The triangle $ABC$ has circumcentre $O$ and circumcircle $\Gamma$. Let $AI$ be a diameter of $\Gamma$. The ray $AI$ extends to intersect the circumcircle $\omega$ of $\triangle BOC$ for the second time at a point $P$. Let $AD$ and $IQ$ be perpendicular to $BC$, with $D$ and $Q$ on $BC$. Let $M$ be the midpoint of $BC$. (a) Prove that $|AD| \cdot |QI| = |CD| \cdot |CQ| = |BD| \cdot |BQ|$. (b) Prove that $IM$ is parallel to $PD$.
Problem
Source: Irish Mathematical Olympiad 2023 Problem 9
Tags: geometry, circumcircle
v4913
15.05.2023 03:16
(a) True by power of a point at D, since the reflection of I over the perpendicular bisector of BC is on AD. (b) Let the line through I parallel to BC intersect OM at J; then it suffices to show that MI || JP, which is equivalent to OM / OJ = OI / OP, which is obvious because if P* is the inverse of P with respect to (O), P* is the intersection of OI and BC, so OM / OJ = OP* / OI = OI / OP as desired.
Chota_don
28.05.2023 13:12
This is the solution of this problem. Credit: me and @geoking
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