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Berkeley Math Circle Monthly Contest 1

Geometry Difficulty 4.7 AIME Prove it United States

Problem:
Let $ABC$ be an acute triangle with orthocenter $H$ , and let $M$ and $N$ denote the midpoints of $AB$ and $AC$ . Rays $MH$ and $NH$ intersect the circumcircle of $ABC$ again at points $X$ and $Y$ . Prove that the four points $M, N, X, Y$ lie on a circle.

Solution

Solution:
Let $\Omega$ be the circumcircle of $\triangle ABC$ . Let $P$ and $Q$ be the reflections of $H$ across $M$ and $N$ .

Since $\angle APB = \angle AHB = 180^\circ - \angle C$ , point $P$ lies on $\Omega$ ; thus so does $Q$ . Moreover, $MN \parallel PQ$ , so
$\angle XMN = \angle XPQ = \angle XYQ$
and we are done.

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