Let's draw the circles that touch the sides and pass through the vertices of a right-angled triangle. What are the lengths of the sides of the triangle if the radius of the first circle is $8 \mathrm{~cm}$ and the radius of the second circle is $41 \mathrm{~cm}$?
Let $ABC$ be the sought triangle, and let the incircle touch the sides of the triangle at $A_1$, $B_1$, and $C_1$. According to the problem,

\[
\begin{gathered}
\overline{BC}=2 \cdot 41=82 \text{ cm}, \text{ and } \overline{B_1 A}=\overline{C_1 A}=8 \text{ cm} \\
\text{If } \overline{B_1 C}=\overline{A_1 C}=y \text{ and } \overline{A_1 B}=\overline{C_1 B}=x, \text{ then } \\
x+y=82
\end{gathered}
\]

and

\[
(x+8)^{2}+(y+8)^{2}=82^{2}
\]

from which

\[
x_{1}=72, x_{2}=10 ; y_{1}=10 \text{ and } y_{2}=72 \text{. }
\]

Thus, the sides of the triangle are:

\[
\overline{AB}=18 \text{ cm}, \overline{AC}=80 \text{ cm} \text{ and } \overline{BC}=82 \text{ cm}.
\]

(Jenő Engler, Pécs.)

The problem was also solved by: Bauer E., Dénes M., Erdős V., Fried E., Grün E., Neumann L., Szántó L., Spitzer L., Viola R.