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 $2 \sqrt{3} \mathrm{~cm}$. Two tangents are drawn from point $C$ to this circle. Find the radius of the circle if the angle between the tangents is $30^{\circ}$, and it is known that $\sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$.

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# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-14.jpg?height=451&width=576&top_left_y=1599&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 $A B C D A C=2 \sqrt{6}$. If radii of the circle are drawn to the points of tangency, a square with side $R$ is formed. Then $O A=R \sqrt{2}$, $O C=\frac{O K}{\sin 15^{\circ}}=\frac{2 \sqrt{2} R}{\sqrt{3}-1}$. As a result, we get the equation $\frac{2 \sqrt{2} R}{\sqrt{3}-1}=R \sqrt{2}+2 \sqrt{6}$, from which $R=\frac{2 \sqrt{6}}{\sqrt{6}+\sqrt{2}-\sqrt{2}}=2$.

Answer: 2.