Subfield: Algebra


Question: Given a positive integer $n$, determine the largest real number $\mu$ satisfying the following condition: for every $4 n$-point configuration $C$ in an open unit square $U$, there exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\mu$.

Solution:
The required maximum is $\frac{1}{2 n+2}$. To show that the condition in the statement is not met if $\mu>\frac{1}{2 n+2}$, let $U=(0,1) \times(0,1)$, choose a small enough positive $\epsilon$, and consider the configuration $C$ consisting of the $n$ four-element clusters of points $\left(\frac{i}{n+1} \pm \epsilon\right) \times\left(\frac{1}{2} \pm \epsilon\right), i=1, \ldots, n$, the four possible sign combinations being considered for each $i$. Clearly, every open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, has area at $\operatorname{most}\left(\frac{1}{n+1}+\epsilon\right) \cdot\left(\frac{1}{2}+\epsilon\right)<\mu$ if $\epsilon$ is small enough.



We now show that, given a finite configuration $C$ of points in an open unit square $U$, there always exists an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$, and has an area greater than or equal to $\mu_{0}=\frac{2}{|C|+4}$.



To prove this, usage will be made of the following two lemmas whose proofs are left at the end of the solution.



Lemma 1. Let $k$ be a positive integer, and let $\lambda<\frac{1}{\lfloor k / 2\rfloor+1}$ be a positive real number. If $t_{1}, \ldots, t_{k}$ are pairwise distinct points in the open unit interval $(0,1)$, then some $t_{i}$ is isolated from the other $t_{j}$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\lambda$.



Lemma 2. Given an integer $k \geq 2$ and positive integers $m_{1}, \ldots, m_{k}$,



$$

\left\lfloor\frac{m_{1}}{2}\right\rfloor+\sum_{i=1}^{k}\left\lfloor\frac{m_{i}}{2}\right\rfloor+\left\lfloor\frac{m_{k}}{2}\right\rfloor \leq \sum_{i=1}^{k} m_{i}-k+2

$$



Back to the problem, let $U=(0,1) \times(0,1)$, project $C$ orthogonally on the $x$-axis to obtain the points $x_{1}<\cdots<x_{k}$ in the open unit interval $(0,1)$, let $\ell_{i}$ be the vertical through $x_{i}$, and let $m_{i}=\left|C \cap \ell_{i}\right|, i=1, \ldots, k$.



Setting $x_{0}=0$ and $x_{k+1}=1$, assume that $x_{i+1}-x_{i-1}>\left(\left\lfloor m_{i} / 2\right\rfloor+1\right) \mu_{0}$ for some index $i$, and apply Lemma 1 to isolate one of the points in $C \cap \ell_{i}$ from the other ones by an open subinterval $x_{i} \times J$ of $x_{i} \times(0,1)$ whose length is greater than or equal to $\mu_{0} /\left(x_{i+1}-x_{i-1}\right)$. Consequently, $\left(x_{i-1}, x_{i+1}\right) \times J$ is an open rectangle in $U$, whose sides are parallel to those of $U$, which contains exactly one point of $C$ and has an area greater than or equal to $\mu_{0}$.



Next, we rule out the case $x_{i+1}-x_{i-1} \leq\left(\left\lfloor m_{i} / 2\right\rfloor+1\right) \mu_{0}$ for all indices $i$. If this were the case, notice that necessarily $k>1$; also, $x_{1}-x_{0}<x_{2}-x_{0} \leq\left(\left\lfloor m_{1} / 2\right\rfloor+1\right) \mu_{0}$ and $x_{k+1}-x_{k}<$ $x_{k+1}-x_{k-1} \leq\left(\left\lfloor m_{k} / 2\right\rfloor+1\right) \mu_{0}$. With reference to Lemma 2 , write



$$

\begin{aligned}

2=2\left(x_{k+1}-x_{0}\right) & =\left(x_{1}-x_{0}\right)+\sum_{i=1}^{k}\left(x_{i+1}-x_{i-1}\right)+\left(x_{k+1}-x_{k}\right) \\

& <\left(\left(\left\lfloor\frac{m_{1}}{2}\right\rfloor+1\right)+\sum_{i=1}^{k}\left(\left\lfloor\frac{m_{i}}{2}\right\rfloor+1\right)+\left(\left\lfloor\frac{m_{k}}{2}\right\rfloor+1\right)\right) \cdot \mu_{0} \\

& \leq\left(\sum_{i=1}^{k} m_{i}+4\right) \mu_{0}=(|C|+4) \mu_{0}=2,

\end{aligned}

$$



and thereby reach a contradiction.







Finally, we prove the two lemmas.



Proof of Lemma 1. Suppose, if possible, that no $t_{i}$ is isolated from the other $t_{j}$ by an open subinterval of $(0,1)$ whose length is greater than or equal to $\lambda$. Without loss of generality, we may (and will) assume that $0=t_{0}<t_{1}<\cdots<t_{k}<t_{k+1}=1$. Since the open interval $\left(t_{i-1}, t_{i+1}\right)$ isolates $t_{i}$ from the other $t_{j}$, its length, $t_{i+1}-t_{i-1}$, is less than $\lambda$. Consequently, if $k$ is odd we have $1=\sum_{i=0}^{(k-1) / 2}\left(t_{2 i+2}-t_{2 i}\right)<\lambda\left(1+\frac{k-1}{2}\right)<1$; if $k$ is even, we have $1<1+t_{k}-t_{k-1}=$ $\sum_{i=0}^{k / 2-1}\left(t_{2 i+2}-t_{2 i}\right)+\left(t_{k+1}-t_{k-1}\right)<\lambda\left(1+\frac{k}{2}\right)<1$. A contradiction in either case.



Proof of Lemma 2. Let $I_{0}$, respectively $I_{1}$, be the set of all indices $i$ in the range $2, \ldots, k-1$ such that $m_{i}$ is even, respectively odd. Clearly, $I_{0}$ and $I_{1}$ form a partition of that range. Since $m_{i} \geq 2$ if $i$ is in $I_{0}$, and $m_{i} \geq 1$ if $i$ is in $I_{1}$ (recall that the $m_{i}$ are positive integers),



$$

\sum_{i=2}^{k-1} m_{i}=\sum_{i \in I_{0}} m_{i}+\sum_{i \in I_{1}} m_{i} \geq 2\left|I_{0}\right|+\left|I_{1}\right|=2(k-2)-\left|I_{1}\right|, \quad \text { or } \quad\left|I_{1}\right| \geq 2(k-2)-\sum_{i=2}^{k-1} m_{i}

$$



Therefore,



$$

\begin{aligned}

\left\lfloor\frac{m_{1}}{2}\right\rfloor+\sum_{i=1}^{k}\left\lfloor\frac{m_{i}}{2}\right\rfloor+\left\lfloor\frac{m_{k}}{2}\right\rfloor & \leq m_{1}+\left(\sum_{i=2}^{k-1} \frac{m_{i}}{2}-\frac{\left|I_{1}\right|}{2}\right)+m_{k} \\

& \leq m_{1}+\left(\frac{1}{2} \sum_{i=2}^{k-1} m_{i}-(k-2)+\frac{1}{2} \sum_{i=2}^{k-1} m_{i}\right)+m_{k} \\

& =\sum_{i=1}^{k} m_{i}-k+2 .

\end{aligned}

$$

Final answer: $\frac{1}{2 n+2}$