[ Touching circles [ $\underline{\text { C }}$ median proportionals in a right triangle ]

In a parallelogram, there are two circles touching each other and three sides of the parallelogram each. The radius of one of the circles is 1. It is known that one of the segments of a side of the parallelogram from a vertex to the point of tangency is $\sqrt{3}$. Find the area of the parallelogram.
Find the segments into which the tangency point of one of the circles divides the smaller side of the parallelogram.

## Solution

The circles are equal. The distance between the points of tangency of the circles with the larger side of the parallelogram is equal to the sum of their radii, i.e., 2. The smaller side of the parallelogram is seen from the center of the circle tangent to it at a right angle. One of the segments of this side from the vertex to the point of tangency is $\sqrt{3}$, so the other is $\frac{1}{\sqrt{3}}$. Therefore, the larger side is

$$
2+\sqrt{3}+\frac{1}{\sqrt{3}}=2+\frac{4}{\sqrt{3}}
$$

Thus, the area of the parallelogram is

$$
2\left(2+\frac{4}{\sqrt{3}}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_93426a107eaf3f76289cg-34.jpg?height=391&width=960&top_left_y=-1&top_left_x=555)

## Answer

$4+\frac{8}{\sqrt{3}}$