Subfield: Geometry


Question: Version 1. Let $n$ be a fixed positive integer, and let $\mathrm{S}$ be the set of points $(x, y)$ on the Cartesian plane such that both coordinates $x$ and $y$ are nonnegative integers smaller than $2 n$ (thus $|\mathrm{S}|=4 n^{2}$ ). Assume that $\mathcal{F}$ is a set consisting of $n^{2}$ quadrilaterals such that all their vertices lie in $S$, and each point in $S$ is a vertex of exactly one of the quadrilaterals in $\mathcal{F}$.

Determine the largest possible sum of areas of all $n^{2}$ quadrilaterals in $\mathcal{F}$.

Solution:
the area of a polygon $P$ will be denoted by $[P]$.

We say that a polygon is legal if all its vertices belong to S. Let $O=\left(n-\frac{1}{2}, n-\frac{1}{2}\right)$ be the centre of $\mathrm{S}$. We say that a legal square is central if its centre is situated at $O$. Finally, say that a set $\mathcal{F}$ of polygons is acceptable if it satisfies the problem requirements, i.e. if all polygons in $\mathcal{F}$ are legal, and each point in $\mathrm{S}$ is a vertex of exactly one polygon in $\mathcal{F}$. For an acceptable set $\mathcal{F}$, we denote by $\Sigma(\mathcal{F})$ the sum of areas of polygons in $\mathcal{F}$.


Each point in $S$ is a vertex of a unique central square. Thus the set $\mathcal{G}$ of central squares is acceptable. We will show that

$$
\Sigma(\mathcal{F}) \leqslant \Sigma(\mathcal{G})=\Sigma(n)
\tag{1}
$$

thus establishing the answer.

We will use the following key lemma.

Lemma 1. Let $P=A_{1} A_{2} \ldots A_{m}$ be a polygon, and let $O$ be an arbitrary point in the plane. Then

$$
[P] \leqslant \frac{1}{2} \sum_{i=1}^{m} O A_{i}^{2}
\tag{2}
$$

moreover, if $P$ is a square centred at $O$, then the inequality (2) turns into an equality.

Proof. Put $A_{n+1}=A_{1}$. For each $i=1,2, \ldots, m$, we have

$$
\left[O A_{i} A_{i+1}\right] \leqslant \frac{O A_{i} \cdot O A_{i+1}}{2} \leqslant \frac{O A_{i}^{2}+O A_{i+1}^{2}}{4}
$$

Therefore,

$$
[P] \leqslant \sum_{i=1}^{m}\left[O A_{i} A_{i+1}\right] \leqslant \frac{1}{4} \sum_{i=1}^{m}\left(O A_{i}^{2}+O A_{i+1}^{2}\right)=\frac{1}{2} \sum_{i=1}^{m} O A_{i}^{2}
$$

which proves (2). Finally, all the above inequalities turn into equalities when $P$ is a square centred at $O$.

Back to the problem, consider an arbitrary acceptable set $\mathcal{F}$. Applying Lemma 1 to each element in $\mathcal{F}$ and to each element in $\mathcal{G}$ (achieving equality in the latter case), we obtain

$$
\Sigma(\mathcal{F}) \leqslant \frac{1}{2} \sum_{A \in S} O A^{2}=\Sigma(\mathcal{G})
$$

which establishes the left inequality in (1).



It remains to compute $\Sigma(\mathcal{G})$. We have

$$
\begin{aligned}
\Sigma(\mathcal{G})=\frac{1}{2} & \sum_{A \in S} O A^{2}=\frac{1}{2} \sum_{i=0}^{2 n-1} \sum_{j=0}^{2 n-1}\left(\left(n-\frac{1}{2}-i\right)^{2}+\left(n-\frac{1}{2}-j\right)^{2}\right) \\
& =\frac{1}{8} \cdot 4 \cdot 2 n \sum_{i=0}^{n-1}(2 n-2 i-1)^{2}=n \sum_{j=0}^{n-1}(2 j+1)^{2}=n\left(\sum_{j=1}^{2 n} j^{2}-\sum_{j=1}^{n}(2 j)^{2}\right) \\
& =n\left(\frac{2 n(2 n+1)(4 n+1)}{6}-4 \cdot \frac{n(n+1)(2 n+1)}{6}\right)=\frac{n^{2}(2 n+1)(2 n-1)}{3}=\Sigma(n) .
\end{aligned}
$$
the area of a polygon $P$ will be denoted by $[P]$.

