#
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.