Task B-1.7. In a right-angled triangle $\triangle A B C$, the length of the hypotenuse is $|A B|=c$, and the legs are $|A C|=\frac{3}{5} c$. Find the distance from vertex $C$ to the incircle of this triangle.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_bde1d6b20cdcb90302c4g-04.jpg?height=670&width=1244&top_left_y=424&top_left_x=451)

Precisely drawn image.

(1 point)

The distance from vertex $C$ to the incircle of the triangle is equal to the distance from vertex $C$ to point $E$. Point $E$ is the intersection of the incircle and the angle bisector of the right angle. If we denote the points of tangency of the circle with the legs as $D$ and $F$, then the quadrilateral $C F S D$ is clearly a square.

Its diagonal is $|C S|=r \sqrt{2}$, and $|C E|=r \sqrt{2}-r=r(\sqrt{2}-1)$. (*) The leg $a$ has length

$$
a=|B C|=\sqrt{c^{2}-b^{2}}=\sqrt{c^{2}-\left(\frac{3}{5} c\right)^{2}}=\frac{4}{5} c
$$

The radius of the incircle of the right triangle is

$$
r=\frac{a+b-c}{2}=\frac{\frac{4}{5} c+\frac{3}{5} c-c}{2}=\frac{1}{5} c
$$

From $(*)$ and $(* *)$ it follows that $|C E|=r(\sqrt{2}-1)=\frac{1}{5}(\sqrt{2}-1) c$.