5. Determine the angles of the isosceles triangle $ABC (AC = BC)$, if the centers $O_1$ and $O$ of the inscribed and circumscribed circles are symmetric with respect to the base $AB$ of the triangle.
Solution. We have $\measuredangle A S C=2 \measuredangle A B C$ (central and peripheral angles in a circle). Furthermore, from $S A=S C=R$ it follows that $\Varangle S A C=\measuredangle S C A$. If we denote $\measuredangle O A C=\varphi$, then due to the symmetry of points $O$ and $S$ with respect to $A B$ and

![](https://cdn.mathpix.com/cropped/2024_06_05_fd071d6ece13c75d5c89g-2.jpg?height=293&width=504&top_left_y=1091&top_left_x=993)
since $A O$ is the angle bisector of $\measuredangle \mathrm{BAC}$, we get $3 \varphi=\measuredangle \mathrm{SAC}=\measuredangle S C A=\frac{\gamma}{2}$. Now, since $\alpha=\beta=2 \varphi$, substituting into $\alpha+\beta+\gamma=180^{\circ}$ gives $2 \varphi+2 \varphi+6 \varphi=180^{\circ}$, i.e., $\varphi=18^{\circ}$. Therefore, the angles of the triangle are $36^{\circ}, 36^{\circ}, 108^{\circ}$.

## Eighth Section