All Russian Math Olympiad 2010

For discussing Olympiad level Geometry Problems
dshasan
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All Russian Math Olympiad 2010

Unread post by dshasan » Sat Apr 01, 2017 1:53 pm

Triangle $ABC$ has perimeter $4$. Points $X$ and $Y$ lies on rays $AB$ and $AC$, respectively, such that $AX = AY = 1$. Segments $BC$ and $XY$ intersect at point $M$. Prove that the perimeter of either $\bigtriangleup ABM$ or $\bigtriangleup ACM$ is $2$.

[I like this problem :) ]
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.

- Charles Caleb Colton

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ahmedittihad
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Re: All Russian Math Olympiad 2010

Unread post by ahmedittihad » Sat Apr 01, 2017 2:26 pm

Assume, $AB<AC$.
Reflect $A$ over $X$ and $Y$ to get $K$ and $L$ respectively. We know that the length of the tangent from $A$ to the $A$-excircle is half the perimeter. So, we get, $K$ and $L$ are the touchpoints of the $A$-excircle with $AB$ and $AC$.
Let the $A$-excircle meet $BC$ at $Z$.
Now, the wow factor.
We show that $MZ=AM$.
As $X$ and $Y$ are the midpoints of $AK$ and $AL$, $XY$ is the radical axis of the $A$-excircle and the zero radius circle at $A$. And $M$ lies on the radical axis. Yielding $MZ=AM$.
Now, $AK=2=AB+BK=AB+BZ=AB+BM+MZ=AB+BM+AM$.
The other cases are similar too.
Frankly, my dear, I don't give a damn.

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Atonu Roy Chowdhury
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Re: All Russian Math Olympiad 2010

Unread post by Atonu Roy Chowdhury » Sun Apr 02, 2017 8:31 pm

Well, I first missed the point that segments $BC$ and $XY$ intersects at $M$. After noticing this the problem is easy. My solution is similar to Ittihad's.

WLOG $AB>AC$. Take the $A$-excircle of $\triangle ABC$. it touches $BC$ at $D$ and the extensions of $AB$ and $AC$ at $E$ and $F$ respectively. $AF = 2$. $XY$ is the rad axis of the excircle and point circle $A$. So, $AM=MD$. The rest is trivial.

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