3. (17 points) In a right triangle, the legs are equal to 3 and 4. Find the distance between the centers of the inscribed and circumscribed circles. In the answer, write the square of this distance.
Answer: 1.25.

Solution. By the Pythagorean theorem, the hypotenuse is 5, the area of the triangle is $S=\frac{1}{2} \cdot 3 \cdot 4=6$. The semiperimeter $p=\frac{3+4+5}{2}=6$, then the radius of the inscribed circle $r=\frac{s}{p}=1$. Let $O_{1}$ be the center of the inscribed circle, $O_{1} P$ and $O_{1} K$ be the radii of the inscribed circle, drawn respectively to the leg and the hypotenuse at the points of tangency of the circle, the legs $B C=4, A C=3$ by the condition. We have $B P=B C-r=4-1=3$. By the property of tangents drawn from one point to a circle, $B K=B P=3$. The center of the circumscribed circle $O_{2}$ around a right triangle lies at the midpoint of the hypotenuse and $B O_{2}=2.5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6b90c54680aad409e6d2g-2.jpg?height=348&width=402&top_left_y=1294&top_left_x=907)

In the right triangle $K O_{1} O_{2}$, we have $K O_{2}=3-2.5=0.5, K O_{1}=r=1, O_{1} O_{2}$ is the desired distance. By the Pythagorean theorem $O_{1} O_{2}^{2}=1+0.25=1.25$.