A rectangle's perimeter is $k$, its area is $t$, what are the lengths of the sides? What is the solvability condition?
I. Solution: Let the two sides of the rectangle be $a$ and $b$, then

$$
a+b=\frac{k}{2}, \quad a b=t
$$

Thus, $a$ and $b$ are the two roots (in any order) of the equation

$$
2 x^{2}-k x+2 t=0
$$

which are:

$$
a=x_{1}=\frac{k+\sqrt{k^{2}-16 t}}{4}, \quad b=x_{2}=\frac{k}{2}-x_{1}=\frac{k-\sqrt{k^{2}-16 t}}{4} .
$$

The solvability condition is:

$$
k^{2} \geq 16 t, \quad \text { or } \quad \frac{k}{4} \geq \sqrt{t}
$$

If the equality holds, then $a=b=\frac{k}{4}$, meaning the rectangle is a square.

Tóth Zsuzsanna (Makó, József A. g. II. o. t.)

II. Solution: The problem can be rephrased as a geometric construction problem as follows: Given a square with area $t$ and a segment of length $k$, construct the sides $x, y$ of a rectangle whose area is $t$ and perimeter is $k$.

Thus, we need to construct segments that satisfy the equations (1) from the previous solution, i.e., divide the segment $\frac{k}{2}$ into two parts such that the geometric mean of these parts is the side of the given square.

For this, we use the well-known theorem that in a right triangle, the altitude to the hypotenuse is the geometric mean of the two segments of the hypotenuse. Therefore, we draw a Thales circle with diameter $\frac{k}{2}$ and intersect it with a line parallel to $\frac{k}{2}$ at a distance of $\sqrt{t}$. The perpendicular projection of any intersection point onto the segment $\frac{k}{2}$ divides it into the desired parts (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_2dae224b71a20cd09485g-1.jpg?height=368&width=612&top_left_y=1388&top_left_x=751)

The solvability condition is that the side of the square, $\sqrt{t}$, should not be larger than the radius of the Thales circle, or

$$
\sqrt{t} \leq \frac{k}{4}
$$

which is the same condition obtained in the first solution.

Dobos Gizella (Szeged, Ságvári g. II. o. t.)