Saudi Arabia Mathematical Competitions

Consider $P^{\prime }$ on $AB$, $Q^{\prime }$ on $AC$ such that $A{P}^{\prime }H{Q}^{\prime }$ is a parallelogram. We shall prove that $P^{\prime }{Q}^{\prime }\perp OM$. Let $T$ be the antipodal point of $A$ in the circumcircle of triangle $ABC$. We have $OM\parallel TH$, hence it suffices to prove that $P^{\prime }{Q}^{\prime }\perp TH$.



Notice that $BH\perp A{Q}^{\prime }$, and $CH\perp A{P}^{\prime }$. Since the diagonals $AH$ and $BC$ of parallelograms $BTCH$ and $A{P}^{\prime }H{Q}^{\prime }$ are perpendicular, it follows that diagonals $P^{\prime }{Q}^{\prime }$ and $HT$ are also perpendicular. Therefore $P=P^{\prime }$ and $Q=Q^{\prime }$, and we are done.



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Alternative solution.

Let $S$ and $T$ be the midpoints of sides $AC$ and $AB$, respectively. Quadrilateral $MQSO$ is cyclic, hence $\widehat{MQO}=\widehat{MSO}$. Quadrilateral $MPTO$ is cyclic, hence we have $\widehat{MPO}=\widehat{MTO}$.



Observe that $MSOT$ is a parallelogram. Indeed, we have $MS\parallel HC$ and $CH\perp AB$ implies $SM\perp AB$, therefore $AM\parallel OT$. Also, we have $MT\parallel HB$ and $BH\perp AC$, implies $TM\perp AC$, therefore $TM\parallel OS$.

Finally, from $\widehat{MQO}=\widehat{MPO}$, it follows that triangle $OQP$ is isosceles, that is $OQ=OP$. Since $OM\perp PQ$, the conclusion follows.