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 $6-2\sqrt{5}$ 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 $36^{\circ}$, and it is known that $\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-26.jpg?height=451&width=576&top_left_y=1442&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 with side $6-2 \sqrt{5}$ cm is formed. Then $O A=\sqrt{2}(6-2 \sqrt{5})$, $O C=\frac{O K}{\sin 18^{\circ}}=\frac{4(6-2 \sqrt{5})}{(\sqrt{5}-1)}$, the diagonal of the square $A B C D \quad A C=\frac{4(6-2 \sqrt{5})}{(\sqrt{5}-1)}-\sqrt{2}(6-2 \sqrt{5})$, and the side of the square $A B=\frac{A C}{\sqrt{2}}=\frac{4(\sqrt{5}-1)^{2}}{\sqrt{2}(\sqrt{5}-1)}-(6-2 \sqrt{5})=(\sqrt{5}-1)(2 \sqrt{2}-\sqrt{5}+1)$.

Answer. $(\sqrt{5}-1)(2 \sqrt{2}-\sqrt{5}+1)$.