Folding Pythagorean Triples with Origami

One-Fold Pythagorean Triple

The fascination with Pythagorean triples continues…so in addition to the method of parametrizing rational points on the unit circle and the matrix method, here’s another way to generate them using origami. To the best of my knowledge, the method was discovered by the late David Gale (the same mathematician who, together with Lloyd Shapley, created the stable marriage algorithm that Emily Riehl discusses in Girls’ Angle’s most recent WIM Video).

Start with a square of origami paper, which we’ll say has sides of unit length. Pick a point on the lower edge of the paper at a distance r from the lower right corner. Fold the upper right corner to this point. The result will contain the right triangle labeled T in the figure above.

If you compute the lengths of the sides of triangle T, you will find that they are:

r, \frac{1}{2}(1 - r^2), and \frac{1}{2}(1 + r^2).

Notice that if r is a rational number, so will be the lengths of the sides of triangle T.

Suppose ra/b, where a and b are relatively prime integers not of the same parity. If we make the fold and then scale by a factor of 2b^2, you will obtain the primitive Pythagorean triple (2ab, b^2 - a^2, b^2 + a^2).

For example, if r = 1/2, you will obtain the famous 3-4-5 triangle. If r = 2/3, you will construct a 5-12-13 triangle, and so on.

Now, here’s a silly question: Because of this, should we regard one-fold origami as rather sophisticated, or, should we regard Pythagorean triples as simple as folding a piece of paper?

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5 Responses to Folding Pythagorean Triples with Origami

  1. Pingback: Getting Pythagorean Triples with Hyperbolic Functions | Girls' Angle

  2. themathpass says:

    I vote for one-fold origami being sophisticated 😉 I love math, and I love origami, so I think this is very cool!

  3. Pingback: Origami-inspired Proof of the Pythagorean Theorem | Girls' Angle

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