A circle $\omega$ with diameter $AK$ is given. The point $M$ lies in the interior of the circle, but not on $AK$. The line $AM$ intersects $\omega$ in $A$ and $Q$. The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to $AK$, at $P$. The point $L$ lies on $\omega$, and is such that $PL$ is tangent to $\omega$ and $L\neq Q$. Show that $K, L$, and $M$ are collinear.
Problem
Source: Netherlands TST for IMO 2017 day 3 problem 1
Tags: geometry
sbealing
01.02.2018 19:48
Let $C=PM \cap AK$ and $O$ be the midpoint of $AK$ then as $\angle OCP=\angle OLP=\angle OQP=90^{\circ}$ we have $OCLPQ$ is cyclic. $$\angle LCM=\angle LCP=\angle LQP=\angle LAQ=\angle LAM$$So $LACM$ is cyclic and hence $\angle ALM=\angle ACM=90^{\circ}$ so $K,L,M$ are colinear as $AK$ is a diameter.
Cycle
12.05.2020 06:58
Redefine $L$ to be the intersection of $KM$ with $\omega$ so that it suffices to show $PL$ is tangent to the circle. Letting $D=PM\cap AK$ and $O$ the center, we have $$\angle PQL=\angle QKM=\angle LAQ=\angle PDL$$by alternate segment and cyclic quadrilaterals $LMDA$, and $ALQK$. Hence $PQDL$ is cyclic. But $\angle QDM=\angle QKM$ implying $\angle LDQ=2\angle QKL=\angle LOQ$ so $LODQP$ is cyclic. Then $\angle OLP=180^\circ-\angle OQP=90^\circ $, proving the claim.
llplp
14.08.2020 01:43
We use complex numbers with $a = 1, k = -1$. We redefine $L,Q = \ell, q$ to be arbitrary points on $\omega$ and $p = \frac{2 \ell q}{\ell + q}$, $M = KL \cap AS$ and we wish to show $PM \perp AK$. From the intersection formula we have $m = \frac{2\ell q + \ell - q}{\ell + q}$. Finally $m - p = \frac{\ell - q}{\ell + q} \in i \mathbb{R}$, so $p-m$ is perpendicular to the real axis, i.e $AK$.