# Problem of the Week Problem E and Solution Three Squares

## Problem

The three squares $$ABCD$$, $$AEFG$$, and $$AHJK$$ overlap as shown in the diagram.

The side length of each square, in centimetres, is a positive integer. The area of square $$AEFG$$ that is not covered by square $$ABCD$$ is $$33\text{ cm}^2$$. That is, the area of the shaded region $$BEFGDC$$ is $$33\text{ cm}^2$$. If $$DG = GK$$, determine all possible side lengths of each square.

## Solution

Let $$AD = x$$ cm and $$DG = y$$ cm. Therefore $$GK = DG = y$$ cm.

Also, since the side length of each square is an integer, $$x$$ and $$y$$ are integers.

The shaded region has area $$33\text{ cm}^2$$. The shaded region is equal to the area of the square with side length $$AG$$ minus the area of the square with side length $$AD$$.

Since $$AD = x$$ and $$AG = AD + DG = x+y$$, we have \begin{align} 33 & = (\text{area of square with side length AG}) - (\text{area of square with side length AD}) \\ &=(x+y)^2 - x^2\\ &= x^2 + 2xy + y^2 - x^2\\ &= 2xy + y^2\\ &= y(2x +y) \end{align} Since $$x$$ and $$y$$ are integers, so is $$2x+y$$. Therefore, $$2x+y$$ and $$y$$ are two positive integers that multiply to give $$33$$. Therefore, we must have $$y = 1$$ and $$2x+y = 33$$, or $$y = 3$$ and $$2x+y = 11$$, or $$y = 11$$ and $$2x+y = 3$$, or $$y = 33$$ and $$2x+y = 1$$. The last two would imply that $$x<0$$, which is not possible. Therefore, $$y=1$$ and $$2x +y = 33$$, or $$y = 3$$ and $$2x+y = 11$$.

When $$y=1$$ and $$2x +y = 33$$, it follows that $$x=16$$. Then square $$ABCD$$ has side length $$x=16$$ cm, square $$AEFG$$ has side length $$x+y=17$$ cm, and square $$AHJK$$ has side length $$x+2y=18$$ cm.

When $$y=3$$ and $$2x +y = 11$$, it follows that $$x=4$$. Then square $$ABCD$$ has side length $$x=4$$ cm, square $$AEFG$$ has side length $$x+y=7$$ cm, and square $$AHJK$$ has side length $$x+2y=10$$ cm.

Therefore, there are two possible sets of squares. The squares are either $$16$$ cm $$\times$$ $$16$$ cm and $$17$$ cm $$\times$$ $$17$$ cm and $$18$$ cm $$\times$$ $$18$$ cm, or $$4$$ cm $$\times$$ $$4$$ cm and $$7$$ cm $$\times$$ $$7$$ cm and $$10$$ cm $$\times$$ $$10$$ cm. Each of these sets of squares satisfies the conditions of the problem.