Proyasov B.Y.

Two circles with radii 1 and 2 have a common center at point $O$. Vertex $A$ of an equilateral triangle $ABC$ lies on the larger circle, and the midpoint of side $BC$ lies on the smaller circle. What can the angle $BOC$ be equal to?
## Author: Oschekina M

Consider the case when points $O$ and $A$ lie in the same half-plane relative to the line $B C$ (see the left figure). Let $K$ be the midpoint of $B C$, and $G$ be the point of intersection of the medians of triangle $A B C$. Extend the segment $O K$ until it intersects the larger circle at point $O_{1}$. Then $K$ is also the midpoint of segment $O O_{1}$, so $B O C O_{1}$ is a parallelogram. $G$ is the point of intersection of the medians of triangle $A O O_{1}$ as well, since $A K$ is a median of this triangle, and $A G: G K=2: 1$. Since this triangle is isosceles, the median $O G$ is also the bisector, which implies the congruence of triangles $A G O$ and $O_{1} G O$. Therefore, $G O_{1}=G A=G B=G C$, meaning that points $A, B, C, O_{1}$ lie on a circle with center $G$. Hence, $\angle B O_{1} C=180^{\circ}-\angle A=60^{\circ}$. Similarly, considering the second case, we get $\angle B O_{1} C=120^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_6c7d8d8bac098e64f10bg-26.jpg?height=948&width=2018&top_left_y=495&top_left_x=16)

## Answer

$60^{\circ}$ or $120^{\circ}$.