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Iranian Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Iran

In triangle $ABC$ let $M$ be the midpoint of $BC$ . Let $\omega$ be a circle inside of $ABC$ and tangent to $AB, AC$ at $E, F$ , respectively. The tangents from $M$ to $\omega$ meet this circle at $P, Q$ such that $P$ and $B$ lie on the same side of $AM$ . Lines $PM, BF$ cut each other at $X$ , and $Y$ is the intersection of $QM, CE$ . If $2PM = BC$ , prove that $XY$ is tangent to $\omega$ .

Solution

Assume that $CP, BQ$ cut each other at point $Z$ . Since $2PM = BC$ , we conclude that points $B, P, Q, C$ lie on the circle $\Omega$ with center $M$ .

We have
$\widehat{PZQ} = \widehat{BPZ} + \widehat{PBZ} = 90^\circ + \widehat{PBQ} = 180^\circ - \widehat{MPQ} = \frac{\widehat{PEQ}}{2}.$
Therefore $Z$ lies on $\omega$ . Note that circles $\omega, \Omega$ are perpendicular to each other, so the polar of point $B$ with respect to $\omega$ , passes through $C$ . This line also passes through $E$ , therefore $CE$ is the polar of $B$ with respect to $\omega$ .

Hence $Y$ lies on the polar of $B$ , which means $B$ also lies on the polar of $O$ with respect to $\omega$ . Again, note that $Q$ is a point on the polar of $Y$ , which leads to the conclusion that $BQ$ is the polar of $Y$ with respect to $\omega$ . Since $Z$ lies on the polar of $Y$ , we conclude that $YZ$ is tangent to $\omega$ . Similarly, $XZ$ is also tangent to $\omega$ , hence $XY$ passes through $Z$ and is tangent to $\omega$ .

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