Determine the sides of two right-angled parallelograms if the sum of their areas is $S$, the sum of their bases is $p$; furthermore, we know that the areas of the individual parallelograms will be $s$ and $s^{\prime}$ if their bases are swapped.
Let the base of the first parallelogram be $x$, its height $y$; the base of the second be $z$ and its height $u$. Then the following equations hold:

$$
\begin{gathered}
x y+z u=S \\
x+z=p \\
z y=s \\
x u=s^{\prime}
\end{gathered}
$$

From 3) and 4), it follows that

$$
\begin{aligned}
& y=\frac{s}{z} \\
& u=\frac{s^{\prime}}{x}
\end{aligned}
$$

Substituting these values into 1) gives:

$$
\frac{x s}{z}+\frac{z s^{\prime}}{x}=S
$$

Eliminating $z$ from 2) and 7), we obtain the following equation:

$$
\frac{x s}{p-x}+\frac{(p-x) s^{\prime}}{x}=S
$$

And if we rearrange this, we get:

$$
\left(s+s^{\prime}+S\right) x^{2}-\left(2 p s^{\prime}+p S\right) x+p^{2} s^{\prime}=0
$$

From this,

$$
\begin{gathered}
x=\frac{p\left(2 s^{\prime}+S\right) \pm \sqrt{p^{2}\left(2 s^{\prime}+S\right)^{2}-4 p^{2} s^{\prime}\left(s+s^{\prime}+S\right)}}{2\left(s+s^{\prime}+S\right)} \\
x=\frac{p\left(2 s^{\prime}+S\right) \pm p \sqrt{4 s^{\prime 2}+4 s^{\prime} S+S^{2}-4 s^{\prime} s-4 s^{\prime 2}-4 s^{\prime} S}}{2\left(s+s^{\prime}+S\right)} \\
\left.x=\frac{p\left(2 s^{\prime}+S \pm \sqrt{\left.S^{2}-4 s s^{\prime}\right)}\right.}{2\left(s+s^{\prime}+S\right.}\right) \\
z=p-x=\frac{p\left(2 s+S \mp \sqrt{S^{2}-4 s s^{\prime}}\right)}{2\left(s+s^{\prime}+S\right.} \\
y=\frac{s}{z}=\frac{2 s+S \pm \sqrt{S^{2}-4 s s^{\prime}}}{2 p} \\
u=\frac{s^{\prime}}{x}=\frac{2 s^{\prime}+S \mp \sqrt{S^{2}-4 s s^{\prime}}}{2 p}
\end{gathered}
$$

(Domokos Berzenczey, 7th grade student, Déva).

The problem was also solved by: Gergely Jorga, Gilád; Sándor Suták, Nyíregyháza; Aladár Visnya and Lipót Weisz, Pécs.