Romanian Mathematical Olympiad

a) Let $A'$ be the point diametrically opposite to $A$ on the circumcircle of the triangle $ABC$. It is known that $H$ and $A'$ are symmetric with respect to $X$, so $H$, $X$ and $A'$ are collinear. Then $\angle ABA' = \angle ACA' = 90^\circ$. Thus, the quadrilaterals $A'BYH$ and $A'CZH$ are cyclic, therefore $\angle YBH = \angle YA'H$ and $\angle ZCH = \angle ZA'H$.
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 $\angle YA'H = \angle ZA'H$, so $A'H$ is both an altitude and an angle bisector in the triangle $YA'Z$, 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 $H'$ 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 $BH'C$, 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 $\overrightarrow{OO'} = 2\overrightarrow{OX} = \overrightarrow{AH}$, since $OX$ is a midline in triangle $HA'A$.
From here it follows $\overrightarrow{AZ} + \overrightarrow{AY} = 2\overrightarrow{OO'}$.