The altitudes of an unequal-sided acute triangle $A B C$ intersect at point $H$. $O$ is the center of the circumscribed circle of triangle $B H C$. The center $I$ of the inscribed circle of triangle $A B C$ lies on the segment $O A$.

Find the angle $A$.

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From the condition, it follows that point $O$ lies at the intersection of the angle bisector of $\angle A$ and the perpendicular bisector of side $BC$. Since these lines intersect on the circumcircle of triangle $ABC$, $O$ lies on this circle and is the midpoint of arc $BC$. Moreover, $\angle BHC = 180^{\circ} - \angle A$, since $H$ is the orthocenter of triangle $ABC$.

$VOS$ is a central angle subtending arc $BHC$ of the circumcircle of triangle $BHC$.

Therefore,

\[
\angle BHC = 180^{\circ} - \frac{1}{2} \angle BOC = 180^{\circ} - \frac{1}{2} \left(180^{\circ} - \angle A\right) = 90^{\circ} + \frac{1}{2} \angle A.
\]

Thus, $\angle 90^{\circ} + \frac{1}{2} \angle A = 180^{\circ} - \angle A$, from which $\angle A = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ccb5665c9ecd710ea159g-48.jpg?height=343&width=326&top_left_y=1226&top_left_x=863)

## Answer

$60^{\circ}$.