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The center of the circumscribed circle of a triangle is symmetric to its incenter with respect to one of the sides. Find the angles of the triangle.

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Prove that the given triangle is isosceles.

## Solution

Let $O$ and $I$ be the centers of the circumcircle and incircle of triangle $ABC$, respectively, and let points $O$ and $Q$ be symmetric with respect to the line $BC$. Denote $\angle OBC = \angle IBC = \alpha$. Since triangle $BOC$ is isosceles, $\angle ICB = \angle OCB = \angle OBC = \alpha$. Since $BI$ is the angle bisector of $\angle B$, $\angle B = 2\alpha$. Similarly, $\angle C = 2\alpha$. Therefore, triangle $ABC$ is isosceles, its bisector $AM$ is an altitude, and points $I$ and $M$ lie on the segment $OA$. Since triangle $AOB$ is also isosceles, $\angle OBA = \angle OAB$, or $3\alpha = 90^\circ - 2\alpha$. Hence, $\alpha = 18^\circ$. Therefore, $\angle C = \angle B = 2\alpha = 36^\circ$.

## Answer

$36^\circ, 36^\circ, 108^\circ$.