Canadian Mathematical Olympiad

Since $\angle C = 90^\circ$, the point $C$ lies on the semicircle with diameter $AB$ which implies that, if $M$ is the midpoint of side $AB$, then $MA = MC = MB$. This implies that triangle $AMC$ is isosceles and hence that $\angle ACM = \angle A$. By definition, $G$ lies on segment $M$ and it follows that $\angle ACG = \angle ACM = \angle A = \angle CPA$. This implies that triangles $APC$ and $ACG$ are similar and hence that $AC^2 = AG \cdot AP$. Now if $D$ denotes the foot of the perpendicular from $C$ to $AB$, it follows that triangles $ACD$ and $ABC$ are similar which implies that $AC^2 = AD \cdot AB$. Therefore $AG \cdot AP = AC^2 = AD \cdot AB$ and, by power of a point, quadrilateral $DGPB$ is cyclic. This implies that $D$ lies on the circumcircle of triangle $BPG$ and, by a symmetric argument, it follows that $D$ also lies on the circumcircle of triangle $AGQ$. Therefore these two circumcircles meet at the point $D$ on side $AB$.

Define $D$ and $M$ as in Solution 1. Let $R$ be the point on side $AB$ such that $AC = CR$ and triangle $ACR$ is isosceles. Since $\angle CRA = \angle A = \angle CPA$, it follows that $CPRA$ is cyclic and hence that $\angle GPR = \angle APR = \angle ACR = 180^\circ - 2\angle A$. As in Solution 1, $MC = MB$ and hence $\angle GMR = \angle CMB = 2\angle A = 180^\circ - \angle GPR$. Therefore $GPRM$ is cyclic and, by power of a point, $AM \cdot AR = AG \cdot AP$. Since $ACR$ is isosceles, $D$ is the midpoint of $AR$ and thus, since $M$ is the midpoint of $AB$, it follows that $AM \cdot AR = AD \cdot AB = AG \cdot AP$. Therefore $DGPB$ is cyclic, implying the result as in Solution 1.