32. (USA 3) The vertex $A$ of the acute triangle $A B C$ is equidistant from the circumcenter $O$ and the orthocenter $H$. Determine all possible values for the measure of angle $A$.
32. Let $C C^{\prime}$ be an altitude, and $R$ the circumradius. Then, since $A H=R$, we have $A C^{\prime}=|R \sin B|$ and hence (1) $C C^{\prime}=|R \sin B \tan A|$. On the other hand, $C C^{\prime}=|B C \sin B|=2|R \sin A \sin B|$, which together with (1) yields $2|\sin A|=|\tan A| \Rightarrow|\cos A|=1 / 2$. Hence, $\angle A$ is $60^{\circ}$. (Without the condition that the triangle is acute, $\angle A$ could also be $120^{\circ}$.)
Second Solution. For a point $X$, let $\bar{X}$ denote the vector $O X$. Then $|\bar{A}|=|\bar{B}|=|\bar{C}|=R$ and $\bar{H}=\bar{A}+\bar{B}+\bar{C}$, and moreover,
$$
R^{2}=(\bar{H}-\bar{A})^{2}=(\bar{B}+\bar{C})^{2}=2 \bar{B}^{2}+2 \bar{C}^{2}-(\bar{B}-\bar{C})^{2}=4 R^{2}-B C^{2} \text {. }
$$

It follows that $\sin A=\frac{R C}{2 R}=\sqrt{3} / 2$, i.e., that $\angle A=60^{\circ}$.
Third Solution. Let $A_{1}$ be the midpoint of $B C$. It is well known that $A H=2 O A_{1}$, and since $A H=A O=B O$, it means that in the rightangled triangle $B O A_{1}$ the relation $B O=2 O A_{1}$ holds. Thus $\angle B O A_{1}=$ $\angle A=60^{\circ}$.