In a parallelogram, there are two circles. One of them, with a radius of 3, is inscribed in the parallelogram, and the second circle touches two sides of the parallelogram and the first circle. The distance between the points of tangency lying on the same side of the parallelogram is 3. Find the area of the parallelogram.
Drop a perpendicular from the center of the smaller circle to the radius of the larger circle, drawn to the point of tangency with one of the sides of the parallelogram.

## Solution

Since a circle is inscribed in the parallelogram $ABCD$, it is a rhombus.

Let $O_{1}$ and $O_{2}$ be the centers of the given circles, $R$ and $r$ be their radii ($R=3$), $M_{1}$ and $M_{2}$ be the points of tangency of the circles with side $AB$ ($M_{2}$ between $M_{1}$ and $A$). Since $M_{1} M_{2}=2 \sqrt{R r}=3$, then $r=\frac{3}{4}$.

Let $K$ be the foot of the perpendicular dropped from $O_{2}$ to $O_{1} M_{1}$. From the similarity of triangles $A M_{2} O_{2}$ and $O_{2} K O_{1}$, we find that $A M^{2}=1$. Therefore,

$$
A M_{1}=A M_{2}+M_{2} M_{1}=1+3=4
$$

Since $\angle A O_{1} B=90^{\circ}$, then $O_{1} M^{2}{ }_{1}=B M_{1} \cdot A M_{1}$. From this, we find that,

$$
B M_{1}=\frac{9}{4}, A B=\frac{25}{4}
$$

Therefore, $S_{\mathrm{ABCD}}=\frac{75}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-04.jpg?height=417&width=809&top_left_y=2330&top_left_x=630)

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-05.jpg?height=112&width=2036&top_left_y=226&top_left_x=16)

Prove that the area of a triangle can be expressed by the formula $S=(p-a) r_{a}$, where $r_{a}$ is the radius of the excircle tangent to the side equal to $a$, and $p$ is the semiperimeter of the triangle.

## Solution

Let $O$ be the center of the circle with radius $r_{a}$, tangent to side $BC=a$ of triangle $ABC$ at point $Q$, and the extensions of sides $AB=c$ and $AC=b$ at points $M$ and $N$ respectively. Then

$$
S_{\triangle A B C}=S_{\triangle A O B}+S_{\triangle A O C}-S_{\triangle B O C}=\frac{1}{2} A B \cdot O M+\frac{1}{2} A C \cdot O N-\frac{1}{2} B C \cdot O Q=
$$

$$
=\frac{1}{2} c \cdot r_{a}+\frac{1}{2} b \cdot r_{a}-\frac{1}{2} a \cdot r_{a}=\frac{c+b-a}{2} r_{a}=(p-a) r_{a} \cdot
$$