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, 2008

Geometry Difficulty 5.9 AIME, harder Prove it India

The in-circle $\Gamma$ of a triangle $ABC$ touches the side $BC$ at $D$ . Let $D'$ be the point which is diametrically opposite to $D$ on the circle $\Gamma$ . The tangent through $D'$ to $\Gamma$ meets $AD$ in $X$ . The tangent to $\Gamma$ through $X$ , other than $XD'$ , touches $\Gamma$ at $N$ . Prove that the circum-circle of triangle $BCN$ touches $\Gamma$ at $N$ .

Solution

Observe that $NX$ is the polar of $Y$ , $EF$ is the polar of $A$ and $BC$ is the polar of $D$ . Since $A$ , $Y$ , $D$ are collinear, it follows that $NX$ , $EF$ , $BC$ are concurrent. Let the point of concurrency be $D'$ . Let $S$ be the point of intersection of $EF$ and $AD$ . Since \{ $E, S, F, D'$ \} form a harmonic range, \{AE, AD, AB, AD'\} is a harmonic pencil. It follows that \{D', B, D, C\} is a harmonic range. We also observe that $\angle DND' = 90^\circ$ . Therefore $ND$ bisects $\angle BNC$ . This implies that $D$ is the mid-point of the minor arc $PQ$ of $\Gamma$ , where $P$ , $Q$ are the points of intersection of $NB$ , $NC$ with $\Gamma$ . Hence $PQ$ is parallel to $BC$ . Now
$\angle YNQ = \angle NPQ = \angle NBC.$
It follows that $YN$ is tangent to the circum-circle of the triangle $BNC$ .

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