Team Selection Test

Let $Q$ be the point where the tangent lines to $\Gamma$ at $A$ and $B$ meet. We will show that $Q$ is the point. Since $\angle MCA=\angle MAP$ and $\angle MCB=\angle MBP$, we have $\angle ACB+\angle APB=180^{\circ }$ and $P$ lies on $\Gamma$. Therefore, if $P^{\prime }$ is the point where $QC$ intersect the circle, it suffices to show that $\angle MCA=\angle BC{P}^{\prime }$ as then it will follow that $A{P}^{\prime }$ is tangent to the circumcircle of the triangle $CAM$ and $P^{\prime }=P$. The triangles $QA{P}^{\prime }$ and $QCA$, and the triangles $QB{P}^{\prime }$ and $QCB$ are similar. Therefore $$\frac{A{P}^{\prime }}{CA}=\frac{QA}{QC}=\frac{QB}{QC}=\frac{B{P}^{\prime }}{CB}.$$ Then by the Ptolemy's Theorem, $$C{P}^{\prime }=\frac{CA\cdot B{P}^{\prime }+CB\cdot A{P}^{\prime }}{BA}=\frac{CA\cdot B{P}^{\prime }}{MA}\text{and}\frac{C{P}^{\prime }}{B{P}^{\prime }}=\frac{CA}{MA}.$$ We conclude that the triangles $C{P}^{\prime }B$ and $CAM$ are similar, and hence $\angle MCA=\angle BC{P}^{\prime }$.