3. A circle with a radius of 26 touches two adjacent sides of a rectangle, the lengths of which are 36 and 60. Find the segments into which the circle divides the sides of the rectangle.
Solution.

Note that the radius of the circle is greater than half the length of the shorter side of the rectangle, while the diameter of the circle is less than its longer side. This means that the circle intersects the longer side of the rectangle, opposite to the touching side, at two points, and does not intersect the second shorter side. The points are marked in the figure. Points $K$ and $M$ are the points of tangency, and the radii $O K$ and $O M$ are perpendicular to the sides of the rectangle. Since $B M O K$ is a square, the segments $B M$ and $B K$ are both 26. In the triangle $O L H$, two sides are known - the hypotenuse $O L$ is 26 and the leg $O H$ is 10. By the Pythagorean theorem, we find the leg $L H$, which is 24. From this, the segment $A L$ is 2, and the segment $L P$ is 48. The circle divides the side $A D$ into segments of lengths $2, 48, 10$.

Answer. 26 and $34 ; 26$ and $10 ; 2, 48$ and 10.

![](https://cdn.mathpix.com/cropped/2024_05_06_d6515d439ee8348f4294g-3.jpg?height=542&width=651&top_left_y=120&top_left_x=725)