9.3. The inscribed circle of triangle $A B C$ with center $O$ touches the sides $A B, B C$ and $A C$ at points $M, N$ and $K$ respectively. It turns out that angle $A O C$ is four times the angle $M K N$. Find angle $B$.
Answer: $108^{\circ}$. Solution. Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Then, in the isosceles triangle $A K M$, the angle at the base is $\angle A K M=\left(180^{\circ}-\alpha\right) / 2$. Similarly, in triangle $C K N$, we find $\angle C K N=\left(180^{\circ}-\gamma\right) / 2$. Therefore, $\angle M K N=180^{\circ}-\angle C K N-\angle A K M=(\alpha+\gamma) / 2=\left(180^{\circ}-\beta\right) / 2=$ $=90^{\circ}-\frac{\beta}{2}$. Next, $\angle A O C=180^{\circ}-\frac{\alpha}{2}-\frac{\gamma}{2}=180^{\circ}-\left(180^{\circ}-\beta\right) / 2=90^{\circ}+\frac{\beta}{2}$. Thus, from the condition of the problem, we have $90^{\circ}+\frac{\beta}{2}=4\left(90^{\circ}-\frac{\beta}{2}\right) \Leftrightarrow \beta=\frac{540^{\circ}}{5}=108^{\circ}$.