From a compact regular tetrahedron with unit edge length, we cut off a regular tetrahedron of edge length $\frac{1}{n}$ at each vertex, where $n>1$ is an integer. For what value of $n$ will the ratio of the surface area to the volume of the remaining solid be the smallest?
We solve the problem for an arbitrary tetrahedron instead of a regular tetrahedron. At each vertex of the tetrahedron, we cut off a small tetrahedron such that any edge emanating from the cut vertex is $1/n$ of the original edge. Consequently, the small tetrahedra can be transformed into the original tetrahedron by a central similarity of ratio $n$. Let the surface area of the tetrahedron be $A$, and its volume be $V$. The same data for the remaining solid are denoted by $A_{n}$ and $V_{n}$. Let the areas of the tetrahedron's faces be $T_{1}, T_{2}, T_{3}, T_{4}$. When cutting, from the face of area $T_{i}$, we cut off an area of $3 \cdot \frac{T_{i}}{n^{2}}$, and each cut, when the vertex opposite to $T_{i}$ is cut off, creates a face of area $\frac{T_{i}}{n^{2}}$. Therefore,

\[
A_{n}=T_{1}+T_{2}+T_{3}+T_{4}-3 \frac{T_{1}+T_{2}+T_{3}+T_{4}}{n^{2}}+\frac{T_{1}+T_{2}+T_{3}+T_{4}}{n^{2}}
\]

which simplifies to

\[
A_{n}=A\left(1-\frac{2}{n^{2}}\right)
\]

Similarly, we find that $V_{n}=V\left(1-\frac{4}{n^{3}}\right)$, and thus

\[
\frac{A_{n}}{V_{n}}=\frac{A}{V} \cdot \frac{1-\frac{2}{n^{2}}}{1-\frac{4}{n^{3}}}=\frac{A}{V} \cdot \frac{n^{3}-2 n}{n^{3}-4}
\]

Since $\frac{A}{V}$ is a positive constant, the minimum of $\frac{A_{n}}{V_{n}}$ is at the same place as the minimum of $\frac{n^{3}-2 n}{n^{3}-4}$. But $\frac{n^{3}-2 n}{n^{3}-4}=\frac{n^{3}-4+4-2 n}{n^{3}-4}=$ $1-\frac{2 n-4}{n^{3}-4}$, which is the smallest when $\frac{2 n-4}{n^{3}-4}$ is the largest, or when $\frac{n^{3}-4}{2 n-4}$ is the smallest. The latter is minimal at the same place where $f(n)=\frac{n^{3}-4}{n-2}$. The function $f(n)$ can be written as:

\[
f(n)=\frac{n^{3}-2^{3}-4}{n-2}=n^{2}+2 n+4+\frac{4}{n-2}
\]

We show that $f(n)$ is strictly increasing for $n \geq 3$. This can be shown as follows:

\[
\begin{gathered}
f(n+1)-f(n)=(n+1)^{2}+2(n+1)+4+\frac{4}{(n+1)-2}-n^{2}-2 n-4-\frac{4}{n-2}= \\
=2 n+3-\frac{4}{(n-1)(n-2)} \geq 2 \cdot 3+3-\frac{4}{(3-1)(3-2)}=9-2=7
\end{gathered}
\]

From this, it follows that for $n \geq 3$, $\frac{A_{n}}{V_{n}}$ is the smallest when $n=3$. From the above form of $\frac{A_{n}}{V_{n}}$, we have $\frac{A_{2}}{V_{2}}=\frac{A}{V}, \frac{A_{3}}{V_{3}}=\frac{A}{V} \cdot \frac{21}{23}$, so for all integers $n>1$, the minimum of $\frac{A_{n}}{V_{n}}$ is at $n=3$.

László Szobonya (Fazekas M. Főv. Gyak. Gimn., IV. o.t.)

Note. Several of our solvers examined the function $f(n)$ or its reciprocal (if $n$ is a real variable greater than 1) using its derivative.