The base edges of a certain right triangular prism are: $a=3 \mathrm{~cm}, \quad b=4 \mathrm{~cm}, \quad c=5 \mathrm{~cm}$. Let's cut through the prism with a plane so that the cross-section is an equilateral triangle. What are the side lengths of this triangle?
Among the equilateral triangles cut from the prism by planes parallel to each other, we select the one whose one vertex coincides with a vertex of the prism. The distances of the other two vertices from the corresponding vertices of the prism are denoted by $x$ and $y$, and the side of the equilateral triangle by $d$. Then:

$$
\begin{gathered}
x^{2}=d^{2}-a^{2} \\
y^{2}=d^{2}-b^{2} \\
(x-y)^{2}=d^{2}-c^{2}
\end{gathered}
$$

The last equation can also be written as:

$$
x^{2}-2 x y+y^{2}=d^{2}-c^{2}
$$

Substituting the values of $x$ and $y$ from (1) and (2) into (4), we get:

$$
d^{2}-a^{2}-2 \sqrt{\left(d^{2}-a^{2}\right)\left(d^{2}-b^{2}\right)}+d^{2}-b^{2}=d^{2}-c^{2}
$$

Given that $c^{2}=a^{2}+b^{2}$, we have:

$$
d^{2}=2 \sqrt{\left(d^{2}-a^{2}\right)\left(d^{2}-b^{2}\right)}
$$

Squaring both sides of this equation, we obtain:

$$
3 d^{4}-4 c^{2} d^{2}+4 a^{2} b^{2}=0
$$

Substituting the given values, we get:

$$
3 d^{4}-100 d^{2}+576=0
$$

The solutions for $d^{2}$ from this equation are 25.9283 and 7.4051.

Of these, only the first one is valid, as $d$ must be greater than 5. Taking the square root of 25.9283, we find that the side of the equilateral triangle is:

$$
d=5.09 \text{ cm}
$$

(Béla Grünhut, real gym. VIII.o. t., Pécs)

The problem was solved by: Mór Feuer, Bernát Friedmann, Károly Fröhlich, Zsigmond Goldstein, Nándor Kántor, Frigyes Riesz, Béla Schneider, István Szabó.