In triangle $ABC, \angle B = 2\angle C$. Points $P$ and $Q$ on the medial perpendicular to $CB$ are such that $\angle CAP = \angle PAQ = \angle QAB = \frac{\angle A}{3}$ . Prove that $Q$ is the circumcenter of triangle $CPB$.
Problem
Source: Sharygin 2011 Final 9.2
Tags: geometry, circumcircle, Angle Chasing, perpendicular bisector