BMO 2017

The straight lines $DO$, $EO$ and $FO$ are the perpendicular bisectors of the sides $BC$, $CA$ and $AB$ respectively. It follows that $[OM]$ is the diameter of the circumscribed circle of the triangle $ODM$ and $MP\perp ON$. The point $O$ is the orthocenter of the triangle $DEF$ (see the picture). Let $O_1$ be the circumcenter of the triangle $DEF$ and $H$ be the diametrically opposite point of $D$. The circumscribed circle of the triangle $DEF$ is the nine points circle of the triangle $ABC$. It follows that $EH\perp DE$, $FH\perp FD$ and $ED\parallel AF$, $DF\parallel AE$. So, the point $H$ is the orthocenter of the triangle $AEF$.

Let $AD\cap EF=\{I\}$ and $R$ is the reflection of the point $N$ with respect to the point $I$, i.e. $R\in (EF)$, $NI=RI$. The point $I$ is the symmetry center of the parallelogram $AEDF$. It follows that the point $I$ is the midpoint of the segment $[OH]$ and the quadrilaterals $AEDF$, $AND$, $HNOR$ are all parallelograms.

Let $Q$ be the reflection of the point $M$ with respect to the midpoint of the segment $DP$. It follows that the quadrilaterals $PQDM$ and $MNRD$ are parallelograms, which imply that the quadrilateral $PQRN$ is a parallelogram. So, $NO\parallel HR$, $NP\parallel RQ$, which imply that the points $H$, $R$ and $Q$ are collinear. We obtain that $m(\angle DQH)=90^{\circ }$, i.e. the point $Q$ belongs on the nine points circle of the triangle $ABC$.