[ Rhombi. Properties and Characteristics ]

In rhombus $A B C D$, the angle $\angle A B C=60^{\circ}$. A circle is tangent to the line $A D$ at point $A$, and the center of the circle lies inside the rhombus. The tangents to the circle, drawn from point $C$, are perpendicular. Find the ratio of the perimeter of the rhombus to the length of the circle.
Apply the Law of Cosines to triangle AOC (where $O$ is the center of the circle).

## Solution

Let $r$ be the radius of the circle, $a$ be the side of the rhombus, $O$ be the center of the circle, $M$ and $N$ be the points of tangency of the circle with the tangents drawn from vertex $C$. Then $A C=a, C M O N$ is a square,

$$
C O=r \sqrt{2}, O A=r, \angle O A C=\angle O A D-\angle C A D=90^{\circ}-60^{\circ}=30^{\circ}
$$

By the Law of Cosines in triangle $A O C$, we find that

$$
O C^{2}=A O^{2}+A C^{2}-2 A O \cdot A C \cos 30^{\circ},
$$

or

$$
r^{2}+a^{2}-a r \sqrt{3}=2 r^{2}, r^{2}+a r \sqrt{3}-a^{2}=0
$$

From this, we find that

$$
r=\frac{a(\sqrt{7}-\sqrt{3})}{2}
$$

Then the desired ratio is

$$
\frac{4 a}{2 \pi r}=\frac{\sqrt{3}+\sqrt{7}}{\pi}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-11.jpg?height=465&width=852&top_left_y=1191&top_left_x=612)

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-11.jpg?height=97&width=172&top_left_y=1735&top_left_x=14)