In an isosceles trapezoid, the bases of which are equal to $a$ and $b(a>b)$, a circle can be inscribed. Find the distance between the centers of the inscribed and circumscribed circles around this trapezoid.
Drop perpendiculars from the centers of the given circles to one of the lateral sides of the trapezoid, connect the center of the inscribed circle with the midpoint of this lateral side, and consider the similar triangles obtained.

## Solution

Let $O$ and $Q$ be the centers of the inscribed and circumscribed circles of trapezoid $ABCD$ with bases $AD = a$ and $BC = b$, $P$ be the point of tangency of the inscribed circle with the lateral side $CD$, and $F$ be the midpoint of $CD$.

The line $OQ$ is the perpendicular bisector of segments $BC$ and $AD$, and triangle $COD$ is a right triangle. Note that $OF = \frac{1}{4}(a + b)$,

$PF = CF - CP = \frac{1}{4}(a + b) - \frac{b}{2} = \frac{1}{2}(a - b)$. Triangles $OPF$ and $FOQ$ are similar by two angles. Therefore, $OQ : OF = PF : OP$, $OP = r = \sqrt{CP \cdot PD} = \frac{\sqrt{ab}}{2}$

$OQ = OF \cdot \frac{PF}{OP} = \frac{(a + b)(a - b)}{8 \sqrt{ab}}$

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-20.jpg?height=407&width=652&top_left_y=1760&top_left_x=703)

## Answer

$\frac{a^2 - b^2}{8 \sqrt{ab}}$.