# Problem of the Week Problem E and Solution A Square in a Triangle

## Problem

In $$\triangle ABC$$, there is a right angle at $$B$$ and the length of $$BC$$ is twice the length of $$AB$$. In other words, $$BC=2AB$$.

Square $$DEFB$$ is drawn inside $$\triangle ABC$$ so that vertex $$D$$ is somewhere on $$AB$$ between $$A$$ and $$B$$, vertex $$E$$ is somewhere on $$AC$$ between $$A$$ and $$C$$, vertex $$F$$ is somewhere on $$BC$$ between $$B$$ and $$C$$, and the final vertex is at $$B$$.

Square $$DEFB$$ is called an inscribed square. Determine the ratio of the area of the inscribed square $$DEFB$$ to the area of $$\triangle ABC$$.

## Solution

First we draw square $$DEFB$$ according to the instructions in the problem. Let $$DB=BF=FE=ED=a$$ and

$$AD=b$$. Since $$BC=2AB$$, it follows that

$$BC=2(AD+DB)=2(a+b)=2a+2b$$. Since $$BC=BF+FC$$, it follows that $$2a+2b=a+FC$$, so $$FC=a+2b$$.

From here we present two solutions. In Solution 1, we solve the problem using similar triangles. In Solution 2, we place the diagram on the $$xy$$-plane and solve the problem using analytic geometry.

Solution 1

Consider $$\triangle ADE$$ and $$\triangle ABC$$. We will first show that $$\triangle ADE \sim \triangle ABC$$.

Since $$DEFB$$ is a square, then $$\angle EDB = 90^{\circ}$$, and so $$\angle EDA = 180^{\circ} - \angle EDB = 180^{\circ} - 90^{\circ} = 90^{\circ}$$. Therefore, $$\angle EDA = \angle ABC$$. Also, $$\angle DAE = \angle BAC$$ since they represent the same angle. Since the angles in a triangle add to $$180^{\circ}$$, then we must also have $$\angle AED = \angle ACB$$.

So $$\triangle ADE \sim \triangle ABC$$, by Angle-Angle-Angle Triangle Similarity.

Since $$\triangle ADE \sim \triangle ABC$$, then corresponding side lengths are in the same ratio. In particular, \begin{aligned} \frac{AD}{DE} &= \frac{AB}{BC}\\ \frac{AD}{DE} &= \frac{AB}{2AB}\\ \frac{b}{a} &= \frac{1}{2}\\ a &= 2b \end{aligned} Since $$BC=2a+2b$$ and $$a=2b$$, then $$BC=2(2b)+2b=6b$$. Since $$AB=a+b$$ and $$a=2b$$, then $$AB=2b+b=3b$$. The area of $$\triangle ABC$$ is $$\frac{1}{2}(BC\times AB) = \frac{1}{2}(6b\times 3b)= 9b^2$$.

The area of square $$DEFB$$ is $$a\times a = a^2 = (2b)^2 = 4b^2$$.

The ratio of the area of inscribed square $$DEFB$$ to the area of $$\triangle ABC$$ is $$4b^2 : 9b^2 = 4:9$$, since $$b>0$$.

Solution 2

First we place the triangle on the $$xy$$-plane with $$B$$ at $$(0,0)$$ and $$BC$$ along the positive $$x$$-axis. The coordinates of $$D$$ are $$(0,a)$$, the coordinates of $$A$$ are $$(0,a+b)$$, the coordinates of $$F$$ are $$(a,0)$$, the coordinates of $$E$$ are $$(a,a)$$, and the coordinates of $$C$$ are $$(2a+2b, 0)$$.

Let’s determine the equation of the line through $$A$$, $$E$$, and $$C$$.

Since this line passes through $$(0,a+b)$$, then we know it has $$y$$-intercept $$a+b$$.

Since it passes through $$(0,a+b)$$ and $$(a,a)$$, then the line has slope $$\frac{a-(a+b)}{a-0}= -\frac{b}{a}$$.

Therefore, the equation of the line through $$A$$, $$E$$, and $$C$$ is $$y = \left(-\frac{b}{a}\right)x+a+b$$.

Since $$C(2a+2b, 0)$$ lies on this line, then substituting $$x = 2a+2b$$ and $$y = 0$$ into $$y = \left(-\frac{b}{a}\right)x+a+b$$ gives \begin{aligned} 0 &= \left(-\frac{b}{a}\right)(2a+2b)+a+b\\ 0 &= (-b)(2a+2b)+(a)(a+b)\\ 0 &= -2ab-2b^2+a^2+ab\\ 0 &= a^2-ab-2b^2\\ 0 &= (a+b)(a-2b) \end{aligned} Thus, $$a=-b$$ or $$a=2b$$. But since $$a,b>0$$, then $$a=-b$$ is inadmissible and we must have $$a = 2b$$.

Since $$BC=2a+2b$$ and $$a=2b$$, then $$BC=2(2b)+2b=6b$$. Since $$AB=a+b$$ and $$a=2b$$, then $$AB=2b+b=3b$$. The area of $$\triangle ABC$$ is $$\frac{1}{2}(BC\times AB) = \frac{1}{2}(6b\times 3b)= 9b^2$$.

The area of square $$DEFB$$ is $$a\times a = a^2 = (2b)^2 = 4b^2$$.

The ratio of the area of inscribed square $$DEFB$$ to the area of $$\triangle ABC$$ is $$4b^2 : 9b^2 = 4:9$$, since $$b>0$$.

Note:

From the equation $$0 = (-b)(2a+2b)+(a)(a+b)$$, we could have instead factored $$(2a+2b)$$ to obtain $$0 = (-2b)(a+b)+a(a+b)$$. Since $$a,b>0$$, $$a+b>0$$, so we could have divided out the common factor of $$(a+b)$$ leaving $$0 = -2b+a$$ which simplifies to $$a=2b$$. Thus, the factoring of $$a^2-ab-2b^2$$ to determine $$a=2b$$ would not have been necessary.

Extension:

If, in the original problem, $$BC=kAB$$, where $$k>0$$, and the square was inscribed as given, what would be the ratio of the area of square $$DEFB$$ to the area of $$\triangle ABC$$?