10.3. Given a triangle $A B C$, point $O$ is the center of the inscribed circle. Find the angle $A$ if it is known that the radii of the circumscribed circles of triangles $A B C$ and $B O C$ are the same.
Answer. $\quad 60^{\circ}$.

Solution. Let $\angle A=\alpha$. Note that $\angle B O C=90^{\circ}+\frac{\alpha}{2}$; indeed, $\angle B O C=180^{\circ}-(\angle O B C+\angle B C O)=180^{\circ}-\frac{1}{2}(\angle B+\angle C)=180^{\circ}-\frac{1}{2}\left(180^{\circ}-\alpha\right)=90^{\circ}+\frac{\alpha}{2}$. Then $B C=2 R_{A B C} \sin \alpha=2 R_{B O C} \sin \left(90^{\circ}+\frac{\alpha}{2}\right)=2 R_{B O C} \cos \frac{\alpha}{2} \Leftrightarrow \quad \sin \alpha=\cos \frac{\alpha}{2} \Leftrightarrow$ $\sin \frac{\alpha}{2}=\frac{1}{2} \Leftrightarrow \frac{\alpha}{2}=30^{\circ}$