In a certain right-angled triangle, the radius of the inscribed circle is $\varrho=8 \mathrm{~cm}$, and the radius of the circumscribed circle is $41 \mathrm{~cm}$. What are the lengths of the sides of the triangle?
I. solution: The center of the circumscribed circle $K$ is the midpoint of the hypotenuse $AB$, so $AK = KB = r = 41$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_c19883d79ecb4eed7495g-1.jpg?height=214&width=409&top_left_y=191&top_left_x=847)

Denoting the center of the inscribed circle by $O$ and the points of tangency on the sides $a, b, c$ by $A_1, B_1, C_1$ respectively, $OA_1CB_1$ is a square with side $\varrho = 8$.

If we denote the segment $C_1K$ by $x$, then

\[
\begin{aligned}
a & = BA_1 + A_1C = BC_1 + A_1C = (r + x) + \varrho = r + \varrho + x \\
b & = AB_1 + B_1C = AC_1 + B_1C = (r - x) + \varrho = r + \varrho - x \\
c & = 2r
\end{aligned}
\]

By the Pythagorean theorem, substituting the values of $r$ and $\varrho$,

\[
(49 + x)^2 + (49 - x)^2 = 82^2
\]

which simplifies to

\[
x^2 = 961
\]

from which

\[
x_1 = 31, \quad x_2 = -31
\]

Thus,

\[
a = 49 \pm 31 = 80 \text{ or } 18, \quad b = 49 \mp 31 = 18 \text{ or } 80
\]

Therefore, the lengths of the legs are $80 \text{ cm}$ and $18 \text{ cm}$, and the hypotenuse is $2r = 82 \text{ cm}$.

Zsombok Zoltán (Bp., IV., Könyves Kálmán g. I. o. t.)

II. solution: If we denote the perimeter of the triangle by $2s$, then the semiperimeter is

\[
s = AC_1 + BA_1 + CB_1 = AC_1 + BC + CB_1 = 2r + \varrho = 82 + 8 = 90
\]

so $2s = 180$, and thus

\[
a + b = 2s - c = 180 - 82 = 98
\]

On the other hand, the area of the triangle $t = \frac{ab}{2} = s\varrho = 90 \cdot 8 = 720$, and thus

\[
8t = 4ab = 5760
\]

Subtracting (2) from the square of (1), we get

\[
(a - b)^2 = 98^2 - 5760 = 3844 = 62^2
\]

Without loss of generality, let $a$ be the larger leg. In this case,

\[
a - b = 62
\]

From (1) and (3),

\[
a = 80 \quad \text{and} \quad b = 18
\]