Problem

Source: Turkey TST 2014 Day 2 Problem 5

Tags: geometry, circumcircle, geometric transformation, homothety, trigonometry, conics, inequalities



A circle $\omega$ cuts the sides $BC,CA,AB$ of the triangle $ABC$ at $A_1$ and $A_2$; $B_1$ and $B_2$; $C_1$ and $C_2$, respectively. Let $P$ be the center of $\omega$. $A'$ is the circumcenter of the triangle $A_1A_2P$, $B'$ is the circumcenter of the triangle $B_1B_2P$, $C'$ is the circumcenter of the triangle $C_1C_2P$. Prove that $AA', BB'$ and $CC'$ concur.