In an isosceles trapezoid, two touching circles of radius $R$ are placed, each touching both bases and one of the lateral sides, and the centers of the circles lie on the diagonals. Find the sides of the trapezoid.
The diagonal of the given trapezoid cuts off an isosceles triangle from it.

## Solution

## First Method.

Let the circle with center \( O_1 \) touch the lateral side \( AB \) and the bases \( AD \) and \( BC \) at points \( M, K \), and \( N \) respectively, and the center of this circle lies on the diagonal \( AC \). Since \( AC \) is the angle bisector of \( \angle BAD \), then \( BC = AB = CD \).

Let \( AK = x \). From the right triangle \( AO_1B \), we find that \( MB = \frac{O_1M^2}{AM} = \frac{R^2}{x} \). Since \( AB = BC \),

\[
x + \frac{R^2}{x} = \frac{2R^2}{x} + 2R, \text{ or } x^2 - 2Rx - R^2 = 0.
\]

From this, we find that \( x = R(1 + \sqrt{2}) \). Therefore,

\[
\begin{gathered}
AB = BC = CD = R(1 + \sqrt{2}) + \frac{R^2}{R(1 + \sqrt{2})} = 2R\sqrt{2} \\
AD = 2x + 2R = 2R(2 + \sqrt{2})
\end{gathered}
\]

## Second Method.

Let \( O_1 \) and \( O_2 \) be the centers of the circles. Then \( O_1O_2 = 2R \), and the segment \( O_1O_2 \) belongs to the midline of the trapezoid. Therefore,

\[
O_1O_2 = \frac{AD - BC}{2}
\]

( \( AD \) and \( BC \) are the bases of the trapezoid, \( AD > BC \) ).

If \( P \) is the projection of vertex \( B \) on \( AD \), then

\[
AP = \frac{AD - BC}{2} = O_1O_2 = 2R
\]

Therefore, \( APB \) is an isosceles right triangle,

\[
AP = BP = 2R, AB = \sqrt{2}BP = 2R\sqrt{2}, BC = CD = AB = 2R\sqrt{2}
\]

\[
AD = BC + 2AP = 2R\sqrt{2} + 4R = 2R(2 + \sqrt{2})
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-04.jpg?height=432&width=1160&top_left_y=2431&top_left_x=449)
\( 2R\sqrt{2}, 2R\sqrt{2}, 2R\sqrt{2}, 2R(2 + \sqrt{2}) \).