A circle with radius $\frac{2}{\sqrt{3}}$ is inscribed in an isosceles trapezoid. The angle between the diagonals of the trapezoid, subtending the larger base, is $2 \operatorname{arctg} \frac{2}{\sqrt{3}}$. Find the segment connecting the points of tangency of the circle with the larger base of the trapezoid and one of its lateral sides.
Denote the halves of the bases of the trapezoid as $x$ and $y$; find $x, y$ and the angle between the lateral side and the larger base.

## Solution

Let $Q$ be the point of intersection of the diagonals $A C$ and $B D$ of trapezoid $A B C D$; $N, M$, and $K$ be the points of tangency of the inscribed circle with the smaller base $A D$, the lateral side $A B$, and the larger base $B C$ respectively; $O$ be the center of the circle. Denote $A N=x, B K=y$. Then,

$$
x y=O M^{2}=\frac{4}{3}, \quad x+y=A N+B K=(N Q+Q K) \operatorname{tg} \angle A N Q=\frac{8}{3}
$$

From the obtained system, we find that $x=\frac{2}{3}$ and $y=2$. Therefore,

$$
\cos \angle A B K=\frac{y-x}{y+x}=\frac{1}{2}, \quad \angle A B K=60^{\circ}, \quad M K=y=2
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_c317f3f1f50d1d30304bg-44.jpg?height=420&width=640&top_left_y=256&top_left_x=709)

## Answer

2.