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Source: Netherlands TST for IMO 2017 day 3 problem 1

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Medjl

#1 h Feb 1, 2018, 2:43 PM • 3 Y

Y by Muradjl, Adventure10, Mango247

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.

This post has been edited 1 time. Last edited by Medjl, Feb 1, 2018, 3:04 PM

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#2 h Feb 1, 2018, 4:48 PM • 3 Y

Y by lolm2k, AlastorMoody, Adventure10

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.

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Cycle

#3 h May 12, 2020, 3:58 AM

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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.

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llplp

#4 h Aug 13, 2020, 10:43 PM

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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 $AK$ .

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