USAMO (GEOMETRY) 1995

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Kazi_Zareer
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USAMO (GEOMETRY) 1995

Unread post by Kazi_Zareer » Mon Jan 30, 2017 8:58 am

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.
We cannot solve our problems with the same thinking we used when we create them.

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ahmedittihad
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Re: USAMO (GEOMETRY) 1995

Unread post by ahmedittihad » Mon Jan 30, 2017 12:26 pm

From the definition of inversions, we get $A_2$ is the inverted $A_1$ wrt circle $ABC$.
So, $AA_2$ is just the $A$-symmedian of $\triangle ABC$. We know that the symmedians of a triangle meet at the symmedian point. The problem is solved.
Frankly, my dear, I don't give a damn.

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