4B. In $\triangle A B C$, the centers of the inscribed and circumscribed circles are symmetric with respect to the side $A B$. Find the angles of the triangle.
Solution. Let $S$ be the center of the inscribed circle, and $O$ be the center of the circumscribed circle. Triangle $ABC$ is isosceles (the line $OS$ is the perpendicular bisector of $AB$). Therefore, $\angle BAC = \angle ABC = \alpha$. If $\{M\} = SO \cap AB$, then $\overline{SM} = \overline{OM}$ and $\triangle AMO \cong \triangle AMS$, so $\angle MAO = \frac{\alpha}{2}$. The inscribed angle $\angle ABC$ subtended by the arc $AC$ is equal to half of the central angle $\angle AOC$ subtended by the same arc. Therefore,

$$
\alpha = \angle ABC = \frac{1}{2} \angle AOC = \frac{1}{2} \left(90^{\circ} - \frac{\alpha}{2}\right) \text{ i.e. } 4 \alpha = 180^{\circ} - \alpha.
$$

Therefore, the angles of the triangle are $\alpha = \beta = 36^{\circ}$ and $\gamma = 108^{\circ}$.

## III year