Problem 4. In a regular tetrahedron \(ABCD\), planes parallel to the faces of the tetrahedron are constructed such that each edge is divided into 6 congruent segments. These planes determine a set of 80 intersection points on the faces, edges, and inside the tetrahedron, denoted by \(V\).

Determine the maximum number of elements of a subset \(W\) of the set \(V \cup \{A, B, C, D\}\) which has the property:

any three distinct points in \(W\) are non-collinear, and the plane determined by them is not parallel to any of the faces of the tetrahedron \(ABCD\).
Solution. We can assume that the height of the tetrahedron $ABCD$ is equal to 6.

Let $k$ be the maximum number of elements in the set $W$ and $M_{1}, M_{2}, \ldots, M_{k}$ be the elements of $W$. For each $i \in \{1,2, \ldots, k\}$, we denote by $a_{i}, b_{i}, c_{i}$, and $d_{i}$ the distances from the point $M_{i}$ to the planes $(BCD), (ACD), (ABD)$, and $(ABC)$, respectively; then $a_{i}, b_{i}, c_{i}, d_{i} \in \{0,1,2, \ldots, 6\}$.

Since the sum of the distances from a point inside a regular tetrahedron to its faces is equal to the height of the tetrahedron, it follows that $a_{i}+b_{i}+c_{i}+d_{i}=6$.

Let $T=\left(a_{1}+a_{2}+\ldots+a_{k}\right)+\left(b_{1}+b_{2}+\ldots+b_{k}\right)+\left(c_{1}+c_{2}+\ldots+c_{k}\right)+\left(d_{1}+d_{2}+\ldots+d_{k}\right)$, then $T=6k$.

For $k$ even, $k=2s$, we have $a_{1}+a_{2}+\ldots+a_{k} \geq 2 \cdot 0+2 \cdot 1+\ldots+2 \cdot(s-1)=s^{2}-s$. We obtain $T=12s \geq 4s^{2}-4s$, thus $s \leq 4$ and $k \leq 8$.

For $k$ odd, $k=2s+1$, we have $a_{1}+a_{2}+\ldots+a_{k} \geq 2 \cdot 0+2 \cdot 1+\ldots+2 \cdot(s-1)+s=s^{2}$. We obtain $T=12s+6 \geq 4s^{2}$, thus $s \leq 3$ and $k \leq 7$.

It remains to construct an example of a set with 8 elements. Using the notation $\left(a_{i}, b_{i}, c_{i}, d_{i}\right)$ instead of the point $M_{i}$, an example is the set $W$ consisting of the elements $(0,1,2,3),(1,0,2,3),(0,1,3,2)$, $(1,0,3,2),(2,3,0,1),(2,3,1,0),(3,2,1,0)$, and $(3,2,0,1)$.