The height of a certain triangular prism is $m$, we extend its lateral edge by $x$ and connect the endpoint to the vertices of the base, thus obtaining a pyramid. Find $x$ such that the part of the prism outside the pyramid is in a constant ratio $k$ to the volume of the prism.
Let the base of the prism be denoted by $S$, then its volume is

$$
V=m S
$$

while the volume of the pyramid is

$$
V'=(m+x) \frac{S}{3}
$$

The common part of the pyramid and the prism is a frustum of a pyramid, whose volume is

$$
V''=(m+x) \frac{S}{3}-x \frac{S'}{3}
$$

where

$$
S': S=x^{2}:(m+x)^{2}
$$

thus

$$
S'=S \frac{x^{2}}{(m+x)^{2}}
$$

The volume of the part of the prism outside the pyramid is finally

$$
V'''=S m-(m+x) \frac{S}{3}+\frac{x^{3} S}{3(m+x)^{2}} S
$$

The desired $x$ can be obtained from the following equation

$$
\begin{gathered}
k=\frac{3 S(m+x)^{2} m-S(m+x)^{3}+S x^{3}}{3 S m(m+x)^{2}} \\
3(m+x)^{2} m k=3 m^{3}+6 m^{2} x+3 m x^{2}-m^{3}-3 m^{2} x-3 m x^{2}-x^{3}+x^{3} \\
2 m^{2}+3 m x=3 m^{2} k+6 m k x+3 k x^{2} \\
3 k x^{2}+(6 m k-3 m) x+3 m^{2} k-2 m^{2}=0 \\
x=\frac{6 m k-3 m \pm \sqrt{(6 m k-3 m)^{2}-12 k\left(3 m^{2} k-2 m^{2}\right)}}{6 k} \\
x=\frac{6 m k-3 m \pm \sqrt{D}}{6 k}
\end{gathered}
$$

where

$$
\begin{gathered}
D=36 k^{2}-36 k+9-36 k^{2}+24 k \\
D=-12 k+9
\end{gathered}
$$

For $x$ to be real, it is necessary that $9-12 k \geq 0$, i.e., that

$$
k \leq \frac{3}{4}
$$

(Bergstein Ignácz, 8th grade student, Nyíregyháza).

The problem was also solved by: Hónig Viktor, 8th grade student, Kaposvár; Seidner Mihály, 8th grade student, Losoncz; Stark Zsigmond, 8th grade student, Pécs. Jorga Gergely, 8th grade student, Arad.