1995 USAMO Problems/Problem 3

Problem

Given a nonisosceles, nonright triangle $ABC,$ let $O$ denote its circumcenter, and let $A_1, \, B_1,$ and $C_1$ be the midpoints of sides $BC, \, CA,$ and $AB,$ respectively. Point $A_2$ is located on the ray $OA_1$ so that $\triangle OAA_1$ is similar to $\triangle OA_2A$ . Points $B_2$ and $C_2$ on rays $OB_1$ and $OC_1,$ respectively, are defined similarly. Prove that lines $AA_2, \, BB_2,$ and $CC_2$ are concurrent.

Solution

LEMMA 1: In $\triangle ABC$ with circumcenter $O$ , $\angle OAC = 90 - \angle B$ .

PROOF of Lemma 1: The arc $AC$ equals $2\angle B$ which equals $\angle AOC$ . Since $\triangle AOC$ is isosceles we have that $\angle OAC = \angle OCA = 90 - \angle B$ . QED

Define $H \in BC$ s.t. $AH \perp BC$ . Since $OA_1 \perp BC$ , $AH \parallel OA_1$ . Let $\angle AA_2O = \angle A_1AO = x$ and $\angle AA_1O = \angle A_2AO = y$ . Since we have $AH \parallel OA_1$ , we have that $\angle HAA_2 = x$ . Also, we have that $\angle A_2AA_1 = y-x$ . Furthermore, $\angle BAH = 90 - \angle B = \angle OAC$ , by lemma 1. Therefore, $\angle A_1AC = 90 - \angle B + x = \angle BAA_2$ . Since $A_1$ is the midpoint of $BC$ , $AA_1$ is the median. However $\angle A_1AC = \angle BAA_2$ tells us that $AA_2$ is just $AA_1$ reflected across the internal angle bisector of $A$ . By definition, $AA_2$ is the $A$ -symmedian. Likewise, $BB_2$ is the $B$ -symmedian and $CC_2$ is the $C$ -symmedian. Since the symmedians concur at the symmedian point, we are done.

QED

Isogonal conjugate

1995 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5
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1995 USAMO Problems/Problem 3

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