Princess Telassim cut a rectangular sheet of paper into 9 squares with sides of 1, 4, 7, 8, 9, 10, 14, 15, and 18 centimeters.

a) What was the area of the sheet before it was cut?

b) What were the dimensions of the sheet before it was cut?

c) Princess Telassim needs to reassemble the sheet. Help her by showing, with a drawing, how to do this reassembly.
Solution

a) The area of the sheet was equal to the sum of the areas of the nine squares, which is (in square centimeters):

$$
1^{2}+4^{2}+7^{2}+8^{2}+9^{2}+10^{2}+14^{2}+15^{2}+18^{2}=1056
$$

b) Let $a$ and $b$ be the dimensions of the sheet, where we assume $a \leq b$. Since the area of a rectangle is the product of its dimensions, we have $a b=1056$. Moreover, since the measures of the sides of the squares into which the sheet was cut are integers, it follows that $a$ and $b$ must be integers. Finally, we observe that $a$ and $b$ must be greater than or equal to 18, as one of the squares into which the sheet was cut has a side with this measure. Since $a$ and $b$ are divisors of 1056, the prime factorization $1056=2^{5} \times 3 \times 11$ shows that $a$ and $b$ are of the form $2^{x} \times 3^{y} \times 11^{z}$, where $x, y$ and $z$ are integers such that $0 \leq x \leq 5,0 \leq y \leq 1$ and $0 \leq z \leq 1$. Remembering that $a b=1056$ and that $a$ and $b$ are greater than 18, we obtain the following possibilities:

| $\mathbf{a}$ | $\mathbf{b}$ |
| :---: | :---: |
| $2 \times 11=22$ | $2^{4} \times 3=48$ |
| $2^{3} \times 3=24$ | $2^{2} \times 11=44$ |
| $2^{5}=32$ | $3 \times 11=33$ |

We now need to decide which of these possibilities can occur as the dimensions of the sheet. Since the largest square has a side of 18, which is less than 22, 24, and 32, we see that no square can touch both sides of length $b$ of the sheet. This means that $b$ can be expressed in two ways as a sum, in which the terms are the measures of the sides of the squares, such that:

- there are no repeated terms in either of the two expressions and
- there are no common terms in the two expressions.

This argument shows that $2 b \leq 1+4+7+8+9+10+14+15+18$ or, in other words, $2 b \leq 86$. Therefore, $b \leq 43$ and the only possibility is $b=33$. It follows that the dimensions of the sheet were $a=32$ and $b=33$.

There are other ways to eliminate the pairs $(22,48)$ and $(24,44)$, using the argument above and showing, for example, that there are no two ways to write 22 and 24 as the sum of the sides of the squares with distinct terms and no common terms. This solution depends on the fact that, in any decomposition of a rectangle into squares, the sides of the squares are necessarily parallel to one of the sides of the rectangle. An intuitive argument to demonstrate this fact consists in selecting a vertex of the rectangle and observing that the square to which this vertex belongs has its sides supported on the sides of the rectangle. Any square that touches this first square (even at just one vertex) has its sides necessarily parallel to the sides of the rectangle, otherwise, we would have angles different from $90^{\circ}$ or $180^{\circ}$ in the decomposition, and these angles cannot be filled with squares.

c) The only possibility (up to rotations and symmetries) is shown below:

|  |  |  |  |
| :---: | :---: | :---: | :---: |
| 14 | 18 |  |  |
| 10 | 4 |  |  |
|  | 7 | 15 |  |
| 1 | 7 |  |  |
| 9 |  |  |  |