Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.3 AIME Prove it

Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with diameter $XY$ . Segments $AY$ and $BX$ meet at $P$ . Point $Z$ is the foot of the perpendicular from $P$ to line $XY$ . Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$ . Let $Q$ be the intersection of segments $AY$ and $XC$ . Prove that $\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.$

Solution

Using the Law of Sines and simplifying, we have $\frac{BY}{XP}+\frac{CY}{XQ}=\frac{\sin \angle PXY \sin \angle XPY+\sin \angle QXY\sin \angle XQY}{\sin \angle AYX}.$
It is easy to see that $APZX$ is cyclic. Also, we are given $XQ\perp AZ$ . Then we have
$\begin{align*} \frac{\sin \angle PXY\sin \angle XPY+\sin \angle QXY\sin \angle XQY}{\sin \angle AYX} &= \frac{\sin \angle YAZ\sin \angle XZA+\cos \angle XZA\cos \angle YAZ}{\sin \angle AYX} \\ &= \frac{\cos(\angle XZA-\angle YAZ)}{\sin \angle AYX} \\ &= \frac{\cos \angle AYX}{\sin \angle AYX} \\ &= \cot AYX \\ &= \frac{AY}{AX}, \end{align*}$ and we are done.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.