Four Triangles, Three Squares, and One Hexagon

Framing the Problem

I recently came across an intriguing problem that I thought worth pursuing. It is represented by the figure below.

We are given a triangle XYZ. On each of its sides, we construct a square so that the squares are outside the triangle. This gives us the squares XYCB, YZED, and ZXAF. Now, we join the segments AB, CD, and EF, thereby forming a hexagon ABCDEF. We have to show that the areas of ΔXYZ, ΔABX, ΔCDY, and ΔEFZ are equal.

Note that this holds true for any initial ΔXYZ. Hence, we cannot assume anything special about the starting triangle. If you wish to explore this, you can access a Geogebra applet I created here. You can click and drag on X, Y, or Z to change the initial triangle. All the other points are automatically generated. You can convince yourself that the four triangles do indeed have the same area.

The solution, once it strikes you, is really quite elegant and does not require making any tedious measurements. Measurements of any given situation of course would not constitute a proof of any kind. I cannot stress this point enough. So please allow me a short diversion while I vent about one particularly troubling issue. I will proceed with the solution after that.

A Reason for Venting

Mathematical proof is something that works quite differently from so-called ‘proofs’ in other areas of human knowledge. In mathematics, proof proceeds by deductive reasoning, while in every other field, what qualifies as ‘proof’ proceeds either by inductive or abductive reasoning. In The Promise and Pitfalls of Mathematical Abduction I had written in length about these modes of reasoning. The image below captures the essential differences between the three primary modes of reasoning quite well.

A Geometric Excursus

Since the image above may be too general, let me reiterate the differences by taking a specific example. Suppose we are attempting to prove that the sum of the angles of a triangle is 180°. We could start with a particular triangle, like the one below.

A quick scratch pad addition will confirm that the sum of the angles is 180°. We could proceed to another triangle shown below.

Once again we can confirm that the sum of the angles is 180°. However, we have actually not proved anything! All we have done is confirm the hypothesis for two triangles. Note that the angles are different in the two triangles. Indeed, there are infinitely man values that could be chosen for angle A and for each of those infinitely many for angle B. Assuming the sum of the angles is 180°, choosing A and B would fix angle C since the sum is a constant. However, we have an infinite number of triangles. And no matter how many triangles we confirm the hypothesis for, there will always be infinitely many more triangles that remain untested.

For example, there is the conjecture that the two numbers a = n¹⁹+ 6 and b = (n + 1)¹⁶ + 9 are relatively prime. However, you have to test till n = 1, 578, 270, 389, 554, 680, 057, 141, 787, 800, 241, 971, 645, 032, 008, 710, 129, 107, 338, 825, 798 (61 digits), before you reach a counterexample with the two numbers having a greatest common divisor equal to 5, 299, 875, 888, 670, 549, 565, 548, 724, 808, 121, 659, 894, 902, 032, 916, 925, 752, 559, 262, 837!¹ We don’t even have names for numbers this large! In contrast, if the universe is 14 billion years old, that is only about 441, 504, 000, 000, 000, 000 seconds! That’s 441 trillion, 504 billion seconds. It’s nameable. My point is that there is no way to prove a conjecture like this by the method of testing since a single counterexample will disprove it and there is no guarantee if or when we will encounter the counterexample.

However, we can prove the result for the sum of the angles of a triangle in a very elegant manner. Consider a general triangle as shown in the figure below.

Now, we draw a line parallel to BC passing through A. This is shown below.

Now since DE is parallel to BC, ∠DAB and ∠ABC are alternate interior angles and are, therefore, equal. By a similar argument, ∠EAC and ∠BCA are equal. Hence, ∠ABC + ∠CAB + ∠BCA = ∠DAC + ∠CAB + ∠EAC. However, ∠DAC, ∠CAB, and ∠EAC add to form a straight line, which by definition is 180°. Hence, we can conclude that the sum ∠ABC + ∠CAB + ∠BCA is 180°, thereby proving the result.

What we have seen here is an example of deductive reasoning where we proceed from a general triangle to a result that is true about any specific triangle. This is the kind of reasoning mathematics engages in and I really wish my fellow mathematics teachers spent time teaching students the difference between ‘proof’ and ‘confirmation’ or ‘verification’.

An Elegant Solution

With that out of the way, we can return to our opening problem for which we will use one of the elements of the preceding proof. Let’s remind ourselves of our problem with the figure once again.

Let us focus on ΔABX and ΔXYZ. If we can prove that they are equal in area, we can prove that all four triangles are equal in area because there is nothing special about ΔABX. We also observe that ∠BXY = ∠AXZ = 90° since they are interior angles of squares. What this means is that ∠AXB + ∠ZXY = 180° since the angles about a point equals 360°.

We now proceed with the realization that moving an object does not change its size. Let us rotate ΔXYZ about vertex X in a counterclockwise direction until line XZ coincides with line XA. This will give us the figure below.

Note that the point A is also the point Z’. However, to avoid confusion, let us refer to it as A. Now, since ∠AXB + ∠ZXY = 180°, it follows that ∠AXB + ∠AXY’ = 180° since rotating ΔXYZ does not change its internal angles. But this means that B, X, and Y’ form a straight line since the angle formed by a straight line is 180°. Further, X is the midpoint of BY’ since BX = XY’. Hence, X divides ΔABY’ into two equal halves. This means that the areas of ΔABX and ΔAXY’ are equal. But since rotating does not change the size of a triangle, this means that ΔABX and ΔXYZ are equal in area.

What we have seen is that this problem can be solved quite elegantly with just the basics of school geometry. In fact, we have not used any geometric results that are taught beyond grade 10. In other words, any high school student should be able to understand the solution. However, what the solution required is a willingness to move things around. There are times when we need to break things down before we can put them together in ways that reveal more than conceal.

1. I obtained this information from Math Stack Exchange. ↩︎