We say that a polygon is legal if all its vertices belong to S. Let $O=\left(n-\frac{1}{2}, n-\frac{1}{2}\right)$ be the centre of $\mathrm{S}$. We say that a legal square is central if its centre is situated at $O$. Finally, say that a set $\mathcal{F}$ of polygons is acceptable if it satisfies the problem requirements, i.e. if all polygons in $\mathcal{F}$ are legal, and each point in $\mathrm{S}$ is a vertex of exactly one polygon in $\mathcal{F}$. For an acceptable set $\mathcal{F}$, we denote by $\Sigma(\mathcal{F})$ the sum of areas of polygons in $\mathcal{F}$.

Let $\mathcal{F}$ be an accessible set of quadrilaterals. For every quadrilateral $A B C D$ in $\mathcal{F}$ write

$$
[A B C D]=\frac{A C \cdot B D}{2} \sin \phi \leqslant \frac{A C^{2}+B D^{2}}{4}
\tag{3}
$$

where $\phi$ is the angle between $A C$ and $B D$. Applying this estimate to all members in $\mathcal{F}$ we obtain

$$
\Sigma(\mathcal{F}) \leqslant \frac{1}{4} \sum_{i=1}^{2 n^{2}} A_{i} B_{i}^{2}
$$

where $A_{1}, A_{2}, \ldots, A_{2 n^{2}}, B_{1}, B_{2}, \ldots, B_{2 n^{2}}$ is some permutation of S. For brevity, denote

$$
f\left(\left(A_{i}\right),\left(B_{i}\right)\right):=\sum_{i=1}^{2 n^{2}} A_{i} B_{i}^{2}
$$

The rest of the solution is based on the following lemma.

Lemma 2. The maximal value of $f\left(\left(A_{i}\right),\left(B_{i}\right)\right)$ over all permutations of $S$ equals $\frac{4}{3} n^{2}\left(4 n^{2}-1\right)$ and is achieved when $A_{i}$ is symmetric to $B_{i}$ with respect to $O$, for every $i=1,2, \ldots, 2 n^{2}$.

Proof. Let $A_{i}=\left(p_{i}, q_{i}\right)$ and $B_{i}=\left(r_{i}, s_{i}\right)$, for $i=1,2, \ldots, 2 n^{2}$. We have

$$
f\left(\left(A_{i}\right),\left(B_{i}\right)\right)=\sum_{i=1}^{2 n^{2}}\left(p_{i}-r_{i}\right)^{2}+\sum_{i=1}^{2 n^{2}}\left(q_{i}-s_{i}\right)^{2}
$$



it suffices to bound the first sum, the second is bounded similarly. This can be done, e.g., by means of the QM-AM inequality as follows:

$$
\begin{aligned}
& \sum_{i=1}^{2 n^{2}}\left(p_{i}-r_{i}\right)^{2}=\sum_{i=1}^{2 n^{2}}\left(2 p_{i}^{2}+2 r_{i}^{2}-\left(p_{i}+r_{i}\right)^{2}\right)=4 n \sum_{j=0}^{2 n-1} j^{2}-\sum_{i=1}^{2 n^{2}}\left(p_{i}+r_{i}\right)^{2} \\
& \leqslant 4 n \sum_{j=0}^{2 n-1} j^{2}-\frac{1}{2 n^{2}}\left(\sum_{i=1}^{2 n^{2}}\left(p_{i}+r_{i}\right)\right)^{2}=4 n \sum_{j=0}^{2 n-1} j^{2}-\frac{1}{2 n^{2}}\left(2 n \cdot \sum_{j=0}^{2 n-1} j\right)^{2} \\
& =4 n \cdot \frac{2 n(2 n-1)(4 n-1)}{6}-2 n^{2}(2 n-1)^{2}=\frac{2 n^{2}(2 n-1)(2 n+1)}{3} .
\end{aligned}
$$

All the estimates are sharp if $p_{i}+r_{i}=2 n-1$ for all $i$. Thus,

$$
f\left(\left(A_{i}\right),\left(B_{i}\right)\right) \leqslant \frac{4 n^{2}\left(4 n^{2}-1\right)}{3}
$$

and the estimate is sharp when $p_{i}+r_{i}=q_{i}+s_{i}=2 n-1$ for all $i$, i.e. when $A_{i}$ and $B_{i}$ are symmetric with respect to $O$.

Lemma 2 yields

$$
\Sigma(\mathcal{F}) \leqslant \frac{1}{4} \cdot \frac{4 n^{2}\left(4 n^{2}-1\right)}{3}=\frac{n^{2}(2 n-1)(2 n+1)}{3}
$$

Finally, all estimates are achieved simultaneously on the set $\mathcal{G}$ of central squares.

Final answer: $\Sigma(n):=\frac{1}{3} n^{2}(2 n+1)(2 n-1)$