## All Russian Math Olympiad 2010

For discussing Olympiad level Geometry Problems
dshasan
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Joined: Fri Aug 14, 2015 6:32 pm

### All Russian Math Olympiad 2010

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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Joined: Mon Mar 28, 2016 6:21 pm

### Re: All Russian Math Olympiad 2010

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.

Atonu Roy Chowdhury
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Joined: Fri Aug 05, 2016 7:57 pm