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Difference between revisions of "2013 Canadian MO Problems/Problem 3"

(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...")
 
(No difference)

Latest revision as of 01:18, 27 November 2023

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

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