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, 2015

Geometry Difficulty 5.2 AIME, harder Prove it Singapore

In an acute-angled triangle $ABC$ , $M$ is a point on the side $BC$ , the line $AM$ meets the circumcircle $\omega$ of $ABC$ at the point $Q$ distinct from $A$ . The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to the diameter $AK$ of $\omega$ at the point $P$ . Let $L$ be the point on $\omega$ distinct from $Q$ such that $PL$ is tangent to $\omega$ at $L$ . Prove that $L$ , $M$ and $K$ are collinear.

Solution

Let $D$ be the foot of the perpendicular from $M$ onto the diameter $AK$ . Since $AK$ is a diameter of $\omega$ , we have $\angle MQK = 90^\circ = \angle MDK$ so that $M$ , $Q$ , $K$ , $D$ are concyclic. Join $QK$ . Then $\angle PMQ = \angle QKD = \angle QKA = \angle PQA = \angle PQM$ . Thus the triangle $PQM$ is isosceles with $PQ = PM$ . Therefore, $PQ = PM = PL$ and $P$ is the circumcentre of the triangle $QML$ . It follows that $\angle QLM = \frac{1}{2}\angle QPM = 90^\circ - \angle PQM = 90^\circ - \angle QKA = \angle QAK = \angle QLK$ . This shows that $L$ , $M$ and $K$ are collinear.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.