# Problem of the Week Problem D and Solution Find the Largest Area

## Problem

Rectangle $$ACEG$$ has $$B$$ on $$AC$$ and $$F$$ on $$EG$$ such that $$BF$$ is parallel to $$CE$$. Also, $$D$$ is on $$CE$$ and $$H$$ is on $$AG$$ such that $$HD$$ is parallel to $$AC$$, and $$BF$$ intersects $$HD$$ at $$J$$. The area of rectangle $$ABJH$$ is $$6\text{ cm}^2$$ and the area of rectangle $$JDEF$$ is $$15\text{ cm}^2$$.

If the dimensions of rectangles $$ABJH$$ and $$JDEF$$, in centimetres, are integers, then determine the largest possible area of rectangle $$ACEG$$. Note that the diagram is just an illustration and is not intended to be to scale.

## Solution

Let $$AB = x$$, $$AH = y$$, $$JD = a$$ and $$JF = b$$.

Then, \begin{align} HJ &= GF = AB = x\\ BJ &= CD = AH= y\\ BC &= FE = JD = a\\ HG &= DE = JF = b \end{align}

Thus, we have \begin{align} \text{area}(ACEG) &= \text{area}(ABJH) + \text{area}(BCDJ) + \text{area}(JDEF) + \text{area}(HJFG)\\ &= 6 + ya + 15 + xb\\ &= 21 + ya + xb \end{align}

Since the area of rectangle $$ABJH$$ is $$6\text{ cm}^2$$ and the side lengths of $$ABJH$$ are integers, then the side lengths must be $$1$$ and $$6$$ or $$2$$ and $$3$$. That is, $$x = 1$$ cm and $$y = 6$$ cm, $$x = 6$$ cm and $$y = 1$$ cm, $$x = 2$$ cm and $$y = 3$$ cm, or $$x = 3$$ cm and $$y = 2$$ cm.

Since the area of rectangle $$JDEF$$ is $$15\text{ cm}^2$$ and the side lengths of $$JDEF$$ are integers, then the side lengths must be $$1$$ and $$15$$ or $$3$$ and $$5$$. That is, $$a = 1$$ cm and $$b = 15$$ cm, $$a = 15$$ cm and $$b = 1$$ cm, $$a = 3$$ cm and $$b = 5$$ cm, or $$a = 5$$ cm and $$b = 3$$ cm.

To maximize the area, we need to pick the values of $$x$$, $$y$$, $$a$$, and $$b$$ which make $$ya + xb$$ as large as possible. We will now break into cases based on the possible side lengths of $$ABJH$$ and $$JDEF$$ and calculate the area of $$ACEG$$ in each case. We do not need to try all $$16$$ possible pairings, because trying $$x = 1$$ cm and $$y=6$$ cm with the four possibilities of $$a$$ and $$b$$ will give the same $$4$$ areas, in some order, as trying $$x=6$$ cm and $$y=1$$ cm with the four possibilities of $$a$$ and $$b$$. Similarly, trying $$x = 2$$ cm and $$y=3$$ cm with the four possibilities of $$a$$ and $$b$$ will give the same $$4$$ areas, in some order, as trying $$x=3$$ cm and $$y=2$$ cm with the four possibilities of $$a$$ and $$b$$. (As an extension, we will leave it to you to think about why this is the case.)

• Case 1: $$x = 1$$ cm, $$y = 6$$ cm, $$a=1$$ cm, $$b=15$$ cm

Then $$\text{area}(ACEG) = 21 + ya + xb = 21 + 6(1) + 1(15) = 42\text{ cm}^2$$.

• Case 2: $$x = 1$$ cm, $$y = 6$$ cm, $$a=15$$ cm, $$b=1$$ cm

Then $$\text{area}(ACEG) = 21 + ya + xb = 21 + 6(15) + 1(1) = 112\text{ cm}^2$$.

• Case 3: $$x = 1$$ cm, $$y = 6$$ cm, $$a=3$$ cm, $$b=5$$ cm

Then $$\text{area}(ACEG) = 21 + ya + xb = 21 + 6(3) + 1(5) = 44\text{ cm}^2$$.

• Case 4: $$x = 1$$ cm, $$y = 6$$ cm, $$a=5$$ cm, $$b=3$$ cm

Then $$\text{area}(ACEG) = 21 + ya + xb = 21 + 6(5) + 1(3) = 54\text{ cm}^2$$.

• Case 5: $$x = 2$$ cm, $$y = 3$$ cm, $$a=1$$, $$b=15$$ cm

Then $$\text{area}(ACEG) = 21 + ya + xb = 21 + 3(1) + 2(15) = 54\text{ cm}^2$$.

• Case 6: $$x = 2$$ cm, $$y = 3$$ cm, $$a=15$$, $$b=1$$ cm

Then $$\text{area}(ACEG) = 21 + ya + xb = 21 + 3(15) + 2(1) = 68\text{ cm}^2$$.

• Case 7: $$x = 2$$ cm, $$y = 3$$ cm, $$a=3$$, $$b=5$$ cm

Then $$\text{area}(ACEG) = 21 + ya + xb = 21 + 3(3) + 2(5) = 40\text{ cm}^2$$.

• Case 8: $$x = 2$$ cm, $$y = 3$$ cm, $$a=5$$, $$b=3$$ cm

Then $$\text{area}(ACEG) = 21 + ya + xb = 21 + 3(5) + 2(3) = 42\text{ cm}^2$$.

We see that the maximum area is $$112\text{ cm}^2$$, and occurs when $$x = 1$$ cm, $$y = 6$$ cm and $$a=15$$ cm, $$b=1$$ cm. It will also occur when $$x = 6$$ cm, $$y = 1$$ cm and $$a=1$$ cm, $$b=15$$ cm.

The following diagrams show the calculated values placed on the original diagram. The diagram given in the problem was definitely not drawn to scale! Both solutions produce rectangles with dimensions $$7\text{ cm}$$ by $$16\text{ cm}$$, and area $$112\text{ cm}^2$$.