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.
Problem
Source: 2015 Singapore Seniors Round 2 p1
Tags: collinear, Tangents, circumcircle, geometry
Ricochet
18.03.2020 22:15
Let $M'=AQ \cap LK, E=AL\cap QK$ and $F=AK\cap LQ$. Note that by Brokard's theorem on $ALQK$ we get that $EM'$ is the polar of $F$ with respect to $(ABC)$, furthermore we know that the circumcenter of $\triangle ABC$ is the orthocenter of $\triangle EM'F$, so $EM' \perp OF$. Now since the polar of $P (LQ)$ passes through $F \implies$ the polar of $F$ passes through $P$, hence $E, P, M'$ are collinear. So in particular we get that $PM' \perp AK$, hence $M\equiv M'$ and the result follows
Attachments:
Kagebaka
19.03.2020 00:57
Fix $\triangle ABC,$ animate $M$ on $BC,$ let $L'$ be the projection of $M$ onto $\omega$ from $K,$ and let $X$ be the foot of the perpendicular to $OP,$ where $O$ is $\omega$'s center. Consider the composition of the following projective maps: Project $\omega$ to $BC$ through $K.$ ($L'\to M$) Project $BC$ to $\omega$ through $A.$ ($M\to Q$) Invert preserving $\omega$ with center $\infty_{XQ}.$ ($Q\to L$) To prove that this is the identity map, it suffices to check that $L=L',$ in other words $L,M,K$ collinear, for 3 cases of $M.$ However, this is trivial for $M=B,C,AK\cap BC$ so we're done. $\blacksquare$
lux73
19.03.2020 08:11
Define $O$ as the center of $\omega$, $Q = AM \cap \omega$, $L = KM \cap \omega$, and $P$ as the intersection of the tangents to $\omega$ at $L$ and $Q$. We will show that $PM \perp AK$. Also, let $X = KQ \cap AL, Y = LQ \cap AK$.
By Pascal's theorem on $LLKQQA$, we have that $P, M, X$ are collinear. But by Brokard's theorem, $O$ is the orthocenter of $\Delta XMY$ so $M$ is the orthocenter of $\Delta XOY$. Thus $XM \perp OY \implies PM \perp AK$.