My solution during the test: Let $\odot (ABC) \cap \odot (PMQN) = X$. Then $\angle MXA = \angle MAX = \angle BAX =\angle BCX =\angle PCX =\angle PMX$ $\Rightarrow PM \parallel AX \Rightarrow$ As $PM \perp QM$, so $AX \perp MQ$. But as $MA = MX$, so $QA = QX$ $\Rightarrow X$ lies on $\odot (Q,QA)$. Similarly, $X$ lies on $\odot (P,PB)$. Thus, $\odot (P,PB), \odot (Q,QA), \odot (ABC)$ concur at $X$.

29 people solved it on the exam

WizardMath wrote: Let $M$ be the midpoint of $AB$ in a right angled triangle $ABC$ with $\angle C = 90^\circ$. A circle passing through $C$ and $M$ meets segments $BC, AC$ at $P, Q$ respectively. Let $c_1, c_2$ be the circles with centers $P, Q$ and radii $BP, AQ$ respectively. Prove that $c_1, c_2$ and the circumcircle of $ABC$ are concurrent. Let $\odot(CPQ)$ meet $\overline{AB}$ at $Y$ and $X$ be the reflection of $Y$ in line $\overline{PQ}$. Now $\angle PXB=\angle MCB=\angle PBX$ so $X$ lies on $c_1$; hence $Y$ lies on $c_1$. Likewise $Y$ lies on $c_2$. Finally, $\angle PYQ=\angle PMQ=180^{\circ}-\angle PCQ=90^{\circ}$ so $Y$ lies on $\odot(CPQ)$. Since $PY=PB$ and $QY=QA$; $\angle BPY=\angle AQY$ we conclude $\triangle YPB \sim \triangle YQA$. By spiral similarity, $Y \in \odot(CAB)$ and we're done.

Construct (ABC) cuts (CPQ)=D. CQD=CPD=CMD=2CAD=2CBD then QD=QA, PD=PA so three circle concurrent at D

Let $X$ be the second intersection of $(ABC)$ with $(CQM)$. It suffices to show $QX=QA$. We have \[\angle QXM = \angle QCM = \angle MAC\]and \[\angle AQM = \angle CQM = \angle CXM = \angle MCX = \angle MQX.\]Hence $\triangle MQX \cong \triangle MQA$ giving $QX = QA$. Similarly $PX= PB$ and we are done.