Isogonal comjugates and equilateral triangles

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Isogonal comjugates and equilateral triangles

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: KoMaL A. 902

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

499 posts

#1 h Apr 14, 2025, 5:30 PM • 2 Y

Y by NO_SQUARES, sami1618

In triangle $ABC$ , interior point $D$ is chosen such that triangle $BCD$ is equilateral. Let $E$ be the isogonal conjugate of point $D$ with respect to triangle $ABC$ . Define point $P$ on the ray $AB$ such that $AP=BE$ . Similarly, define point $Q$ on the ray $AC$ such that $AQ=CE$ . Prove that line $AD$ bisects segment $PQ$ .

Proposed by Áron Bán-Szabó, Budapest

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

913 posts

#2 h Apr 15, 2025, 12:11 AM • 1 Y

Y by geometry6

Let $F$ be the reflection of $D$ about $BC$ and let $E'$ be the point such that $BECE'$ is a parallelogram. Since $\angle ABE=\angle ACE=60^{\circ}$ , by the Parallelogram Isogonality Lemma, $A$ , $D$ , and $E'$ are collinear. Construct the points $X$ and $Y$ inside $BCF$ such that $\triangle BE'C\cong \triangle FXB\cong \triangle CYF$ . It is easy to see that $BEX$ and $CEY$ are also equilateral triangles.
$[asy] import geometry; size(12cm); pair A=dir(175); pair B=dir(330); pair C=dir(390); pair D=.5(B+C)+rotate(90)*sqrt(3)*.5*(C-B); pair F=.5(B+C)+rotate(90)*sqrt(3)*.5*(B-C); triangle t=triangle(A,B,C); pair E=isogonalconjugate(t,D); pair X=B+rotate(300)*(E-B); pair Y=C+rotate(60)*(E-C); pair Q=A+Y-E; pair P=A+X-E; pair Ep=B+C-E; fill(A--P--Q--cycle,lightred); fill(C--Y--F--cycle, palered); fill(C--E--B--Ep--cycle, palered); fill(B--X--F--cycle, palered); fill(E--X--Y--cycle, lightred); draw(F--C--A--B--C--D--B--F--Y--C--Ep--B--X--F--B--E--C); draw(A--Ep,dashed); draw(P--Q); draw(X--Y); draw(Y--E--X); draw(E--F); dot(.5(X+Y)); dot(.5(P+Q)); dot("A",A,dir(A)); dot("B",B,dir(B)); dot("C",C,dir(C)); dot("D",D,dir(100)); dot("E",E,dir(180)); dot("F",F,dir(0)); dot("X",X,dir(300)); dot("Y",Y,dir(60)); dot("P",P,dir(240)); dot("Q",Q,dir(120)); dot("E'",Ep,dir(-10)); [/asy]$

Notice that $EX\parallel AP$ and $EX=BE=AP$ . Similarly, $EY\parallel AQ$ and $EY=CE=AQ$ , so triangles $APQ$ and $EXY$ are shifted copies of each other. Therefore, the $A$ -median of $APQ$ is parallel to the $E$ -median of $EXY$ . Since corresponding sides have equal measure, $EXFY$ is a parallelogram. Thus line $EF$ is the $E$ -median of $EXY$ , so it suffices to show that $EF\parallel AD$ . But by symmetry, $EF\parallel E'D=AD$ , as required.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

143 posts

#3 h Apr 15, 2025, 10:26 AM

Y by

Since $\angle ACE = \angle ABE = 120^\circ$ , we have
$\frac{[ACE]}{[ABE]} = \frac{AC \cdot CE \cdot \sin(120^\circ)}{AB \cdot BE \cdot \sin(120^\circ)} = \frac{AC \cdot AQ}{AB \cdot AP}.$ Hence,
$\frac{[AQE]}{[APE]} = \frac{[ACE]}{[ABE]} \cdot \frac{AQ/AC}{AP/AB}= \frac{AQ^2}{AP^2},$ implying that $AE$ is the $A$ -symmedian of $\triangle APQ$ . Thus, $AD$ is the $A$ -median of $\triangle APQ$ as desired.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a