I just saw an absolutely charming problem on Numberphile that I was shocked never to have seen before. They call it the Three Square Problem (featuring Professor Zvezdelina Stankova).

## Three Square Problem

- Prove that $latex \alpha + \beta + \gamma = 90^\circ$

It feels, as Prof. S says in the video, like a beautiful conjecture.

I highly recommend trying to come up with a proof. There are many (54!), and I’ve come up with about seven since I saw the video. More intriguing, though, is the question of whether this type of thing happens in any other cases. Let’s look at a picture that summarizes the most surprising part of this problem.

It is not particularly surprising that $latex \alpha = 45^\circ$. The shock is that $latex \gamma + latex = 45^\circ$ as well. Are there other right triangles we can draw on a grid that have this property? The answer, it turns out, is yes.

*Prove that $latex \gamma + \beta = 45^\circ$.*

Let’s go into even greater generality. Suppose we have two rectangles with integer side lengths. (Everything that comes later will refer to the picture below.)

**Big Question 1**: For what collections A, B, C, and D will the $latex x + y = 45^\circ$?**Big Question 2**: Given any A and B, does there always exist a C and D so that $latex x + y = 45^\circ$?

Here are some of the answers I’ve discovered so far:

A = 2, B = 3, C = 5, D = 1.

A = 3, B = 4, C = 7, D = 1.

A = 4, B = 5, C = 9, D = 1.

*Prove the examples above all satisfy $latex x + y = 45^\circ$.**Define a pattern in the numbers above. Will always work?*

Here’s another observation about the list of numbers above. $latex A^2 + B^2 = \frac{C^2 + D^2}{2}$! What could that have to do with things?

**Wild Conjecture**: If $latex A^2 + B^2 = \frac{C^2 + D^2}{2}$, then $latex x + y = 45^\circ$.

*Prove or disprove the wild conjecture.*

Here’s another sequence of solutions where $latex x + y = 45^\circ$:

A = 3, B = 5, C = 4, D = 1

A = 5, B = 7, C = 6, D = 1

*Prove that these sets of A,B,C,D satisfy $latex x + y = 45^\circ$.*

What happens if we sum the squares? Once again, $latex 3^2 + 5^2 = 9 + 25 = 34$ is double $latex 4^2 + 1^2 = 17$. This is evidence in favor of the Wild Conjecture. Not nearly a proof.

Notice what’s happened here. We began with an isolated question about a cool relationship between specific angles. By asking the next natural questions, we have very surprising variations of that original relationship, and an entirely different pattern emerging in the squares of the sides of the rectangles. True understanding comes not from solving one problem, but in solving families of problems in multiple ways, and following the natural questions as far as we can.

I’ll leave you with a suggestive diagram for a proof of the original three square problem.Here we have the original 1 by 2 right triangle (AEB) next to a scaled up 1 by 3 right triangle (CDB). Can you see why their two small angles (the ones at B) sum to 45 degrees? Do you see how this picture could generalize?

## Comments 8

Author

I’m burying one of the leads a little… let me put another provocation out there:

Pick two numbers, say 7 and 4. We’ll have one of our angles come from the small side of a right triangle with legs of length 7 and 4. Now build the next right triangle with legs of length 7+4 and 7-4, i.e., 11 and 3.

Then both of the following is true:

1. The angles from these triangles sum to 45 degress

2. The sum of 11 squared and 3 squared is double the sum of 7 squared and 4 squared.

Does this trick always work?

With a little modification you can use this problem to prove arctan(1) + arctan(2) + arctan(3) = pi.

Visual proofs are awesome!

Author

I agree! What’s interesting is that you get a slew of other identities, like:

arctan(2/3) + arctan(1/5) = pi/4, or

arctan(4/7) + arctan(3/11) = pi/4.

In fact, if a

I just proved arctan(a/b)+arctan((b-a)/(b+a))=45degrees using just a picture inspired by your final diagram.

It was easy to use that equation to prove the tangent angle sum formula.

I used one of my favorite approaches to proofs: do the proof backwards first.

I don’t know what her 54 proofs are. I just came up with one that was not like hers in the video. Made me happy.

Author

I posed this problem to a group of four middle schoolers, and they came up with three methods of solving it. One I hadn’t thought of ever remotely. It was maybe the most elegant, too.

It’s a really satisfying problem.

Your conjecture that A^2+B^2=(C^2+D^2)/2 is false simply because of commutativity. If A=5, B=7, C=1 and D=6 works (and it does) then so must A=1, B=6, C=5, B=7. In which case A^2+B^2=2*(C^2+D^2). Furthermore, if arctan(5/7) + arctan(1/6) = 45°, then arctan(5/7) + arctan(100/600) = 45°. But 5^2 + 7^2 =/= (100^2+600^2)/2. So, there’s still some interesting things there if you insist on reduced fractions, but your conjecture as stated is false.

I think that the sum and difference formulas for sines and cosines are interesting to use here. The only place where sinX=cosX is at 45°. So, sin(x+y) = cos(x+y). Crossing out common denominators you end up with AD+BC = BD-AC. From that, if you give me a triangle with integer legs, I can give you another pair of legs to make the sum of the angles equal to 45°.

Well putting it algebraic terms the side of the square or the height of the triangle is x. x could be anything but it disappears at a certain point so you can keep it there. If that is true then the bases of the triangles are x, 2x, and 3x. So we all know (a) is 45 but that’s easy. Now we can add the reverse tangent of 1/2 and the reverse tangent of 1/3 third. (b) has x/2x so it turns into 1/2. Then take the inverse tangent. Do the exact same with (c). x/3x so it is the inverse tangent of 1/3. This I believe is a close solution.