74th Romanian Mathematical Olympiad

a) Let $A^{\prime }$ be the point diametrically opposite to $A$ on the circumcircle of the triangle $ABC$. It is known that $H$ and $A^{\prime }$ are symmetric with respect to $X$, so $H$, $X$ and $A^{\prime }$ are collinear. Then $\angle AB{A}^{\prime }=\angle AC{A}^{\prime }=90^{\circ }$. Thus, the quadrilaterals $A^{\prime }BYH$ and $A^{\prime }CZH$ are cyclic, therefore $\angle YBH=\angle Y{A}^{\prime }H$ and $\angle ZCH=\angle Z{A}^{\prime }H$. On the other hand, from the right triangles $BEA$ and $CFA$ we have $\angle ABE=\angle ACF=90^{\circ }-\angle BAC$ ($E$ and $F$ are the feet of the altitudes from $B$ and $C$ respectively). Consequently $\angle Y{A}^{\prime }H=\angle Z{A}^{\prime }H$, so $A^{\prime }H$ is both an altitude and an angle bisector in the triangle $Y{A}^{\prime }Z$, so it is also a median, that is $HY=HZ$.

b) From a) it follows that $\overrightarrow{AZ}+\overrightarrow{AY}=2\overrightarrow{AH}$. It is known that the symmetrical $H^{\prime }$ of the orthocenter $H$ with respect to the side $BC$ is on the circumcircle of the triangle $ABC$, so the symmetrical of the circumcircle of the triangle $BHC$ with respect to $BC$ is the circumcircle of the triangle $B{H}^{\prime }C$, i.e. the circumcircle of the triangle $ABC$. Thus, the center of the circumcircle of the triangle $BHC$ is symmetrical to $O$ with respect to $BC$. We deduce that $\overrightarrow{O{O}^{\prime }}=2\overrightarrow{OX}=\overrightarrow{AH}$, since $OX$ is a midline in triangle $H{A}^{\prime }A$. From here it follows $\overrightarrow{AZ}+\overrightarrow{AY}=2\overrightarrow{O{O}^{\prime }}$.