32. In an acute triangle $ABC$, the distance from vertex $A$ to the circumcenter $O$ is equal to the distance from $A$ to the orthocenter $H$. Find all possible values of $\angle A$.
32. Let $C^{\prime} C^{\prime}$

be the height. Since
$A H=R$ (the radius of the circumcircle, so
$$
\begin{array}{l}
A C^{\prime}=R \\
\cdot \sin \angle A H C^{\prime} \\
=R \sin B .
\end{array}
$$

Thus, $C C^{\prime}=R \sin B \operatorname{tg} A$.
Also, $C C^{\prime}=B C \sin B=2 R \sin A \sin B$,
comparing the two equations we get $2 \sin A=\operatorname{tg} A$.
Therefore, $A=60^{\circ}$.
(Single Zhen, Liu Yaqiang, provided solution)