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Geometry Difficulty 6.4 National Olympiad Prove it Taiwan

Let $O$ , $H$ be the circumcenter and orthocenter of scalene triangle $ABC$ respectively, and let $P$ be a point inside triangle $AHO$ satisfying $\angle AHP = \angle POA$ , with $M$ being the midpoint of $\overline{OP}$ . Let $BM$ , $CM$ intersect the circumcircle of triangle $ABC$ again at points $X$ , $Y$ respectively.
Prove that line $XY$ passes through the circumcenter of triangle $APO$ .

Solution

Let $AP$ meet $\odot(ABC)$ at $D$ , and let $O_1, O_2$ be the circumcenters of $\triangle APO$ , $\triangle DPO$ respectively. Then
$\angle PO_1O = 2 \cdot \angle PAO = 2 \cdot \angle ODP = \angle OO_2P.$
Hence $PO_1OO_2$ is a rhombus centered at $M$ .

Claim. The point $O_2$ lies on $BC$ .

Proof of Claim. Take $O'$ such that $M$ is the midpoint of $AO'$ , and translate $\triangle AHP$ to $\triangle OO'P'$ . From
$\angle OP'P = \angle PAO = \angle ODP$
we know that $D, O, P, P'$ lie on a common circle $\Gamma$ . Also,
$\angle OO'P' = \angle AHP = \angle POA = \angle OPP',$
so $O'$ also lies on $\Gamma$ , and therefore $O_2$ lies on the perpendicular bisector of $OO'$ , that is, on $BC$ .

Returning to the original problem, since $O_1O_2$ is perpendicular to $OM$ , by the Butterfly Theorem the reflection of $O_2$ about $M$ , namely $O_1$ , lies on $XY$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.