$A B C$ is an isosceles triangle; the angle at vertex $A$ is $120^{\circ}$; $A B=A C=a$. Raise two perpendiculars to the plane of the triangle at points $B$ and $C$ and take two points, $P$ and $Q$, on these such that the triangle $P A Q$ is right-angled at $A$ and that the volume of the pyramid $A B C P Q$ is a given value $\left(m^{3}\right)$.
Let $B P=x, C Q=y, B C=c$. According to the problem:

$$
\overline{P Q}^{2}=\overline{P A}^{2}+\overline{A Q}^{2}
$$

But

$$
\begin{gathered}
\overline{P Q}^{2}=c^{2}+(x-y)^{2}=3 a^{2}+x^{2}-2 x y+y^{2} \\
\overline{P A}^{2}=a^{2}+x^{2} \\
\overline{A Q}^{2}=a^{2}+y^{2}
\end{gathered}
$$

$$
a^{2}-2 x y=0
$$

from which

$$
x y=\frac{a^{2}}{2}
$$

According to the second condition, the volume of the pyramid is $m^{3}$; the base of the pyramid is the trapezoid $P Q C B$, whose area is:

$$
t=\frac{(x+y) c}{2}=\frac{a(x+y) \sqrt{3}}{2}
$$

and its height is the height of the isosceles triangle $A B C$:

$$
h=a \sin 30^{\circ}=\frac{a}{2}
$$

and thus the volume of the pyramid is:

$$
m^{3}=\frac{(x+y) a^{2} \sqrt{3}}{12}
$$

from which

$$
(x+y)=\frac{4 m^{3} \sqrt{3}}{a^{2}}
$$

According to (5) and (6), $x$ and $y$ are the roots of the following equation:

$$
z^{2}-\frac{4 m^{3} \sqrt{3}}{a^{2}} z+\frac{a^{2}}{2}=0
$$

Solving the equation, we get:

$$
x=\frac{4 m^{3} \sqrt{3}+\sqrt{48 m^{6}-2 a^{6}}}{2 a^{2}}, y=\frac{4 m^{3} \sqrt{3}-z \sqrt{48 m^{6}-2 a^{6}}}{2 a^{2}}
$$

The roots are real if:

$$
48 m^{6} \geqq 2 a^{6}
$$

or

$$
\left(4 \sqrt{3} m^{3}+a^{3} \sqrt{2}\right)\left(4 \sqrt{3} m^{3}-a^{3} \sqrt{2}\right) \geqq 0
$$

which requires that:

$$
m^{3} \geqq \frac{a^{3} \sqrt{2}}{4 \sqrt{3}}
$$

or

$$
m^{3} \geqq \frac{a^{3} \sqrt{6}}{12}
$$

If

$$
m^{3}=a^{3} \frac{\sqrt{6}}{12}
$$

then:

$$
x=y=\frac{4 m^{3} \sqrt{3}}{2 a^{2}}=\frac{2 \sqrt{3} a^{3} \sqrt{6}}{12 a^{2}}=\frac{a \sqrt{18}}{6}
$$

or

$$
x=y=\frac{a}{2} \sqrt{2}
$$

(Frigyes Riesz, Gymn. VIII. year, Györ.)

The problem was also solved by: Bernát Friedmann, Zsigmond Goldstein, Béla Grünhut, Ervin Hofbauer, Nándor Kántor, Mór Klein.