ , $ CN$ respectively obtain that $SG\parallel AC$ and $SF\parallel AB$ , i.e. $AGSF$ is a parallelogram.

$\boxed {\Longleftarrow}$ Suppose that $S$ is the foot of the $A$-symmedian. Using the point $(1)$ from the lower lemma obtain $GF$ is antiparallel to $BC$ $\implies$ $GBFC$ is cyclically $\implies$

$\widehat{GBF}\equiv\widehat{GCF}$ $\implies$ $ \widehat{ABE}\equiv\widehat{ACD}=\phi$ . Observe that $\frac {c^2}{b^2}=$ $\frac {SB}{SC}=$ $\frac {FA}{FC}=$ $\frac {BA}{BC}\cdot\frac {\sin \widehat {FBA}}{\sin\widehat{FBC}}=$ $\frac {c\cdot \sin\phi}{a\cdot \sin (B-\phi )}$ $\implies$ $ac\cdot \sin (B-\phi )=b^2\cdot\sin\phi$ $\implies$

$ac\cdot\sin B-ac\cdot\cos B\tan\phi =$ $b^2\cdot\tan\phi$ $\implies$ $\tan\phi=\frac {ac\cdot \sin B}{ac\cdot\cos B+b^2}$ $\implies$ $\boxed{\tan\phi =\frac {4S}{a^2+b^2+c^2}}$ .

Using the remarkable inequality $a^2+b^2+c^2\ge 4S\sqrt 3$ obtain that and $\phi\le 30^{\circ}$ .

$\boxed{\Longleftrightarrow}$ Denote $m(\widehat{ABE})=\phi$ , $m(\widehat{ACD})=\psi$ and $\frac {SB}{SC}=m$ . Observe that $\left\|\begin{array}{ccc} m=\frac {FA}{FC}=\frac ca\cdot \frac{\sin \phi}{\sin(B-\phi )} & \implies & \tan\phi =\frac {ma\sin B}{c+ma\cos B}\\\\ m=\frac {GB}{GA}=\frac ab\cdot \frac{\sin (C-\psi )}{\sin\psi} & \implies & \tan\psi =\frac {a\sin C}{mb+a\cos C}\end{array}\right\|$ .

Therefore, $\phi =\psi$ $\iff$ $\tan\phi =\tan\psi$ $\iff$ $b\sin B\cdot \underline m^2+a\sin (B-C)\cdot \underline m-c\sin C=0$ . Observe that $(-1)$ verifies

the previous equation in $m$ because $b\sin B+a\sin (B-C)-c\sin C=0$ $\iff$ $\sin^2B-\sin^2C=\sin A\sin (B-C)$ ,

what is truly. In conclusion, $m=\frac {c\sin C}{b\sin B}$ , i.e. $m=\frac {c^2}{b^2}$ $\iff$ the point $S$ is the foot of the $A$-symmedian.

$\blacktriangleright$ Lemma (well-known). Consider in $\triangle ABC$ the points $M\in (BC)$ , $N\in (CA)$ and $P\in (AB)$ . Find the point $M$ for which :

$(1).\ MN\parallel AB\ ,\ MP\parallel AC\ \implies\ NP$ is antiparallel to $BC$ . Answer : $M$ is the foot of $A$-symmedian.

$(2).\ MN$ is antiparallel to $AB\ ,\ MP$ is antiparallel to $AC\ \implies\ NP\parallel BC$ . Answer : $M$ is the foot of $A$-symmedian.

$(3).\ MN$ is antiparallel to $AB\ ,\ MP$ is antiparallel to $AC\ \implies\ NP$ is antiparallel to $BC$ . Answer : $M$ is the foot of $A$-altitude.

Note. Say that in $\triangle ABC$ the line $MN$ is antiparallel to $AB$ $\iff$ $M\in CA$ , $N\in CB$ and $AMNB$ is cyclically.

Generalization 2. Consider the points $S\in (BC)$ and $M\in (CA)$ , $N\in (AB)$ so that $\frac {MA}{MC}\cdot\frac {NA}{NC}=1$ . Denote

$D\in BM\cap AS$ , $E\in CN\cap AS$ . Prove that $ m\left(\widehat{ACD}\right)=m\left(\widehat{ABE}\right)$ $\iff$ $\frac {SB}{SC}=\left(\frac {AB}{AC}\right)^2\cdot \frac {MA}{MC}$ .

