1. Two circles of equal radii touch each other externally. Tangents are drawn from the center of one circle to the other circle. Determine the ratio of the area of the quadrilateral whose vertices are the centers of the circles and the points of tangency to the area of the region between the tangents and the circles determined by the given circles (shaded in the figure).

![](https://cdn.mathpix.com/cropped/2024_06_03_5d6b13b2bd1879787abdg-4.jpg?height=532&width=736&top_left_y=1136&top_left_x=657)
1. Two circles of equal radii touch each other externally. Tangents are drawn from the center of one circle to the other circle. Determine the ratio of the area of the quadrilateral whose vertices are the centers of the circles and the points of tangency to the area of the region between the tangents and the arcs of the circles (shaded in the figure).

![](https://cdn.mathpix.com/cropped/2024_05_30_ba00de53bdad84e80dffg-20.jpg?height=557&width=782&top_left_y=835&top_left_x=634)

## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_ba00de53bdad84e80dffg-20.jpg?height=620&width=985&top_left_y=1512&top_left_x=524)

Let the centers of the circles be $H$ and $I$, and the points of tangency be $K$ and $J$, as shown in the figure.

The area of the quadrilateral $HKIJ$ is equal to the area of an equilateral triangle with side length $2r$, i.e.,

$P_{H K I J}=\frac{(2 r)^{2} \sqrt{3}}{4}=r^{2} \sqrt{3}$.

The area of the region between the tangents and the arcs of the circles is equal to the difference between the area of the quadrilateral $HKIJ$ and the sum of the areas of two circular segments.

Since $\Delta H I J$ is a right triangle with acute angles of $30^{\circ}$ and $60^{\circ}$, the central angle of one of these segments is $60^{\circ}$, and the other is $120^{\circ}$, with radii of $r$.

$$
P_{I_{1}}+P_{I_{2}}=60 \cdot \frac{r^{2} \pi}{360}+120 \cdot \frac{r^{2} \pi}{360}=\frac{1}{2} r^{2} \pi
$$

Thus, the area of the desired region is

$$
P_{d}=P_{H K I J}-\left(P_{I_{1}}+P_{I_{2}}\right)=r^{2} \sqrt{3}-\frac{1}{2} r^{2} \pi=r^{2}\left(\sqrt{3}-\frac{1}{2} \pi\right)
$$

The ratio of the areas is

$\frac{r^{2} \sqrt{3}}{r^{2}\left(\sqrt{3}-\frac{1}{2} \pi\right)}=\frac{\sqrt{3}}{\sqrt{3}-\frac{1}{2} \pi}=\frac{2 \sqrt{3}}{2 \sqrt{3}-\pi}$.