A circle is circumscribed around an isosceles trapezoid. The lateral side of the trapezoid is $a$, and the segment connecting the points of tangency of the lateral sides with the circle is $b$. Find the diameter of the circle.
The triangle, whose vertices are the center of the circle, the point of tangency with the lateral side of the trapezoid, and the midpoint of this side, is similar to the triangle, whose vertices are the center of the circle, the point of tangency with the lateral side of the trapezoid, and the midpoint of the segment with endpoints at the specified points of tangency.

## Solution

Let the inscribed circle touch the lateral side $AB$ of trapezoid $ABCD$ at point $M$, and the lateral side $CD$ at point $N$. The center $O$ of this circle is located on the midline $PQ$ of the trapezoid (where $Q$ is the midpoint of $CD$), and the projection $K$ of point $O$ onto $MN$ is the midpoint of $MN$.

By the property of a circumscribed quadrilateral, $PQ = 1/2(AD + BC) = 1/2(AB + CD) = a$. Therefore, $OQ = a/2$. From the similarity of triangles $OKN$ and $QNO$, it follows that $KN : ON = ON : OQ$, hence $ON^2 = OQ \cdot KN = ab/4$. Since $ON$ is the radius of the inscribed circle, and the height of the trapezoid is equal to the diameter, this height is equal to $2ON$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-06.jpg?height=377&width=643&top_left_y=2547&top_left_x=702)

## Problem