Two circles with radii $1$ and $2$ have a common center at the point $O$. The vertex $A$ of the regular triangle $ABC$ lies on the larger circle, and the middpoint of the base $CD$ lies on the smaller one. What can the angle $BOC$ be equal to?
Let $M$ be the midpoint of $\overline{CB}$ and $O'$ be the reflection of $O$ with respect to $M$. Let $G$ be the centroid of $\triangle ACB$ then $\frac{\overline{AG}}{\overline{AM}} = \frac{2}{3}$, also we have $|MO|=|MO'|$ so $G$ is the centroid of $\triangle AOO'$ Hence $OG$ passes through the midpoint of $\overline{AO'}$
combined with the fact that $|OO'|=|OA|$ we get that $O$ is equidistant from $A,O$ so $G$ is the center of $(ACBO')$ which clearly means $ACBO'$ is cyclic. We have $\measuredangle CAB \equiv \measuredangle CO'B(mod 180^{\circ})$ so $\angle COB(=\angle CO'B$) is either $60^{\circ}$ or $120^{\circ}$. To see that both of these angles are really answers consider two different constructions, in both $BC \perp AO$ and $G$ be one the intersections of $AO$ with the smaller circle.