2013 Canadian MO Problems/Problem 3

Revision as of 00:18, 27 November 2023 by Tomasdiaz (talk | contribs) (Created page with "==Problem == Let <math>G</math> be the centroid of a right-angled triangle <math>ABC</math> with <math>\angle BCA = 90^\circ</math>. Let <math>P</math> be the point on ray <ma...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Let $G$ be the centroid of a right-angled triangle $ABC$ with $\angle BCA = 90^\circ$. Let $P$ be the point on ray $AG$ such that $\angle CPA = \angle CAB$, and let $Q$ be the point on ray $BG$ such that $\angle CQB = \angle ABC$. Prove that the circumcircles of triangles $AQG$ and $BPG$ meet at a point on side $AB$.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.