For a certain right-angled triangle, the perimeter is $k$ and the radius of the inscribed circle is $\varrho$. What are the lengths of the sides? (Numerical example: $k=40 \mathrm{~cm}, \varrho=3 \mathrm{~cm})$.
I. Solution: The points of tangency of the inscribed circle divide the sides into 2-2 segments (see figure). Among these segments, $\varrho$ is known, while $x$ and $y$ are unknown.

![](https://cdn.mathpix.com/cropped/2024_05_02_e4c18d1b29720e1759dag-1.jpg?height=322&width=439&top_left_y=224&top_left_x=840)

Let the semiperimeter of the triangle be denoted by $s$, then

$$
\varrho + x + y = s
$$

and thus

$$
x + y = s - \varrho
$$

Furthermore, by the Pythagorean theorem,

$$
(\varrho + x)^{2} + (\varrho + y)^{2} = (x + y)^{2}
$$

which simplifies to

$$
x y = \varrho (\varrho + x + y) = \varrho s
$$

From (1) and (2), $x$ and $y$ are the two roots of the quadratic equation

$$
z^{2} - (s - \varrho) z + \varrho s = 0
$$

Thus,

$$
\begin{aligned}
& x_{1} = y_{2} = \frac{(s - \varrho) + \sqrt{(s - \varrho)^{2} - 4 \varrho s}}{2} \\
& x_{2} = y_{1} = \frac{(s - \varrho) - \sqrt{(s - \varrho)^{2} - 4 \varrho s}}{2}
\end{aligned}
$$

Accordingly,

$$
\begin{aligned}
& a = x + \varrho = \frac{(s + \varrho) + \sqrt{(s - \varrho)^{2} - 4 \varrho s}}{2} \\
& b = y + \varrho = \frac{(s + \varrho) - \sqrt{(s - \varrho)^{2} - 4 \varrho s}}{2} \\
& c = x + y = s - \varrho
\end{aligned}
$$

In our example, $s = 20 \, \text{cm}$ and $\varrho = 3 \, \text{cm}$, so

$$
\begin{aligned}
& a = \frac{23 + \sqrt{289 - 240}}{2} = \frac{23 + 7}{2} = 15 \, \text{cm} \\
& b = \frac{23 - 7}{2} = 8 \, \text{cm} \\
& c = 20 - 3 = 17 \, \text{cm}
\end{aligned}
$$

Gaál István (Csorna, Latinka S. g. II. o. t.)

II. Solution: Draw one of the circles inscribed in the triangle and denote its radius by $r$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e4c18d1b29720e1759dag-1.jpg?height=466&width=634&top_left_y=2236&top_left_x=735)

According to the figure,

$$
(a - \varrho) : \varrho = (a + r) : r
$$

Since the sum of the two tangents drawn from point $B$ to the circle with radius $r$ is equal to the perimeter of the triangle, we have

$$
a + r = s \quad \text{and thus} \quad r = s - a
$$

Substituting these values into (1),

$$
(a - \varrho) : \varrho = s : (s - a)
$$

which simplifies to the equation

$$
a^{2} - (s + \varrho) a + 2 \varrho s = 0
$$

This equation yields the same result for $a$ as the first solution.

Kántor Sándor (Debrecen, Ref. g. III. o. t.)