10.412 The area of a right triangle is equal to $2 r^{2} / 3$, where $r$ is the radius of the circle that touches one leg and the extension of the other leg and the hypotenuse. Find the sides of the triangle.
## Solution.

B
A

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-249.jpg?height=256&width=213&top_left_y=1434&top_left_x=222)
K
0
Let the circle with center $O$ and radius $r$ touch the leg $A C$ and the extensions of the leg $A B$ and the hypotenuse $B C$ of the given $\triangle A B C$ at points $M, K, N$ respectively. Let $A B=x, A C=y$. Then $A K=A M=r \quad(A M O K-$ square), $C N=M C=y-r$ (Theorem 3) $B N=B K=x+r$ (Theorem 3) $\Rightarrow x+r=B C+C N=B C+y-r \Rightarrow$
$\Rightarrow B C=\sqrt{x^{2}+y^{2}}=2 r+x-y \Rightarrow\left\{\begin{array}{l}x^{2}+y^{2}=(2 r+x-y)^{2}, \\ \frac{x y}{2}=\frac{2 r^{2}}{3}\end{array} \Rightarrow\left\{\begin{array}{l}y-x=\frac{r}{3}, \\ x y=\frac{4 r^{2}}{3}\end{array} \Rightarrow\left\{\begin{array}{l}y=\frac{4}{3} r, \\ x=r .\end{array}\right.\right.\right.$

Answer: $r, \frac{4}{3} r, \frac{5}{3} r$.