Given the task, if the adjacent sides of the rectangle are $a$ and $b$,

\[
\begin{gathered}
a+b=\frac{k}{2} \\
\frac{a b}{a^{2}+b^{2}}=\mu
\end{gathered}
\]

Squaring the first equation,

\[
a^{2}+2 a b+b^{2}=\frac{k^{2}}{4}
\]

Substituting the value of \(a^{2}+b^{2}\) from (2) into (3), we get:

\[
\frac{a b}{\mu}+2 a b=\frac{k^{2}}{4}
\]

From which,

\[
a b=\frac{\mu k^{2}}{4(1+2 \mu)}
\]

Considering (1) and (4), \(a\) and \(b\) are the roots of the following quadratic equation:

\[
z^{2}-\frac{k}{2} z+\frac{\mu k^{2}}{4(1+2 \mu)}=0
\]

Therefore,

\[
\begin{aligned}
& a=\frac{k}{4}\left(1 \pm \sqrt{\frac{1-2 \mu}{1+2 \mu}}\right) \\
& b=\frac{k}{4}\left(1 \mp \sqrt{\frac{1-2 \mu}{1+2 \mu}}\right)
\end{aligned}
\]

Substituting the given numerical values,

\[
a_{1}=b_{2}=3, \quad a_{2}=b_{1}=1
\]

(Dezső Schönfeld, Győr.)

Number of solutions: 52.
Let's calculate the sides of the rectangle, given the perimeter $k$, and the ratio of its area to the area of the square that can be inscribed over its diagonal $(\mu)$. For example, $k=8 \mathrm{~cm}, \mu=\frac{3}{10}$.