10.362 The hypotenuse of a right triangle is equal to $m$, the radius of the inscribed circle is $r$. Determine the legs. Under what relation between $r$ and $\boldsymbol{m}$ does the problem have a solution?
Solution. Let $x, y$ be the legs of the given triangle. Then, by Theorem 1 and the Pythagorean Theorem, we have:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ r = \frac { x + y - m } { 2 } } \\
{ m ^ { 2 } = x ^ { 2 } + y ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 2 r + m = x + y , } \\
{ m ^ { 2 } = ( x + y ) ^ { 2 } - 2 x y }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+y=2 r+m, \\
x y=2 r^{2}+2 r m
\end{array} \Leftrightarrow\right.\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x=\frac{2 r+m+\sqrt{m^{2}-4 r^{2}-4 r m}}{2}, \\
y=\frac{2 r+m-\sqrt{m^{2}-4 r^{2}-4 r m}}{2}, \\
m^{2} \geq 4 r^{2}+4 r m .
\end{array} \quad m^{2} \geq 4 r^{2}+4 r m \Leftrightarrow r \leq \frac{m(\sqrt{2}-1)}{2} .\right.
\end{aligned}
$$

Answer: $\frac{2 r+m \pm \sqrt{m^{2}-4 r^{2}-4 r m}}{2}$ when $r \leq \frac{m(\sqrt{2}-1)}{2}$.