Three Square Problem and Variations

September 22, 2014

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?

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