We need to shorten the legs of a right triangle with equal segments so that

a) we get the legs of a right triangle with half the area?

b) we get the adjacent sides of a rectangle with the same perimeter?
I. Solution: a) Let the legs of the right triangle be denoted by $a, b$, the hypotenuse by $c$, and the sought distance by $x$.

According to the problem,

$$
\frac{(a-x)(b-x)}{2}=\frac{a b}{4}
$$

or, rearranged,

$$
2 x^{2}-2(a+b) x+a b=0
$$

from which

$$
\begin{aligned}
x_{1,2}= & \frac{2(a+b) \pm \sqrt{4\left(a^{2}+2 a b+b^{2}\right)-8 a b}}{4}= \\
& =\frac{a+b \pm \sqrt{a^{2}+b^{2}}}{2}=\frac{a+b \pm c}{2}
\end{aligned}
$$

Thus,

$$
x_{1}=\frac{a+b-c}{2}=s-c=\varrho
$$

where $s$ is the semiperimeter of the triangle and $\varrho$ is the radius of the inscribed circle.

The lengths of the shortened legs are:

$$
\begin{aligned}
& a-\varrho=a-\frac{a+b-c}{2}=\frac{a-b+c}{2}=s-b \\
& b-\varrho=b-\frac{a+b-c}{2}=\frac{b-a+c}{2}=s-a
\end{aligned}
$$

We arrive at the same result if we consider the root $x_{2}=\frac{a+b+c}{2}=s$ and take the absolute values of $a-s$ and $b-s$, i.e., $(s-a)$ and $(s-b)$.

b) Let the sought distance be denoted by $y$. According to the problem,

$$
a+b+c=2(a-y)+2(b-y)
$$

from which

$$
4 y=2 a+2 b-a-b-c=a+b-c
$$

and thus

$$
y=\frac{a+b-c}{4}=\frac{s-c}{2}=\frac{\varrho}{2}
$$

Tibor Behringer (Bp. III., Árpád g. III. o. t.)

II. Solution: The labeling is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_5036c080e8b12aab6eb0g-1.jpg?height=304&width=469&top_left_y=1732&top_left_x=839)

Let the area of the original triangle be denoted by $t$. In Exercise 251, we proved that

$$
c_{1} c_{2}=t, \quad \text { or } \quad \frac{c_{1} c_{2}}{2}=\frac{t}{2}
$$

But $c_{1}=a-\varrho, c_{2}=b-\varrho$, and it is clear that the legs of the right triangle must be shortened by $\varrho$ to obtain the legs of a right triangle with half the area.

b) The perimeter of the right triangle is

$$
2 c_{1}+2 c_{2}+2 \varrho
$$

and thus the semiperimeter of the rectangle, or the sum of two adjacent sides,

$$
c_{1}+c_{2}+\varrho=\left(c_{1}+\frac{\varrho}{2}\right)+\left(c_{2}+\frac{\varrho}{2}\right)=\left(a-\frac{\varrho}{2}\right)+\left(b-\frac{\varrho}{2}\right) .
$$

Therefore, the legs must be shortened by $\frac{\varrho}{2}$ to obtain the adjacent sides of a rectangle with the same perimeter.