3. In triangle $ABC$, the center of the inscribed circle $S$ and the center of the circumscribed circle $O$ are symmetric with respect to side $AB$. Calculate the angles of the triangle.
Solution. Point $O$ lies on the perpendicular bisector of side $AB$, so point $S$ must also lie on it. Since $\overline{AS}=\overline{BS}$, it follows that $\frac{\alpha}{2}=\frac{\beta}{2}$, and thus $\alpha=\beta$, meaning $\triangle ABC$ is isosceles with $\overline{AC}=\overline{BC}$. From $\triangle AOC$ and $\triangle BOC$, we conclude that $\frac{\gamma}{2}=3 \frac{\alpha}{2}=3 \frac{\beta}{2}$, i.e., $\gamma=3\alpha=3\beta$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_06_05_d112afdaa6f4bb49aaeeg-3.jpg?height=509&width=548&top_left_y=1303&top_left_x=932)

$$
\alpha+\beta+\gamma=5\alpha=180^{\circ}
$$

so $\alpha=\beta=36^{\circ}$ and $\gamma=108^{\circ}$.