Given a right-angled triangle with the leg $a$ and the radius $\varrho_{a}$ of the circle inscribed in it, let's calculate the perimeter of the triangle.





















The given information includes the leg $a$ and the radius $\varrho_{a}$ of the inscribed circle. To find the perimeter of the right-angled triangle, we need to determine the lengths of the other leg $b$ and the hypotenuse $c$.

The formula for the radius of the inscribed circle in a right-angled triangle is:
\[
\varrho_{a} = \frac{a + b - c}{2}
\]

The perimeter $P$ of the triangle is:
\[
P = a + b + c
\]

We also know from the Pythagorean theorem that:
\[
c = \sqrt{a^2 + b^2}
\]

To find $b$, we can use the relationship between the radius and the sides of the triangle. Rearranging the formula for $\varrho_{a}$, we get:
\[
2\varrho_{a} = a + b - c
\]
\[
b = 2\varrho_{a} + c - a
\]

Substituting $c = \sqrt{a^2 + b^2}$ into the equation, we get:
\[
b = 2\varrho_{a} + \sqrt{a^2 + b^2} - a
\]

This equation is a bit tricky to solve directly for $b$, so we can use the fact that the perimeter can also be expressed in terms of the area and the semiperimeter. The area $A$ of the triangle is:
\[
A = \frac{1}{2}ab
\]

The semiperimeter $s$ is:
\[
s = \frac{a + b + c}{2}
\]

The radius of the inscribed circle is also given by:
\[
\varrho_{a} = \frac{A}{s}
\]

Substituting the expressions for $A$ and $s$, we get:
\[
\varrho_{a} = \frac{\frac{1}{2}ab}{\frac{a + b + c}{2}} = \frac{ab}{a + b + c}
\]

Rearranging this equation, we get:
\[
\varrho_{a}(a + b + c) = ab
\]

Since we are looking for the perimeter $P = a + b + c$, we can solve for $P$:
\[
P = \frac{ab}{\varrho_{a}}
\]

However, we need to express $b$ in terms of $a$ and $\varrho_{a}$. Using the relationship:
\[
b = 2\varrho_{a} + \sqrt{a^2 + b^2} - a
\]

We can approximate or solve this equation numerically if needed. For a more straightforward approach, we can use the known relationship:
\[
P = a + b + c = a + b + \sqrt{a^2 + b^2}
\]

Given the complexity, we can use the simplified form:
\[
P = a + b + \sqrt{a^2 + b^2}
\]

Finally, substituting the value of $b$ from the earlier equation, we can find the perimeter. If we have specific values for $a$ and $\varrho_{a}$, we can plug them in to get the numerical result.
I. Solution: The labeling is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_189485f71aef6386c31fg-1.jpg?height=556&width=640&top_left_y=221&top_left_x=729)

It is known that two tangents drawn from a given point to a given circle are equal, so

$$
A D=A E \quad \text { and } \quad B E=B F
$$

which means

$$
b+\varrho_{a}=c+a-\varrho_{a}
$$

from which we get

$$
c-b=2 \varrho_{a}-a
$$

On the other hand, by the Pythagorean theorem,

$$
c^{2}-b^{2}=a^{2}
$$

Dividing (2) by (1), we obtain

$$
c+b=\frac{a^{2}}{2 \varrho_{a}-a}
$$

and thus the desired perimeter is

$$
2 s=a+b+c=a+\frac{a^{2}}{2 \varrho_{a}-a}=\frac{2 \varrho_{a} a}{2 \varrho_{a}-a}
$$

Lajos Fekete (Debrecen, Reformed Secondary School II.)

II. Solution: Construct the incircle of the triangle outside the circle inscribed to side $a$. The labeling is shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_189485f71aef6386c31fg-1.jpg?height=618&width=843&top_left_y=1816&top_left_x=627)

The centers of the two circles $\left(O_{a}\right.$ and $\left.O\right)$ lie on the angle bisector starting from vertex $A$.

It is known that (denoting the semiperimeter of the triangle by $s$)

$$
A D=s \quad \text { and } \quad A E=s-a
$$

and thus

$$
\begin{gathered}
s-(s-a)=a=D E= \\
=D C+C E=\varrho_{a}+\varrho
\end{gathered}
$$

from which we get

$$
\varrho=a-\varrho_{a}
$$

Since $A O_{a} D_{\triangle} \sim A O E_{\triangle}$, the ratio of corresponding sides is equal, that is,

$$
\varrho_{a}: \varrho=\varrho_{a}:\left(a-\varrho_{a}\right)=s:(s-a)
$$

from which we get

$$
s=\frac{\varrho_{a} a}{2 \varrho_{a}-a}, \quad \text { and thus } \quad 2 s=\frac{2 \varrho_{a} a}{2 \varrho_{a}-a} .
$$

János Pakuts (Győr, Bencés Secondary School I.)