Task 2. (10 points) A circle touches the extensions of two sides $AB$ and $AD$ of square $ABCD$, and the point of tangency cuts off a segment of length $2+\sqrt{5-\sqrt{5}}$ cm from vertex $A$. Two tangents are drawn from point $C$ to this circle. Find the side of the square if the angle between the tangents is $72^{\circ}$, and it is known that $\sin 36^{\circ}=\frac{\sqrt{5-\sqrt{5}}}{2 \sqrt{2}}$.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-42.jpg?height=451&width=576&top_left_y=1465&top_left_x=817)

Fig. 1

The segment cut off from vertex $A$ by the point of tangency of the circle is equal to the radius of this circle. If radii of the circle are drawn to the points of tangency, a square is formed with a side length of $2+\sqrt{5-\sqrt{5}}$ cm.

Then $O A=\sqrt{2}(2+\sqrt{5-\sqrt{5}}), O C=\frac{O K}{\sin 36^{\circ}}=\frac{2 \sqrt{2}(2+\sqrt{5-\sqrt{5}})}{(\sqrt{5-\sqrt{5}})}$, so the diagonal of the square $A B C D A C=\frac{2 \sqrt{2}(2+\sqrt{5-\sqrt{5}})}{(\sqrt{5-\sqrt{5}})}-\sqrt{2}(2+\sqrt{5-\sqrt{5}})$, and the side of the square

$$
A B=\frac{A C}{\sqrt{2}}=\frac{2 \sqrt{2}(2+\sqrt{5-\sqrt{5}})}{\sqrt{5-\sqrt{5}}}-(2+\sqrt{5-\sqrt{5}})=\frac{(2+\sqrt{5-\sqrt{5}})(2-\sqrt{5-\sqrt{5}})}{\sqrt{5-\sqrt{5}}}=
$$

$$
=\frac{\sqrt{5}-1}{\sqrt{5-\sqrt{5}}}=\frac{\sqrt{\sqrt{5}-1}}{\sqrt[4]{5}}=\frac{\sqrt{\sqrt{5}-1} \cdot \sqrt[4]{125}}{5}
$$

Answer. $\frac{\sqrt{\sqrt{5}-1} \cdot \sqrt[4]{125}}{5}$.