It is known that the point symmetric to the center of the inscribed circle of triangle $ABC$ with respect to side $BC$ lies on the circumcircle of this triangle. Find angle $A$.

#
If $O$ is the center of the inscribed circle of triangle $ABC$, then $\angle BOC = 90^{\circ} + \frac{1}{2} \angle BAC$.

## Solution

Let $\angle BAC = \alpha$. Suppose $O$ is the center of the inscribed circle of triangle $ABC$. Then $BO$ and $CO$ are the angle bisectors of $\angle ABC$ and $\angle ACB$. Therefore,

$$
\angle BOC = 90^{\circ} + \frac{1}{2} \angle BAC = 90^{\circ} + \frac{\pi}{2}.
$$

Let $O_1$ be the point symmetric to point $O$ with respect to the line $BC$. Then

$$
\angle BO_1C = \angle BOC = 90^{\circ} + \frac{\pi}{2}.
$$

By the problem's condition, point $O_1$ lies on the circumcircle of triangle $ABC$, so the quadrilateral $ABO_1C$ is inscribed in this circle. Therefore,

$$
\angle BO_1C + \angle BAC = 180^{\circ}
$$

or

$$
\left(90^{\circ} + \frac{\alpha}{2}\right) + \alpha = 180^{\circ}
$$

From this equation, we find that $\alpha = 60^{\circ}$.

## Answer

$60^{\circ}$.