A square-based, straight frustum has a height of $m$; the edge of the base is $a$, and the edge of the top is $b$; at what distance from the base should this frustum be cut by a plane parallel to the base so that the volumes of the resulting parts are equal?
We cut the truncated pyramid with a plane that is perpendicular to the base and bisects two parallel edges of the base and the top face. The parallel sides of this section are: \(AB = a, DC = b\). The plane parallel to the base at a distance \(x\) from the base of the pyramid intersects the first cutting plane at the line \(GF = x\). The plane parallel to the base at a distance \(x\) from the base of the pyramid intersects the first cutting plane at the line \(GF = y\). According to the problem:

\[
\frac{x}{3}\left(a^{2}+a y+y^{2}\right)=\frac{m-x}{3}\left(y^{2}+y b+b^{2}\right)
\]

If the perpendicular dropped from \(D\) to the base intersects \(GF\) at \(K\), then based on the similarity of triangles \(AHG\) and \(GKD\):

\[
x: m-x = AH: GK
\]

or

\[
x: m-x = a-y: y-b
\]

Considering (2), from (1) we get:

\[
(a-y)\left(a^{2}+a y+y^{2}\right)=(y-b)\left(y^{2}+y b+b^{2}\right)
\]

which simplifies to

\[
a^{3}-y^{3}=y^{3}-b^{3}
\]

from which

\[
y=\sqrt[3]{\frac{a^{3}+b^{3}}{2}}
\]

However, (2) can also be written as:

\[
x: m = a-y: b
\]

from which

\[
x=\frac{m}{a-b}(a-y)
\]

Substituting the value of \(y\) from (3) into this:

\[
x=\frac{m}{a-b}\left(a-\sqrt[3]{\frac{a^{3}+b^{3}}{2}}\right)
\]

(Bernát Friedmann.)

The problem was also solved by: Dénes A., Goldstein Zs., Goldziher K., Hofbauer E., Kántor N., Krátky Gy., Probst E., Spitzer Ö., Szabó I.