, 2022

Solution:

Every point of $K$ is a vertex of a uniquely defined central square. Hence the set $Q$ of all central squares is admissible. We show $S(V) \leq S(Q) = S(k)$, from which the answer follows.

We use the following

Lemma 1. For every quadrilateral $V = A_{1}A_{2}A_{3}A_{4}$ and any point $O$ in the plane, we have
$$[V] \leq \frac{1}{2} \sum_{i=1}^{4} OA_{i}^{2}$$
where equality holds if and only if $V$ is a square with center $O$.

Proof: For $i = 1, \ldots, 4$ and $A_{5} = A_{1}$, we have
$$[OA_{i}A_{i+1}] \leq \frac{OA_{i} \cdot OA_{i+1}}{2} \leq \frac{OA_{i}^{2} + OA_{i+1}^{2}}{4}.$$
Hence
$$[V] \leq \sum_{i=1}^{4} [OA_{i}A_{i+1}] \leq \frac{1}{4} \sum_{i=1}^{4} (OA_{i}^{2} + OA_{i+1}^{2}) = \frac{1}{2} \sum_{i=1}^{4} OA_{i}^{2},$$
which proves (2). Indeed, equality holds throughout when $V$ is a square with center $O$.

Now we consider an arbitrary admissible set $V$. Applying Lemma 1 to each element of $V$ and each element of $Q$ yields
$$S(V) \leq \frac{1}{2} \sum_{A < K} OA^{2} = S(Q),$$
which proves the left-hand side of (1).

Now we compute
$$S(Q) = \frac{1}{2} \sum_{A \in K} OA^{2} = \frac{1}{2} \sum_{i=0}^{2k-1} \sum_{j=0}^{2k-1} \left(\left(k-\frac{1}{2}-i\right)^{2} + \left(k-\frac{1}{2}-j\right)^{2}\right)$$
$$= \frac{1}{8} \cdot 4 \cdot 2k \sum_{i=0}^{k-1} (2k-2i-1)^{2} = k \sum_{j=0}^{k-1} (2j+1)^{2} = k\left(\sum_{j=1}^{2k} j^{2} - \sum_{j=1}^{k} (2j)^{2}\right)$$
$$= k\left(\frac{2k(2k+1)(4k+1)}{6} - 4 \cdot \frac{k(k+1)(2k+1)}{6}\right) = \frac{k^{2}(2k+1)(2k-1)}{3} = S(k).$$

For every quadrilateral $ABCD$ in an admissible set $V$, we have
$$[ABCD] = \frac{AC \cdot BD}{2} \cdot \sin \varphi \leq \frac{AC^{2} + BD^{2}}{4}$$
with $\varphi = \measuredangle(AC, BD)$. Applying (3) to all elements of $V$, we obtain
$$S(V) \leq \frac{1}{4} \sum_{i=1}^{2k^{2}} A_{i}B_{i}^{2},$$
where $\left(A_{1}, A_{2}, \ldots, A_{2k^{2}}, B_{1}, B_{2}, \ldots, B_{2k^{2}}\right)$ is a permutation of $K$.

With the abbreviation $S := \sum_{i=1}^{2k^{2}} A_{i}B_{i}^{2}$, we formulate the following

Lemma 2. The greatest possible value of $S$ over all permutations of $K$ is $\frac{4}{3}k^{2}(4k^{2}-1)$ and is attained when $A_{i}$ and $B_{i}$ are symmetric with respect to $O$ for all $i=1,2,\ldots,2k^{2}$.

Proof: Let $A_{i} = (p_{i}, q_{i})$ and $B_{i} = (r_{i}, s_{i})$ for $i=1,2,\ldots,2k^{2}$. Then
$$S = \sum_{i=1}^{2k^{2}} (p_{i} - r_{i})^{2} + \sum_{i=1}^{2k^{2}} (q_{i} - s_{i})^{2}.$$
Using the QM-AM inequality, we estimate the first sum
$$\sum_{i=1}^{2k^{2}} (p_{i} - r_{i})^{2} = \sum_{i=1}^{2k^{2}} (2p_{i}^{2} + 2r_{i}^{2} - (p_{i} + r_{i})^{2}) = 4k \sum_{j=0}^{2k-1} j^{2} - \sum_{i=1}^{2k^{2}} (p_{i} + r_{i})^{2}$$
$$\leq 4k \sum_{j=0}^{2k-1} j^{2} - \frac{1}{2k^{2}} \left(2k \cdot \sum_{j=0}^{2k-1} j\right)^{2}$$
$$= 4k \cdot \frac{2k(2k-1)(4k-1)}{6} - 2k^{2}(2k-1)^{2} = \frac{2k^{2}(2k-1)(2k+1)}{3},$$
where equality holds if and only if $p_{i} + r_{i} = 2k-1$ for all $i$. Since the second sum can be estimated in a completely analogous manner, it follows that
$$S \leq \frac{4}{3}k^{2}(4k^{2}-1)$$
with equality when $p_{i} + r_{i} = q_{i} + s_{i} = 2k-1$ for all $i=1,\ldots,2k^{2}$, i.e., when $A_{i}$ and $B_{i}$ are symmetric with respect to $O$ for all $i=1,2,\ldots,2k^{2}$.

Using the result of Lemma 2, we obtain
$$S(V) \leq \frac{1}{4} \cdot \frac{4k^{2}(4k^{2}-1)}{3} = \frac{k^{2}(2k-1)(2k+1)}{3},$$
where the estimate is sharp for the set $Q$. ㅁ.