For a fixed positive integer $k$, let $K$ be the set of all lattice points $(x, y)$ in the plane, both of whose coordinates $x$ and $y$ are nonnegative integers less than $2k$. Thus, $|K|=4k^2$.
A set $V$ consists of $k^2$ non-degenerate quadrilaterals with the following properties:
i) The vertices of all these quadrilaterals are elements of $K$.
ii) Each point in $K$ is a vertex of exactly one of the quadrilaterals in $V$.

Determine the maximum possible value that the sum of the areas of all $k^2$ quadrilaterals in $V$ can take.

Answer: The maximum possible value of the area sum is $S(k):=\frac{1}{3} k^2 (2k+1)(2k-1)$.
Notations: The area of a polygon $X$ is denoted by $[X]$. A quadrilateral is called legal if all its vertices belong to $K$. Furthermore, let $O=\left(k-\frac{1}{2}, k-\frac{1}{2}\right)$ be the center of $K$. A legal quadrilateral is called central if it is point-symmetric with $O$ as the center. Any set $V$ of quadrilaterals that satisfies the conditions of the problem is called acceptable. For any acceptable set $V$, the sum of the areas of its quadrilaterals is denoted by $S(V)$.
Every point in $K$ is a vertex of a uniquely defined central square. Therefore, the set $Q$ of all central squares is acceptable. 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
$[V] \leq \frac{1}{2} \sum_{i=1}^{4} O A_{i}^{2}$
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 $\left[O A_{i} A_{i+1}\right] \leq \frac{O A_{i} \cdot O A_{i+1}}{2} \leq \frac{O A_{i}^{2}+O A_{i+1}^{2}}{4}$. Therefore, $[V] \leq \sum_{i=1}^{4}\left[O A_{i} A_{i+1}\right] \leq \frac{1}{4} \sum_{i=1}^{4}\left(O A_{i}^{2}+O A_{i+1}^{2}\right)=\frac{1}{2} \sum_{i=1}^{4} O A_{i}^{2}$, which proves (2). Indeed, equality holds in each case if $V$ is a square with center $O$. Now consider any acceptable set $V$. Applying Lemma 1 to each element of $V$ and each element of $Q$ yields $S(V) \leq \frac{1}{2} \sum_{A} MATHEMATICAL COMPETITIONS

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