## Task 6 - 251236

For any natural number $n \geq 2$, let $2 n$ points $P_{1}, \ldots, P_{2 n}$ be arranged in space such that no four of these points lie on the same plane.

Let $T$ denote the set of all tetrahedra whose four vertices belong to the set $M=$ $\left\{P_{1}, \ldots, P_{2 n}\right\}$.

For every plane $\epsilon$ that does not contain any point of $M$, let $t_{\epsilon}$ be the number of all those tetrahedra from $T$ that have a quadrilateral as a common intersection with $\epsilon$.

Determine, for every natural number $n \geq 2$, the largest possible value that $t_{\epsilon}$ can take!
For each $2 n$ points in the given position and for every plane $\epsilon$ that does not contain any point of $M$, the following holds:

The plane $\epsilon$ divides the space into two half-spaces, and with a natural number $k \leq n$, one of these half-spaces contains a set $M_{1}$ of $k$ points from $M$, and the other contains the set $M_{2}$ of the remaining $2 n-k$ points.

Furthermore, for every tetrahedron from $T$: With a natural number $h \leq 2$, $h$ vertices of the tetrahedron lie in one of these half-spaces, and $4-h$ vertices lie in the other. 

If $h=0$, then $\epsilon$ does not intersect the tetrahedron; if $h=1$, then $\epsilon$ intersects the tetrahedron in a triangle; if $h=2$, then $\epsilon$ intersects the tetrahedron in a quadrilateral.

Thus, one obtains exactly a tetrahedron from $T$ that is intersected by $\epsilon$ in a quadrilateral if one takes as the vertex set the union of an arbitrarily chosen two-element subset $Z_{1}$ of $M_{1}$ and an independently chosen two-element subset $Z_{2}$ of $M_{2}$. Two selection possibilities for $Z_{1}$ and $Z_{2}$ result in the same tetrahedron if and only if they coincide both in $Z_{1}$ and in $Z_{2}$.

The number of all such selection possibilities is thus the number $t_{\epsilon}$ explained in the problem. It is obtained by multiplying the number of all two-element subsets of $M_{1}$ with the number of all two-element subsets of $M_{2}$, i.e., it holds that

$$
\begin{aligned}
t_{\epsilon} & =\binom{k}{2} \cdot\binom{2 n-k}{2}=\frac{k(k-1)}{2} \cdot \frac{(2 n-k)(2 n-k-1)}{2} \\
& =\frac{1}{4}(n-(n-k))(n-1-(n-k))(n+(n-k))(n-1+(n-k)) \\
& =\frac{1}{4}\left(n^{2}-(n-k)^{2}\right)\left((n-1)^{2}-(n-k)^{2}\right)
\end{aligned}
$$

For all natural numbers $k \leq n$, it holds that $0 \leq n-k \leq n$, thus

$$
\begin{array}{r}
0 \leq n^{2}-(n-k)^{2} \leq n^{2} \\
(n-1)^{2}-(n-k)^{2} \leq(n-1)^{2} \quad \text { and } \quad(n-1)^{2} \geq 0
\end{array}
$$

From this, it follows that

$$
t_{\epsilon} \leq \frac{1}{4} n^{2} \cdot(n-1)^{2}
$$

For $k=n$, the equality sign results from (1) in (2). Thus, $\frac{1}{4} n^{2}(n-1)^{2}$ is the desired maximum value of $t_{\epsilon}$.

Taken from $[5]$

### 9.27.4 IV. Round 1985, Class 12