IMO 2013 Problem 4

Question: Let $ABC$ be an acute triangle with orthocenter $H$ , and let $W$ be a point on the side $BC$ , lying strictly between $B$ and $C$ . The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$ , respectively. Denote by $\omega_1$ is the circumcircle of $BWN$ , and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$ . Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$ , and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$ . Prove that $X,Y$ and $H$ are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand

My Solution: Let $P$ be the second intersection of $\omega_1, \omega_2$ . Then $P$ is the Miquel point of $\triangle MNW$ w.r.t $\triangle ABC$ , so $AMPHN$ is cyclic. $BX \parallel AH$ , hence $X, H, P$ are collinear by Reim’s theorem; similarly, $H,P,Y$ are collinear, and the result follows.