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Rectangles in a Square

Statement

You have to completely fill a \(1 \times 1\) square and you can only use rectangles of side ratio \(1:2\) - you can have rectangles of any size, as long as the ratio of their side lengths is \(1:2\). Find all \(n \in\mathbb N\) such that you can use \(n\) rectangles to make such a square.

Solution

\(2\) is possible.

+-------+-------+
|       |       |
|       |       |
|       |       |
|       |       |
+-------+-------+

And we can split one rectangle into \(4\) equal rectangles of equal size.

This means that if \(n\) is possible, then \(n+3\) is also possible.

+---+---+-------+
|   |   |       |
|   |   |       |
+---+---+       |
|   |   |       |
|   |   |       |
+---+---+-------+

If we have \(n\) rectangles, then \(n+4\) is possible. That is, if you can fill the square with \(n\), then resize them as shown and add four rectangles.

+--------------+------+
|              |      |
|              |      |
+------+---+---+      |
|      |   |   |      |
|      |   |   |      |
|      +---+---+------+
|      |              |
|      |              |
+------+--------------+

From these two facts, we conclude: if \(n\) is possible, then \(n+3\) and \(n+4\) are possible.

\[2 \rightarrow 5,6 \rightarrow 8,9,10 \rightarrow \text{all } n \geq 11, \quad n \in \mathbb{Z}\]

This means that if \(2\) is possible, then \(5\) and \(6\) are possible. From \(5,6\) we get \(8,9,10\). After this, we can just add \(3\) to \(8\) to get \(11\), etc.

This leaves only \(1, 3, 4,\) and \(7\). We already know \(1\) is not possible.

For \(3\), we observe that at least one rectangle should touch \(2\) vertices of the square.

+-------+-------+
|       |       |
|       |       |
|       |       |
|       |       |
+-------+-------+

From this we see that the other \(2\) rectangles have to fill the remaining space, which is not possible.

In case of \(4\), two things are possible:

1. Each rectangle touches only \(1\) vertex, then it is not possible to fill the square

+-------------+------+
|             |      |
|             |      |
+------+------+      |
|      | gap  |      |
|      |      |      |
|      +------+------+
|      |             |
|      |             |
+------+-------------+

In this image, we see that there is a space left in the center of the square

2. One of the rectangles touches \(2\) vertices, the other \(3\) have to fill the remaining space, which is not possible.

+-------+-------+
|       |       |
|       |       |
|       |       |
|       |       |
+-------+-------+

For \(7\):

+---+------+---------+
|   |      |         |
|   +------+         |
|   |      |         |
+---+      |         |
|   |      |         |
|   |      |         |
|   |      |         |
+---+      |         |
|   |      |         |
|   +------+         |
|   |      |         |
+---+------+---------+

This is a possible placement of rectangles.

So, except \(1, 3\), and \(4\), all natural numbers are possible.

\[\Huge\varnothing\]