Problem 1630

The idea for the solution starts from the observation that the requirement $|AN|=|AC|=|BC|$ remains unchanged if one swaps $A$ and $C$ as well as $N$ and $B$ with each other. This suggests that one should also consider the circumcircle of the triangle $AMB$. The second intersection point of the circumcircles of $AMB$ and $CMN$ can finally be identified as the incenter of $AMC$.

Let $I$ be the incenter of the triangle $AMC$. As is well known, $\measuredangle MIC=\frac{\pi}{2}+\frac{1}{2} \measuredangle MAC=\frac{\pi}{2}+\frac{1}{2} \measuredangle NAC$, and because the triangle $ANC$ is isosceles at $A$, we have $\measuredangle NAC=\pi-2 \measuredangle CNA$, so $\measuredangle MIC=\pi-\measuredangle CNA=\pi-\measuredangle CNM$, and because $I$ and $N$ lie in different half-planes with respect to the line $CM$, it follows from the theorem on cyclic quadrilaterals that $IMNC$ is a cyclic quadrilateral, i.e., that $I$ lies on the circumcircle of $CMN$. Due to the symmetry mentioned at the beginning, $I$ also lies on the circumcircle of $ABM$. Since $I$, as the incenter, is an interior point of the triangle $AMC$, in particular $I \neq M$, and thus the line $IM$ is the radical axis of the circumcircles of $ABM$ and $CMN$. Since furthermore the line $AB$ is the radical axis of the circumcircles of $ABC$ and $ABM$ and the line $CP$ is the radical axis of the circumcircles of $ABC$ and $CMN$, the lines $IM, AB$ and $CP$ meet at a point. This point is, by definition, $Q$, i.e., $Q$ lies on the angle bisector $IM$ of the angle $CMA$, but this is identical with the angle bisector of the angle $BMN$, consequently $\measuredangle BMQ=\measuredangle QMN$.