# Problem of the Week Problem D and Solution The Largest Square

## Problem

Three squares are placed side by side with the smallest square on the left and the largest square on the right. The bottom sides of the three squares form a horizontal line.

The side length of the smallest square is $$5$$ units, and the side length of the medium-sized square is $$8$$ units. If the top-left corner of each square all lie on a straight line, determine the side length of the largest square.

## Solution

First we draw a line segment connecting the top-left corner of each square and label the vertices as shown in the diagram.

Let $$a$$ represent the side length of the largest square. From here we present three different solutions.

In Solution 1, we solve the problem by calculating the slope of $$BH$$. In Solution 2, we solve the problem using similar triangles. In Solution 3, we place the diagram on the $$xy$$-plane and solve the problem using analytic geometry.

Solution 1

The slope of a line is equal to its rise divided by its run. If we look at the line segment from $$B$$ to $$E$$, $$BC = 5$$ and $$CE = DE - DC = 8-5 = 3$$. Therefore, slope $$BE= \frac{CE}{BC} = \frac{3}{5}$$.

If we look at the line segment from $$E$$ to $$H$$, $$EF = 8$$ and $$FH = GH - GF = a - 8$$. Therefore, slope $$EH= \frac{FH}{EF} = \frac{a-8}{8}$$.

Since $$B$$, $$E$$, and $$H$$ lie on a straight line, the slope of $$BE$$ must equal the slope of $$EH$$. Therefore, \begin{aligned} \frac{3}{5} &= \frac{a-8}{8}\\ 5(a-8) &= 3(8)\\ 5a-40 &= 24\\ 5a &= 64\\ a &= \frac{64}{5} \end{aligned} Therefore, the side length of the largest square is $$\dfrac{64}{5}$$ units.

Solution 2

Consider $$\triangle BCE$$ and $$\triangle EFH$$. We will first show that $$\triangle BCE \sim \triangle EFH$$.

Since $$ABCD$$ is a square, $$\angle BCD = 90^{\circ}$$. Therefore, $$\angle BCE = 180^{\circ} - \angle BCD = 180^{\circ} - 90^{\circ} = 90^{\circ}$$. Since $$DEFG$$ is a square, $$\angle EFG = 90^{\circ}$$. Therefore, $$\angle EFH = 180^{\circ} - \angle EFG = 180^{\circ} - 90^{\circ} = 90^{\circ}$$. Thus, $$\angle BCE = \angle EFH$$.

Since $$ABCD$$ and $$DEFG$$ are squares and $$AG$$ is a straight line, $$BC$$ is parallel to $$EF$$. Therefore, $$\angle EBC$$ and $$\angle HEF$$ are corresponding angles and so $$\angle EBC = \angle HEF$$.

Since the angles in a triangle add to $$180^{\circ}$$, then we must also have $$\angle BEC = \angle EHF$$.

Therefore, $$\triangle BCE \sim \triangle EFH$$, by Angle-Angle-Angle Triangle Similarity.

Since $$\triangle BCE \sim \triangle EFH$$, corresponding side lengths are in the same ratio. In particular, \begin{aligned} \frac{EC}{BC} &= \frac{HF}{EF}\\ \frac{DE - DC}{BC} &= \frac{GH - GF}{EF}\\ \frac{8 - 5}{5} &= \frac{a - 8}{8}\\ \frac{3}{5} & = \frac{a-8}{8}\\ 5(a-8) &= 3(8)\\ 5a-40 &= 24\\ 5a &= 64\\ a &= \frac{64}{5} \end{aligned} Therefore, the side length of the largest square is $$\dfrac{64}{5}$$ units.

Solution 3

We start by placing the diagram on the $$xy$$-plane with $$A$$ at $$(0,0)$$ and $$AL$$ along the $$x$$-axis.

The coordinates of $$B$$ are $$(0,5)$$, the coordinates of $$D$$ are $$(5,0)$$, the coordinates of $$E$$ are $$(5,8)$$, the coordinates of $$G$$ are $$(13,0)$$, and the coordinates of $$H$$ are $$(13,a)$$.

Letâ€™s determine the equation of the line through $$B$$, $$E$$, and $$H$$.

Since this line passes through $$(0,5)$$, it has $$y$$-intercept $$5$$. Since the line passes through $$(0,5)$$ and $$(5,8)$$, it has a slope of $$\frac{8-5}{5-0}= \frac{3}{5}$$. Therefore, the equation of the line through $$B$$, $$E$$, and $$H$$ is $$y = \frac{3}{5}x + 5$$.

Since $$H(13,a)$$ lies on this line, substituting $$x = 13$$ and $$y = a$$ into $$y = \frac{3}{5}x + 5$$ gives $a = \frac{3}{5}(13) + 5 = \frac{39}{5} + 5 = \frac{39+25}{5} = \frac{64}{5}$ Therefore, the side length of the largest square is $$\dfrac{64}{5}$$ units.