Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 5.8 AIME, harder Find the answer United States

Problem:

The diagram below is an example of a rectangle tiled by squares:

Each square has been labeled with its side length. The squares fill the rectangle without overlapping.

In a similar way, a rectangle can be tiled by nine squares whose side lengths are $2$ , $5$ , $7$ , $9$ , $16$ , $25$ , $28$ , $33$ , and $36$ . Sketch a possible arrangement of those squares. They must fill the rectangle without overlapping. Label each square in your sketch by its side length, as in the picture above.

Solution

Solution:

To tile a rectangle, the areas of the squares must add to match the area of the rectangle. The total area of the 9 squares is:
$2^{2}+5^{2}+7^{2}+9^{2}+16^{2}+25^{2}+28^{2}+33^{2}+36^{2}=4209.$
When we factor $4209$ we obtain $4209=3 \times 23 \times 61$ .

To fit the largest square, the rectangle has to be at least $36$ units wide and high, and the only way to do that with these three factors is a rectangle of size $61 \times 69$ .

It's easy to get started by noticing that $33+36=69$ and with so few squares the two largest squares must lie on one edge of the rectangle of length $69$ . From there it becomes easy to place the other ones, and the following figure illustrates a possible solution (the unlabeled square has side length of $2$ ). There are actually $3$ additional solutions that are obtained from this one by either a rotation of $180^{\circ}$ or taking a mirror image across either the horizontal or vertical axis.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.