Argentine National Olympiad 2016

We show that $M$ coincides with the incenter $I$ of the triangle. Since $\hat{A}+\hat{B}=90^{\circ }$, this implies $\widehat{AM}\hat{B}=\widehat{AI}\hat{B}=180^{\circ }-\frac{1}{2}(\hat{A}+\hat{B})=135^{\circ }$.



By hypothesis $AD=AC$, meaning that $D$ is the reflection of $C$ in the bisector $AI$ of $\hat{A}$. Likewise $Q$ is the reflection of $D$ in the bisector $BI$ of $\hat{B}$. It follows that $CI=DI=QI$. Analogously $E$ is the reflection of $C$ in the bisector $BI$ and $P$ is the reflection of $E$ in the bisector $AI$, hence $CI=EI=PI$. We obtain $CI=PI=QI$. Also $P\hat{C}I=Q\hat{C}I=45^{\circ }$ since $CI$ bisects $\hat{C}=90^{\circ }$. Therefore $C\hat{I}P=C\hat{I}Q=90^{\circ }$. In conclusion $P$, $Q$ and $I$ are collinear, and $I$ is the midpoint of $PQ$ as $PI=QI$. Thus $M$ and $I$ coincide, as stated. The solution is complete.