3 [Radii of the inscribed, circumscribed, and exscribed circles (other).]

Let $O$ be the center of the circle circumscribed around triangle $A B C, \angle A O C=60^{\circ}$. Find the angle $A M C$, where $M-$ is the center of the circle inscribed in triangle $A B C$.
If points $O$ and $B$ lie on opposite sides of line $A C$ (Fig.1), then the degree measure of arc $A C$, not containing point $B$, is $360^{\circ}-60^{\circ}=300^{\circ}$, therefore

$$
\angle A B C=\frac{1}{2} \cdot 300^{\circ}=150^{\circ}
$$

The sum of the angles at vertices $A$ and $C$ of triangle $A B C$ is $180^{\circ}-150^{\circ}=30^{\circ}$, and since $A M$ and $C M$ are the angle bisectors of triangle $A B C$, the sum of the angles at vertices $A$ and $C$ of triangle $A M C$ is $15^{\circ}$. Therefore,

$$
\angle A M C=180^{\circ}-15^{\circ}=165^{\circ}
$$

If points $O$ and $B$ lie on the same side of line $A C$ (Fig.2), then similarly we get $\angle A M C=105^{\circ}$.

## Answer

$165^{\circ}$ or $105^{\circ}$.