Task 2. (10 points) A circle touches the extensions of two sides $A B$ and $A D$ of square $A B C D$ with side length $6+2 \sqrt{5}$ cm. Two tangents are drawn from point $C$ to this circle. Find the radius of the circle 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-08.jpg?height=454&width=576&top_left_y=1638&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. The diagonal of the square $A B C D A C=\sqrt{2}(6+2 \sqrt{5})$. If radii of the circle are drawn to the points of tangency, a square with side $R$ is formed. Then $O A=R \sqrt{2}, \quad O C=\frac{O K}{\sin 18^{\circ}}=\frac{4 R}{\sqrt{5}-1}$. As a result, we get the equation $\frac{4 R}{\sqrt{5}-1}=R \sqrt{2}+8$, from which $R=\frac{\sqrt{2}(6+2 \sqrt{5})}{\frac{4}{\sqrt{5}-1}-\sqrt{2}}=\frac{(\sqrt{5}+1)^{2}(\sqrt{5}-1)}{2 \sqrt{2}-\sqrt{5}+1}=2(2 \sqrt{2}+\sqrt{5}-1)$.

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