Let's construct a rectangle whose area is equal to that of a given right-angled triangle and whose semi-perimeter is equal to the hypotenuse of the latter triangle.
I. solution: Let the legs of the given right triangle be denoted by $a$ and $b$, its hypotenuse by $c$, and the sides of the rectangle to be constructed by $x$ and $y$.

According to the problem,

$$
\begin{aligned}
& x+y=c \\
& x y=\frac{a b}{2}
\end{aligned}
$$

Thus, $x$ and $y$ are the two roots of the following quadratic equation:

$$
x^{2}-c x+\frac{a b}{2}=0
$$

From this (using the Pythagorean theorem),

$$
x_{1,2}=\frac{c \pm \sqrt{c^{2}-2 a b}}{2}=\frac{c \pm \sqrt{a^{2}+b^{2}-2 a b}}{2}=\frac{c \pm(a-b)}{2}
$$

Therefore, the sides of the desired rectangle are

$$
x_{1}=y_{2}=\frac{c+a-b}{2}=s-b, \quad x_{2}=y_{1}=\frac{c-a+b}{2}=s-a
$$

which can be easily constructed. There is always one and only one solution.

Katalin Pásztor (Tatabánya, Rákosi Mátyás g. II. o. t.)

II. solution: According to the problem, a given segment $c$ must be divided into two parts such that the rectangle constructed with these segments as sides has a given area, specifically the area of the given right triangle.

![](https://cdn.mathpix.com/cropped/2024_05_02_de3349842c0c79570a08g-1.jpg?height=702&width=653&top_left_y=1302&top_left_x=714)

First, transform the area of the given triangle into a square, for example, by constructing the geometric mean between $a$ and $\frac{b}{2}$, which will be the side of the desired square: $p$ (see the figure). Then, construct a right triangle $A B C^{*}$ over $A B \stackrel{2}{=} c$ such that the altitude to the hypotenuse is the previously constructed $p$. The altitude to the hypotenuse of this triangle divides the hypotenuse into the desired segments $x$ and $y$, according to the well-known geometric mean theorem for right triangles.

Zsófia Jäger (Kiskunhalas, Szilády Áron g. II. o. t.)

III. solution: In problem 251, we proved that the points of tangency of the incircle of a right triangle divide the hypotenuse into segments whose product is equal to the area of the right triangle. Therefore, we only need to construct the perpendicular projection of the center $O$ of the incircle onto the hypotenuse $A B$ (see the figure).