Track / Stage 6 / 49 of 400 #1049 of 1964

Problem 1049

National Olympiad, first round

Geometry Difficulty 6.0 Prove it Taiwan IMO Selection Camp · Taiwan

平面上給定三角形 $ABC$ 及一點 $P$ 。令 $\triangle ABC$ , $\triangle BPC$ , $\triangle CPA$ , $\triangle APB$ 的外接圓圓心分別為點 $O$ , $D$ , $E$ , $F$ 。設直線 $BC$ 與 $EF$ 交於點 $T$ ，而點 $O$ 對直線 $EF$ 的對稱點為 $X$ 。證明： $PT \perp DX$ 。

Given a triangle $ABC$ and a point $P$ in the plane. Let the circumcenters of $\triangle ABC$ , $\triangle BPC$ , $\triangle CPA$ , $\triangle APB$ be $O$ , $D$ , $E$ , $F$ respectively. Let line $BC$ meet $EF$ at point $T$ , and let $X$ be the reflection of point $O$ about line $EF$ . Prove that $PT \perp DX$ .

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 $O$ in $EF$ and $Y$ , $Z$ be the reflections of $P$ in $EX$ , $FX$ , respectively. Since $A$ , $B$ , $C$ are the reflections of $P$ in $EF$ , $FD$ , $DE$ , respectively, so $O$ , $P$ are isogonal conjugate with respect to $\triangle DEF$ and hence
$\angle FEX = \angle OEF = \angle DEP, \quad \angle EFX = \angle OFE = \angle DFP,$
$\Rightarrow D$ , $X$ are isogonal conjugate with respect to $\triangle EPF$ . Since the reflection of $P$ in the bisector of $\angle PEF$ , $\angle PFE$ lies on $EF$ , so $Y$ , $Z$ are the reflections of $C$ , $B$ in $EF$ , respectively $\Rightarrow T$ lies on the radical axis of $\odot(BPC)$ , $\odot(YPZ)$ $\Rightarrow PT \perp DX$ .

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