# rectangularizing the square

There is a square that I want to divide to n people, such that each person gets a rectangular piece with an equal area.

An obvious option is to cut 1-by-n rectangles of size n-by-1, but the people don't want such rectangles, they say they are too skinny. They want to get R-balanced rectangles, which are defined as axis-aligned rectangles whose width-height ratio is between $R$ and $1/R$ (where $R \geq 1$).

What is the minimum R for which I can guarantee that, for every n, there is a division of a square to n equal-area R-balanced rectangles?

Some notes and sub-questions:

• A 1-balanced rectangle is just a square. So, for $R=1$, a division is possible only when n is a square number.
• For $R=2$, I managed to find a division for $n=1, 2, 4, 5, 6, 7, 8, 9$, but I haven't found a division for $n=3$, and also haven't managed to prove that it is impossible. What do you think?
• What are all the numbers $n$ for which there is a division of a square into $n$ 2-balanced rectangles? Obviously all square and double-square numbers are included, but there are other numbers, such as 5, 6 and 7.
• For $R=3$, I haven't found a counter-example, but also couldn't prove that for every n it is possible to find a division to 3-balanced rectangle. What do you think?

(Note that Erich's packing center contains a nice summary of packing rectangles in a square, but it is limited to identical rectangles. I allow different rectangles, as long as they are the same area, and with an R-balanced width-height ratio).

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A possible way to divide a square into rectangles is to first divide it into shelves (horizontal strips), and then divide each shelf into identical adjacent rectangles. In order to have rectangles of equal area, the number of rectangles in each shelf should be proportional to the height of the shelf. For concreteness, if the total number of rectangles is $N$, and the square side length is $L$, and the height of a particular shelf is $h$, then the number of rectangles in this shelf should be $Nh/L$ (h must be chosen so that this is a whole number). Thus the width of each rectangle is $L^2/Nh$, and the height/width ratio is $Nh^2/L^2 = N(h/L)^2$.

In order to have a balanced width/height ratio, we should make this number as close to 1 as possible. Therefore we should choose values of $h$ such that $L/h$ is as close as possible to $sqrt(N)$. If $N$ is a square number, then the division is trivial. So let's assume N is not a square number, and define:

• $S=sqrt(N)$,
• $S_-=floor(S)$,
• $S_+=ceiling(S) = S_-+1$,
• $h_-=L S_- / N$,
• $h_+=L S_+ / N$,

It is always possible to write $N$ as a sum of several $S_-$'s and several $S_+$'s. Similarly, it is possible to divide the square to several shelves of height $h_-$ and several shelves of height $h_+$'s. The height/width ratios in the two types of shelves are:

• $N(h_-/L)^2 = S_-^2/N$
• $N(h_+/L)^2 = S_+^2/N$

By the definition of S, the following relations hold:

• $S-1 < S_- < S < S_+ < S+1$
• $(S^2-2S+1) < S_-^2 < S^2=N < S_+^2 < (S^2+2S+1)$
• $(1-2S/N+1/N) < S_-^2/N < 1 < S_+^2/N < (1+2S/N+1/N)$

Thus, the height/width ratios are bounded by $(1-2S/N+1/N)$ and $(1+2S/N+1/N)$. These numbers become closer to 1 as N increases. Therefore, it is enough to find a threshold N for each R. Also, it is enough to find a threshold for the lower bound, since the upper bound is always geometrically closer to 1. Here are some examples:

• For $N=6$, the lower bound is 0.35+, which is more than 1/3. Therefore, for all $N \geq 6$, it is possible to rectangularize a square to N equal-area 3-balanced rectangles. It is easy to check that this is possible also for $N<6$. Thus we have proven that For all N, it is possible to rectangularize a square to N equal-area 3-balanced rectangles.
• For $N=12$, the lower bound is 0.51-, which is more than 1/2. Therefore, for all $N \geq 12$, it is possible to rectangularize a square to N equal-area 2-balanced rectangles. I leave the cases of $N=10$, $N=11$ as an exercise to the readers :-)
• For $N=30$, the lower bound is 0.668+, which is more than 2/3. Therefore, for all $N \geq 30$, it is possible to rectangularize a square to N equal-area 1.5-balanced rectangles.
• And so on...

To complete the answer, here is a proof that this is not possible to rectangluarize a square to 3 equal-area R-balanced rectangles with $R<3$.

The proof is by contradiction. Suppose a square is divided to 3 equal-area R-balanced rectangles with $R<3$. Then none of the rectangles can touch two opposite sides of the square (otherwise its height would be more than $L/3$, and its area would be more than $L^2/3$). Then all square sides must be touched by at least two rectangles, and each rectangle may touch at most 2 square sides. Then the number of rectangles must be at least as large as the number of square sides. Thus $3 \geq 4$, a contradiction.

Thus, have shown that, the smallest $R \geq 1$ such that, For all N, it is possible to rectangularize a square to N equal-area R-balanced rectangles, is 3.

Thus, the answer to my question is: 3.

I love questions that can be answered with a single digit :-)

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I'm awarding this answer a +100 bounty because I cannot award two bounties for answers to this other question. There will be a 24 hour delay before the bounty is awarded. – Jim Belk Nov 10 '13 at 1:10