8,9 [ Inscribed Circle in an Angle

In an isosceles trapezoid, there are two circles. One of them, with a radius of 1, is inscribed in the trapezoid, and the second circle touches two sides of the trapezoid and the first circle. The distance from the vertex of the angle formed by the two sides of the trapezoid that touch the second circle to the point of tangency of the circles is twice the diameter of the second circle. Find the area of the trapezoid.

#
Find the values of the trigonometric functions of the angle between the line connecting the centers of two given circles and the lateral side of the trapezoid.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the first and second circles, respectively, $M_{1}$ and $M_{2}$ be their points of tangency with the lateral side $A B$ of trapezoid $A B C D$, $N$ be the point of tangency of the first circle with the base $A D$, $K$ be the point of tangency of the circles, and $x$ be the radius of the second circle.

In triangle $A O_{2} M_{2}$, the leg $O_{2} M_{2}=x$, the hypotenuse $A O_{2}=3 x$,

$$
\sin \angle B A O_{2}=\frac{1}{3}, \operatorname{ctg} \angle B A O_{2}=2 \sqrt{2}
$$

From the right triangle $A O_{1} M_{1}$, we find that

$$
M_{1} A=O_{1} M_{1} \operatorname{ctg} \angle B A O_{1}=2 \sqrt{2}
$$

Since $O_{1} M^{2}_{1}=B M_{1} \cdot M_{1} A$, then

$$
B M_{1}=\frac{O_{1} M_{1}^{2}}{M_{1} A}=\frac{\sqrt{2}}{4}
$$

Therefore, the bases of the trapezoid are $4 \sqrt{2}$ and $\frac{\sqrt{2}}{2}$. Then

$$
S_{\mathrm{ABCD}}=\frac{4 \sqrt{2}+\frac{\sqrt{2}}{2}}{2} \cdot 2=\frac{9 \sqrt{2}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_8c2fdb390ca80df56784g-30.jpg?height=392&width=719&top_left_y=1150&top_left_x=675)

Answer

$\frac{9 \sqrt{2}}{2}$.