3 [Trigonometric ratios in a right triangle]

Given a rhombus $A B C D$. The radii of the circles circumscribed around triangles $A B C$ and $B C D$ are 1 and 2. Find the distance between the centers of these circles.

#
Segment $B C$ is the common chord of the circles centered at $O 1$ and $O 2$, circumscribed around triangles $A B C$ and $B C D$ respectively, so line $O 1 O 2$ is perpendicular to segment $B C$ and bisects it.

Let $M$ be the midpoint of $B C$. Then $O 1 M$ and $B D$ are the perpendicular bisectors of sides $B C$ and $A C$ of triangle $A B C$. Denote $\angle C B O 2=\angle B C O 2=\angle B O 1 M=\alpha$. From the right triangles $B M O 1$ and $B M O 2$, we find that

$$
B M=B O 1 \sin \alpha=1 \cdot \sin \alpha=\sin \alpha, B M=O 2 B \cos \alpha=2 \cos \alpha
$$

from which $\operatorname{tg} \alpha=2$. Then $\cos \alpha=\frac{1}{\sqrt{5}}, \sin \alpha=\frac{2}{\sqrt{5}}$, so,

$$
O 1 M=B O 1 \cos \alpha=\frac{1}{\sqrt{5}}, O 2 M=B O 2 \sin \alpha=\frac{t}{\sqrt{5}} \text {. }
$$

Therefore,

$$
O 1 O 2=O 2 M-O 1 M=\frac{t}{\sqrt{5}}-\frac{1}{\sqrt{5}}=\frac{3}{\sqrt{5}}
$$

## Answer

$\frac{3 \sqrt{5}}{5}$.