An easy geometry problem that looks impossible

Subtitle: Applications of similar triangles and the Pythagorean Theorem

The previous post The Prediction Machine, Part I - Protocols and OpenBB was supposed to have been followed by a post on a machine learning method called Long Short-Term Memory (LSTM), but it’s taking longer than I thought, so I’m putting this up instead and will get back to predictions later. I see a great future for having the ability to predict things!

The problem

I saw this problem recently and was intrigued by how it seemed at first to be nearly impossible to solve, and yet was easy to understand. The problem is to calculate the area of a square where the only information given is the length of two apparently unrelated line segments. Here’s the setup:


You’re supposed to calculate the area of the beige square, but all you have are some lines drawn through it, and only two lengths. The length of the long line at the bottom of the square, extending out to the left, is 6. We’re also given the length 3\sqrt{3} for the line segment in the upper left corner of the square. One more bit of information given is that the 3\sqrt{3} segment intersects the long segment going from lower left to upper right at right angles.

It turns out the solution is pretty easy once you see the trick of using similar triangles.

Similar triangles

First I’ll add some labels to the figure. The three triangles are 1,2, and 3, and the short side and hypotenuse of triangle 1 are aa and bb.


Next, we’ll show that these three triangles are similar. The definition of similar triangles is that their angles must be the same, and the lengths of their sides must be in proportion. There are several ways to show that two triangles are similar, but one easy way is to show that two of the three angles are the same. Since the sum of the angles of any triangle is 180°=π radian180 \degree = \pi \text{ radian}, showing two angles gives the third for free.

The problem says that the angle between the two shorter sides of triangle 1 is a right angle (90°90 \degree). The adjacent angle in triangle 2 must be a right angle too since they’re joined by a straight line. Triangle 3 also has a right angle in the lower right corner since it’s part of the square. So each of the three triangles has a right angle, and we only need to show that one other angle matches in all three to prove similarity.

Labeling the unknown angles in each triangle helps to identify them. I’ll call 1T1\angle 1_{T1} the first angle in triangle 1, and similarly for the others.


Since the sum of the angles is 180°180 \degree, then 1+2=90°\angle 1 + \angle 2 = 90 \degree for each of the three triangles. But, 2T1+1T2=90°\angle 2_{T1} + \angle 1_{T2} = 90 \degree because they form a corner of the square, which means 1T1=1T2\angle 1_{T1} = \angle 1_{T2} and therefore 2T1=2T2\angle 2_{T1} = \angle 2_{T2}, so T1T_1 is similar to T2T_2. Using the same reasoning, T2T_2 is similar to T3T_3, so all three are similar triangles. The lengths of the sides aren’t the same, but the ratios hold, and that will give the solution.

The area of the square

The hypotenuse in T2T_2 is one side of the square, which we can call xx. Then the ratio of the sides aa and bb in T1T_1 is the same as the ratio of 3\sqrt{3} to xx in T2T_2, or

x3=ba\frac{x}{\sqrt{3}} = \frac{b}{a}


x=3ba.x = \sqrt{3} \frac{b}{a}.

Comparing the sides of T1T_1 and T3T_3, the ratio of 3\sqrt{3} to aa is the same as 66 to xx , so

6x=3a\frac{6}{x} = \frac{\sqrt{3}}{a}


6=x3a=3(3ba)a=3ba2.6 = x \frac{\sqrt{3}}{a} = \frac{\sqrt{3} \left( \sqrt{3}\frac{b}{a} \right)}{a} = 3 \frac{b}{a^2}.

Solving for bb from this gives 2a2=b2a^2 = b. We can substitute back into the first equation to get

x=3ba=32a2a=23a.x = \sqrt{3}\frac{b}{a} = \sqrt{3}\frac{2a^2}{a} = 2\sqrt{3}a.

Using the Pythagorean theorem in T1T_1,

a2+(3)2=b2a^2 + (\sqrt{3})^2 = b^2

and since 2a2=b2a^2 = b then a2=12ba^2 = \frac{1}{2}b so

12b+3=b2.\frac{1}{2}b + 3 = b^2.


b212b3=0.b^2 - \frac{1}{2}b - 3 = 0.

or 2b2b6=02b^2 - b - 6 = 0.

From the quadratic formula,

b=14(1±14(2)(6))=14(1±1+48)=14(1±7).\begin{aligned} b &= \frac{1}{4} \left( 1 \pm \sqrt{1 - 4(2)(-6)} \right) \\ &= \frac{1}{4} \left( 1 \pm \sqrt{1 + 48} \right) \\ &= \frac{1}{4} \left( 1 \pm 7 \right). \\ \end{aligned}

This means that either b=6/4b = -6/4 or b=2b = 2. Since the length of a side must be positive, then b=2b = 2, and a2=1a^2 = 1 meaning a=1a=1. Substituting for xx we get

x=321x=23.x = \sqrt{3} \frac{2}{1} \Rightarrow x = 2 \sqrt{3}.

Then, the area of the square is x2=12x^2 = 12.

So, with the insight of using similar triangles and an application of the Pythagorean Theorem, we can solve for the area of the square even though it seems at first that we don’t have enough information for a solution.