The center of the circle circumscribed around the triangle coincides with the center of the inscribed circle. Find the angles of the triangle.

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Let $O$ be the common center of the inscribed and circumscribed circles of triangle $ABC$. Then triangles $AOB$, $AOC$, and $BOC$ are isosceles, and $CO$, $AO$, and $BO$ are the angle bisectors of angles $C$, $A$, and $B$.

## Solution

Let $O$ be the common center of the inscribed and circumscribed circles of triangle $ABC$. Triangle $AOC$ is isosceles. Therefore, $\angle OAC = \angle OCA$. Since $AO$ and $CO$ are the angle bisectors of angles $BAC$ and $BCA$, we have $\angle BAC = \angle BCA$. Similarly, $\angle BAC = \angle ABC$.

![](https://cdn.mathpix.com/cropped/2024_05_06_02603d12f3437ea52c37g-11.jpg?height=372&width=535&top_left_y=2415&top_left_x=767)

Problem