16th Turkish Mathematical Olympiad

Since $H$ is the orthocenter, $\angle BHC=180^{\circ }-\angle BAC=\angle B{A}_{0}C$. As $A_1$ is the midpoint of $BC$, this is possible only if $BHC{A}_0$ is a parallelogram. Then $\angle AC{A}_0=90^{\circ }$, and $A{A}_0$ is a diameter of the circumcircle of $ABC$. Therefore, the reflection across $O$ takes $ABC$ to $A_0{B}_0{C}_0$, and takes $H$ to $H_0$. In particular, $O$ is the midpoint of $H{H}_0$.