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 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 $30^{\circ}$, 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-37.jpg?height=442&width=563&top_left_y=1498&top_left_x=815)

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 a side of 2 cm is formed. Then $O A=2 \sqrt{2}$, $O C=\frac{O K}{\sin 15^{\circ}}=\frac{4 \sqrt{2}}{(\sqrt{3}-1)}$, then the diagonal of the square $A B C D \quad A C=\frac{4 \sqrt{2}}{\sqrt{3}-1}-2 \sqrt{2}$, and the side of the square $A B=\frac{A C}{\sqrt{2}}=\frac{4}{\sqrt{3}-1}-2=2 \sqrt{3}$.

Answer. $2 \sqrt{3}$.