[ Rhombi. Properties and Characteristics [

The side of the rhombus $ABCD$ is 6. The distance between the centers of the circles circumscribed around triangles $ABC$ and $BCD$ is 8. Find the radii of these circles.
Let $O_1$ and $O_2$ be the centers of the circumcircles of triangles $ABC$ and $BDC$ respectively, and $R_1$ and $R_2$ be their radii. Suppose that $\angle BCD \geqslant 90^{\circ}$. The line connecting the centers of two intersecting circles is perpendicular to their common chord and passes through its midpoint, i.e., the line $O_1 O_2$ intersects side $BC$ at its midpoint $M$.

Denote $\angle ACB = \angle BO_1M = \alpha$. From the right triangles $CO_2M$ and $BO_1M$, we find that $O_2M = CM \operatorname{tg} \alpha = 3 \operatorname{tg} \alpha$, and $O_1M = BM \operatorname{ctg} \alpha = 3 \operatorname{ctg} \alpha$.

According to the problem, $O_2M - O_1M = 8$, or $3 \operatorname{tg} \alpha - 3 \operatorname{ctg} \alpha = 8$. From this equation, we find that $\operatorname{tg} \alpha = 3$. Then,

$$
\cos \alpha = \frac{1}{\sqrt{10}}, \sin \alpha = \frac{3}{\sqrt{10}}
$$

From the same right triangles, we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_1f91526468b1b4d32730g-32.jpg?height=100&width=514&top_left_y=2351&top_left_x=772)

$$
\begin{aligned}
& R_2 = O_2C = \frac{CM}{\cos \alpha} = \frac{3}{\sqrt{10}} = 3 \sqrt{10}.
\end{aligned}
$$

Similarly for the case $\angle BCD < 90^{\circ}$.