Track / Stage 5 / 66 of 400 #666 of 1964

Problem 666

AIME late

Geometry Difficulty 5.1 Prove it Iranian Mathematical Olympiad · Iran

The Euler circle of the acute-angled triangle $ABC$ is reflected with respect to the altitude from $A$ to $BC$ and intersected the circumcircle of the triangle $ABC$ at distinct points $X$ and $Y$ ( $X \neq Y$ ). Let $H$ be the orthocenter of triangle $ABC$ . Prove that $AH$ is the external angle bisector of $\angle XHY$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $X'$ be the reflection of $X$ with respect to $AH$ and $X''$ be the reflection of $H$ with respect to $X'$ . Notice that $X'$ lies on the nine point circle (Euler circle) and $X''$ lies on the circumcircle.

Lines $X''H$ and $XH$ intersect the circumcircle of triangle $ABC$ for the second time at $Y'$ and $D$ . Let $M$ be the midpoint of $HD$ , so
$XH \cdot HD = X''H \cdot HY' \implies HY' = \frac{1}{2}HD = HM$
Since the point $M$ lies on the nine point circle, and $\angle XHA = \angle X''HA$ , $Y'$ is the reflection of $M$ with respect to $AH$ , so $Y' \equiv Y$ and the result follows.

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