Proof. Denote $\frac {SB}{SC}=s\ ,\ \frac {MA}{MC}=m\ ,\ \frac {NA}{NB}=n$ , where $mn=1$ and $F\in AC\cap BE\ ,\ G\in AB\cap CD$ . Apply the Ceva's theorem

to the points $D\ ,\ E$ in the $\triangle ABC$ and obtain that $\frac {FA}{FC}=ns\ ,\ \frac {GA}{GB}=\frac ms$ . Denote $m(\widehat{ABE})=\phi\ ,\ m(\widehat{ACD})=\psi$ . Thus,

$\left|\begin{array}{ccc} ns=\frac {FA}{FC}=\frac ca\cdot\frac {\sin\phi}{\sin (B-\phi )} & \implies & \tan\phi =\frac {ans\sin B}{c+ans\cos B}\\\\ \frac ms=\frac {GA}{GB}=\frac ba\cdot \frac {\sin\psi}{\sin (C-\psi )} & \implies & \tan\psi =\frac {am\sin C}{bs+am\cos C}\end{array}\right|$ . Therefore, $\phi =\psi$ $\iff$ $\tan\phi =\tan\psi$ $\iff$ $\frac {ns\sin B}{c+ans\cos B}=\frac {m\sin C}{bs+am\cos C}$ $\iff$

$\boxed {\ bn\sin B\cdot\underline s^2+a\sin (B-C)\cdot \underline s-cm\sin C=0\ }$ . Using $mn=1$ prove easily that $\Delta =4R^2\left(\sin ^2B+\sin^2C\right)^2$ and obtain that

$s=\frac {-2R\sin (B+C)\sin (B-C)+2R\left(\sin^2B+\sin^2C\right)}{4nR\sin^2B}=$ $\frac {-\left(\sin^2B-\sin^2C\right)+\left(\sin^2B+\sin^2C\right)}{2n\sin^2B}$ , i.e. $s=m\cdot\left(\frac cb\right)^2$ because $mn=1$ and

$\frac {\sin C}{\sin B}=\frac cb$ . In conclusion, $\boxed{\phi =\psi\ \iff\ s=m\cdot\left(\frac cb\right)^2}$ . Prove easily that in this case $\tan\phi =\tan\psi =\frac {4S}{a^2+b^2+c^2}$ (constant) and $\phi\le\frac {\pi}{6}$ .
This post has been edited 22 times. Last edited by Virgil Nicula, Nov 22, 2015, 5:27 PM

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62. Nice and easy problem with a right triangle.
61. IMO Shortlist 2009 (very nice problem and its proof).
60. Mediteranean M.C. 2010 Problem 3.
59. IMO 2010, problem 4.
58. A cevian and two incircles.
57. Opinii si comentarii relativ la inv. rom. II
56. A extremum problem with areas.
55. Two important circles - mixtilinear incircle/excircle.
54. Cinci grade de libertate.
53. An remarkable concurrency.
52. Harmonical division.
51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.
51. A nice problem with perpendicularity from Jaime.
50. IMAC 2010 seniors, the first day.
49. The midpoint of a cevian in an acute triangle.
+ June 2010
48. An inequality in a triangle : h_a+max{b,c} ...
47. Interesting and difficult identities in a triangle.
46. Outside of a quadrilateral.
45. \cos B+\cos C=1 <==> nice property.
44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.
43. A "slicing" problem with a parameter.
42. IMAC 2010 Problema 2.
41. ONM 2010 Calarasi Problema 3.
+ May 2010
40. Lema lui Haruki.
39. A nice and remarkable problem with Brocard's point.
38. Minimum area of triangle touching with semicircle.
37. Own extension of the Steiner-Lehmus' problem.
36. ONM (juniori) - 2010 Iasi, problema 4.
35. Nice problem from Arqady (Eduard_Tur).
34. IMO Shortlist 2002 geometry problem 7.
+ April 2010
32. Linkuri diverse.
31.Some nice metrical problems.
29. Inegalitate integrala de la Kunny.
28. A fost odată ...
27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

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    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

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    by OlympusHero, May 14, 2020, 8:43 PM

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  • Joined: Jun 22, 2005
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  • Total entries: 456
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