75th Romanian Mathematical Olympiad

Let $H$ be the orthocenter of the triangle $ABC$. Quadrilaterals $BIHK$ and $CIHJ$ are cyclic, so $\angle HIK=\angle HBK=\angle ABJ=\angle ACK=\angle JCH=\angle HIJ$.

Therefore, $IA$ is the angle bisector of $JIK$. Since $IA\perp BC$ and $MP\parallel BC$, it follows that $IA\perp MP$. Because $IA$ is both an angle bisector and an altitude in $△IMP$, this triangle is isosceles and $A$ is the midpoint of $MP$. ()

Let $S$ be the reflection point of $A$ with respect to point $D$. Since $D$ is the midpoint of $BC$, it follows that $ABSC$ is a parallelogram. Consequently, $BS\parallel AC$ and, since $BJ\perp AC$, we deduce that $BJ\perp BS$. From $\angle SBN=\angle SAN=90^{\circ }$ it follows that $SBAN$ is cyclic, therefore $\angle NSA=\angle ABN=90^{\circ }-\angle A$. Similarly, the quadrilateral $SCAQ$ is cyclic, hence $\angle QSA=\angle ACQ=90^{\circ }-\angle A=\angle NSA$. In the triangle $SNQ$, the altitude $SA$ is also an angle bisector, therefore $A$ is the midpoint of $NQ$. Since the diagonals of the quadrilateral $MNPQ$ bisect each other, it follows that $MNPQ$ is a parallelogram.

Alternative solution for (). Let $E$ be the projection of $D$ onto $AC$. Since $\angle DEA=\angle AJN=90^{\circ }$ and $\angle DAE=90^{\circ }-\angle NAJ=\angle ANJ$, it follows that $△DAE\sim △ANJ$. Thus, $\frac{NA}{AD}=\frac{AJ}{DE}=2\cdot \frac{AJ}{BJ}$. Similarly, we obtain $\frac{QA}{AD}=2\cdot \frac{AK}{CK}$. The triangles $ABJ$ and $ACK$ are similar (AA), thus $\frac{AJ}{BJ}=\frac{AK}{CK}$, and consequently $AQ=AN$.