3. In a right-angled triangle $ABC\left(\gamma=90^{\circ}\right)$, the center of the inscribed circle is at a distance of $\sqrt{5}, \sqrt{10}, \sqrt{2} \mathrm{~cm}$ from the vertices $A, B, C$ respectively. Determine the radius of the circumscribed circle of this triangle.
## Solution.

Let $S$ be the center of the inscribed circle of the triangle, $r$ the radius of the inscribed circle, $R$ the radius of the circumscribed circle, and let $A_{1}, B_{1}$, and $C_{1}$ be the points of tangency of the sides of the triangle and the inscribed circle.

![](https://cdn.mathpix.com/cropped/2024_05_30_d793ff9f9da41a4dd093g-06.jpg?height=505&width=920&top_left_y=1592&top_left_x=705)

First, note that the hypotenuse of the right triangle is the diameter of the circumscribed circle of the triangle. Therefore, the desired radius is half the length of the hypotenuse.

Additionally, it is a known fact that $\left|A C_{1}\right|=\left|A B_{1}\right|,\left|B C_{1}\right|=\left|B A_{1}\right|$, and $\left|C A_{1}\right|=\left|C B_{1}\right|$.

Furthermore, because the angle $\angle A C B$ is a right angle, the quadrilateral $C B_{1} S A_{1}$ must be a square, so $\left|C A_{1}\right|=\left|C B_{1}\right|=r$.

Let $x=\left|A C_{1}\right|=\left|A B_{1}\right|$ and $y=\left|B C_{1}\right|=\left|B A_{1}\right|$.

From the right triangles $A B_{1} S, B C_{1} S$, and $C A_{1} S$, we have respectively

$$
x^{2}+r^{2}=5, \quad y^{2}+r^{2}=10, \quad 2 r^{2}=2
$$

From the last equation, it follows that $r=1$, and then from the first two, $x=2$ and $y=3$.

![](https://cdn.mathpix.com/cropped/2024_05_30_d793ff9f9da41a4dd093g-07.jpg?height=479&width=894&top_left_y=483&top_left_x=701)

Finally, $R=\frac{x+y}{2}=\frac{5}{2}$.