Difference between revisions of "1995 USAMO Problems/Problem 3"

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== Solution ==

Revision as of 19:58, 4 July 2012

Given a nonisosceles, nonright triangle $ABC,$ let $O$ denote the center of its circumscribed circle, 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, i.e. these three lines intersect at a point.



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.


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