Subfield: Combinatorics


Question: Let $n$ points be given inside a rectangle $R$ such that no two of them lie on a line parallel to one of the sides of $R$. The rectangle $R$ is to be dissected into smaller rectangles with sides parallel to the sides of $R$ in such a way that none of these rectangles contains any of the given points in its interior. Prove that we have to dissect $R$ into at least $n+1$ smaller rectangles.

Solution:
Let $k$ be the number of rectangles in the dissection. The set of all points that are corners of one of the rectangles can be divided into three disjoint subsets:

- $A$, which consists of the four corners of the original rectangle $R$, each of which is the corner of exactly one of the smaller rectangles,
- $B$, which contains points where exactly two of the rectangles have a common corner (T-junctions, see the figure below),
- $C$, which contains points where four of the rectangles have a common corner (crossings, see the figure below).
<img_3599>

Figure 1: A T-junction and a crossing

We denote the number of points in $B$ by $b$ and the number of points in $C$ by $c$. Since each of the $k$ rectangles has exactly four corners, we get

$$
4 k=4+2 b+4 c
$$

It follows that $2 b \leqslant 4 k-4$, so $b \leqslant 2 k-2$.

Each of the $n$ given points has to lie on a side of one of the smaller rectangles (but not of the original rectangle $R$ ). If we extend this side as far as possible along borders between rectangles, we obtain a line segment whose ends are T-junctions. Note that every point in $B$ can only be an endpoint of at most one such segment containing one of the given points, since it is stated that no two of them lie on a common line parallel to the sides of $R$. This means that

$$
b \geqslant 2 n \text {. }
$$

Combining our two inequalities for $b$, we get

$$
2 k-2 \geqslant b \geqslant 2 n
$$

thus $k \geqslant n+1$, which is what we wanted to prove.
Let $k$ denote the number of rectangles. In the following, we refer to the directions of the sides of $R$ as 'horizontal' and 'vertical' respectively. Our goal is to prove the inequality $k \geqslant n+1$ for fixed $n$. Equivalently, we can prove the inequality $n \leqslant k-1$ for each $k$, which will be done by induction on $k$. For $k=1$, the statement is trivial.

Now assume that $k>1$. If none of the line segments that form the borders between the rectangles is horizontal, then we have $k-1$ vertical segments dividing $R$ into $k$ rectangles. On each of them, there can only be one of the $n$ points, so $n \leqslant k-1$, which is exactly what we want to prove.

Otherwise, consider the lowest horizontal line $h$ that contains one or more of these line segments. Let $R^{\prime}$ be the rectangle that results when everything that lies below $h$ is removed from $R$ (see the example in the figure below).

The rectangles that lie entirely below $h$ form blocks of rectangles separated by vertical line segments. Suppose there are $r$ blocks and $k_{i}$ rectangles in the $i^{\text {th }}$ block. The left and right border of each block has to extend further upwards beyond $h$. Thus we can move any points that lie on these borders upwards, so that they now lie inside $R^{\prime}$. This can be done without violating the conditions, one only needs to make sure that they do not get to lie on a common horizontal line with one of the other given points.

All other borders between rectangles in the $i^{\text {th }}$ block have to lie entirely below $h$. There are $k_{i}-1$ such line segments, each of which can contain at most one of the given points. Finally, there can be one point that lies on $h$. All other points have to lie in $R^{\prime}$ (after moving some of them as explained in the previous paragraph).

<img_3717>

Figure 2: Illustration of the inductive argument

We see that $R^{\prime}$ is divided into $k-\sum_{i=1}^{r} k_{i}$ rectangles. Applying the induction hypothesis to $R^{\prime}$, we find that there are at most

$$
\left(k-\sum_{i=1}^{r} k_{i}\right)-1+\sum_{i=1}^{r}\left(k_{i}-1\right)+1=k-r
$$

points. Since $r \geqslant 1$, this means that $n \leqslant k-1$, which completes our induction.