Saudi Arabian Mathematical Competitions

Consider $P'$ on $A B$, $Q'$ on $A C$ such that $A P' H Q'$ is a parallelogram. We shall prove that $P' Q' \perp O M$. Let $T$ be the antipodal point of $A$ in the circumcircle of triangle $A B C$. We have $O M \parallel T H$, hence it suffices to prove that $P' Q' \perp T H$.

Notice that $B H \perp A Q'$, and $C H \perp A P'$. Since the diagonals $A H$ and $B C$ of parallelograms $B T C H$ and $A P' H Q'$ are perpendicular, it follows that diagonals $P' Q'$ and $H T$ are also perpendicular. Therefore $P = P'$ and $Q = Q'$, and we are done.

Let $S$ and $T$ be the midpoints of sides $A C$ and $A B$, respectively. Quadrilateral $M Q S O$ is cyclic, hence $\widehat{M Q O} = \widehat{M S O}$. Quadrilateral $M P T O$ is cyclic, hence we have $\widehat{M P O} = \widehat{M T O}$.

Observe that $M S O T$ is a parallelogram. Indeed, we have $M S \parallel H C$ and $C H \perp A B$ implies $S M \perp A B$, therefore $A M \parallel O T$. Also, we have $M T \parallel H B$ and $B H \perp A C$, implies $T M \perp A C$, therefore $T M \parallel O S$.

Finally, from $\widehat{M Q O} = \widehat{M P O}$, it follows that triangle $O Q P$ is isosceles, that is $O Q = O P$. Since $O M \perp P Q$, the conclusion follows.