Find the sides of parallelogram $ABCD$, in which the radii of the circumcircles of triangles $ABC$ and $ABD$ are 5 and $\sqrt{13}$, respectively, and the distance between the centers of these circles is 2.
Let $O$ be the center of parallelogram $ABCD$, $O_1$ and $O_2$ be the centers of the circumcircles of triangles $ABC$ and $ABD$ respectively, and the line $O_1O_2$ intersects segment $AB$ at point $E$. Since $AB$ is a common chord of the circles, point $E$ is the midpoint of $AB$ and $O_1O_2 \perp AB$. Let $AB = 2x$, $BC = 2y$, $\angle ACB = \angle CAD = \alpha$, $\angle ADB = \angle DBC = \beta$. Then $AO_1 = 5$, $AO_2 = \sqrt{13}$, $AE = BE = x$. From the right triangles $AO_1E$ and $AO_2E$, we find that $O_1E = \sqrt{25 - x^2}$ and $O_2E = \sqrt{13 - x^2}$. Since $O_1O_2 = O_1E - O_2E$, we get the equation

$$
\sqrt{25 - x^2} - \sqrt{13 - x^2} = 2
$$

from which we find that $x = 3$. Therefore, $AB = 2x = 6$. By the Law of Sines in triangles $ABC$ and $ABD$, we find that

$$
\sin \alpha = \frac{AB}{2 \cdot 5} = \frac{3}{5}, \sin \beta = \frac{AB}{2 \sqrt{13}} = \frac{3}{\sqrt{13}}.
$$

Then $\cos \alpha = \frac{4}{5}$, $\cos \beta = \frac{2}{\sqrt{13}}$,

$$
\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \frac{3}{5} \cdot \frac{2}{\sqrt{13}} + \frac{4}{5} \cdot \frac{3}{\sqrt{13}} = \frac{18}{5 \sqrt{13}}
$$

By the Law of Sines in triangle $BOC$, we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_b86fbfb53af157d03a0ag-42.jpg?height=141&width=1121&top_left_y=1833&top_left_x=471)

The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides, i.e., $(2BO)^2 + (2CO)^2 = 2AB^2 + 2BC^2$, or

$$
4\left(\frac{\sqrt{13}}{3} y\right)^2 + 4\left(\frac{5}{3} y\right)^2 = 2 \cdot 36 + 2 \cdot 4y^2
$$

from which $y^2 = \frac{81}{10}$. Therefore,

$$
BC = 2y = \frac{15}{\sqrt{10}} = 9 \sqrt{\frac{2}{5}}
$$

## Answer

$AB = 6$, $AD = 9 \sqrt{\frac{2}{5}}$.