Task 2. (10 points) A circle with a radius of 15 touches two adjacent sides $AB$ and $AD$ of the square $ABCD$. On the other two sides, the circle intersects at points, cutting off segments of 6 cm and 3 cm from the vertices, respectively. Find the length of the segment that the circle cuts off from vertex $B$ at the point of tangency.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_b2236406d6db6e8875c1g-22.jpg?height=497&width=488&top_left_y=1348&top_left_x=858)

Fig. 1

Let $X$ be the desired segment, then $X+15$ is the side of the square. The segment $K L=15+X-6-3=X+6$. Consider $\triangle O N L$. By the Pythagorean theorem, the following equality holds:

$$
15^{2}=X^{2}+\left(\frac{X+6}{2}\right)^{2}
$$

This is a quadratic equation in terms of $X$, which has only one positive root 12.

Answer. 12.