Circles with centers $O_{1}$ and $O_{2}$ have a common chord $A B, \angle A O_{1} B=60^{\circ}$. The ratio of the length of the first circle to the length of the second is $\sqrt{2}$. Find the angle $A O_{2} B$.
Express the sides of triangle $A O_{2} B$ in terms of the radius of the smaller circle.

## Solution

Let $R$ and $r$ be the radii of the circles with centers $O_{1}$ and $O_{2}$, respectively. According to the problem,

$$
\frac{2 \pi R}{2 \pi r}=\frac{R}{r}=\sqrt{2} \text {. }
$$

Therefore, $R=r \sqrt{2}$. Triangle $A O_{1} B$ is equilateral, so

$$
A B=O_{1} A=R=r \sqrt{2}
$$

The lateral sides of the isosceles triangle $A O_{2} B$ are equal to $r$, and the base $A B=r \sqrt{2}$. Then

$$
O_{2} A^{2}+O_{2} B^{2}=r^{2}+r^{2}=2 r^{2}=A B^{2}
$$

Therefore, triangle $A O_{2} B$ is a right triangle and $\angle A O_{2} B=90^{\circ}$.

## Answer

$90^{\circ}$.