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Taiwan IMO Selection Camp, 2022

Geometry Difficulty 5.1 AIME, harder Prove it Taiwan

平面上 $ABC$ 為銳角三角形，其外心為 $O$ ，外接圓為 $\Omega$ 。分別在線段 $AB, AC$ 上各取一點 $D, E$ ，並作過 $A$ 與 $DE$ 垂直的直線 $\ell$ 。設 $\ell$ 分別與三角形 $ADE$ 的外接圓及 $\Omega$ 再交於點 $P, Q$ 。令直線 $OQ$ 與 $BC$ 交於點 $N$ ，直線 $OP$ 與 $DE$ 交於點 $S$ ，且點 $W$ 為三角形 $AOS$ 的垂心。
試證： $S, N, O, W$ 四點共圓。

Solution

Let $D', E'$ be points on lines $AB, AC$ , respectively, such that $D'E'$ is parallel to $DE$ , and passes through $N$ . From
$\angle ND'B = 90^\circ - \angle BAQ = \angle OQB = \angle NQB,$
we see that $B, D', N, Q$ are concyclic. Since
$\angle QD'E' = \angle QBC = \angle QAE',$
$A, D', Q, E'$ are concyclic. Therefore, $\triangle ADE \cup P \stackrel{+}{\sim} \triangle AD'E' \cup Q$ .
Let $S', T \neq A$ be the intersections of $AS$ and $D'E'$ , $\Omega$ , respectively. Then
$\angle QTS' = \angle QTA = \angle(QO, \perp AQ) = \angle QNS'$
gives the concyclicity of $Q, T, N, S'$ . From the similarity $\triangle ADE \cup \{P, S\} \stackrel{+}{\sim} \triangle AD'E' \cup \{Q, S'\}$ , we see that $PS$ is parallel to $QS'$ . According to Reim's theorem, $S, N, O, T$ are concyclic. Thus,
$\angle OWS = \angle SAO = \angle OTA = \angle ONS,$
as desired.

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