, 2020

Geometry Difficulty 8.6 Shortlist Prove it Taiwan

Let point $H$ be the orthocenter of a scalene triangle $ABC$ . Line $AH$ intersects with the circumcircle $\Omega$ of triangle $ABC$ again at point $P$ . Line $BH, CH$ meets with $AC, AB$ at point $E$ and $F$ , respectively. Let $PE, PF$ meet $\Omega$ again at point $Q, R$ , respectively. Point $Y$ lies on $\Omega$ so that lines $AY, QR, EF$ are concurrent. Prove that $PY$ bisects $EF$ .

Solution

Consider $ACRQPY$ . By Pascal's theorem, the points $E = AC \cap QP$ , $CR \cap PY$ , $QR \cap AY$ are collinear, and similarly the points $F = AB \cap RP$ , $BQ \cap PY$ , $QR \cap AY$ are collinear. Therefore $BQ, CR, PY, EF$ are concurrent at a point $M$ .

Take a point $B' \neq B$ on $\Omega$ such that $AB = AB'$ . Then
$\angle AFE = \angle BCA = \angle BB'A = \angle ABB' \Rightarrow BB' \parallel EF$
and
$\angle APB' = \angle ABB' = \angle BB'A = \angle BCA = \angle AHB \Rightarrow BE \parallel B'P$
Extend $BE$ to meet $\Omega$ at $K$ . Then we have
$-1 = (H, K; E, \infty_{BE}) \stackrel{P}{\cong} (A, K; Q, B') \stackrel{B}{\cong} (F, E; M, \infty_{EF})$
This means $M$ is the midpoint of $EF$ , that is, $PY$ bisects $EF$ , 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.