## Origami-inspired Proof of the Pythagorean Theorem

While working with math teachers Richard Chang and Randi Currier of the Buckingham Browne & Nichols Middle School to come up with problems that use polynomials and rational expressions, we stumbled upon an origami-inspired proof of the Pythagorean theorem that I’d like to share.

I haven’t seen this proof before, but there are so many proofs of the Pythagorean theorem that it wouldn’t surprise me if this has already been discovered. If you’ve seen this proof published elsewhere, please send email to Girls’ Angle and let me know.

I am aware of other “origami” proofs of the Pythagorean theorem, although the ones I’ve seen don’t particularly use origami so much as that they take some diagram used to illustrate a proof and then render the lines in the diagram as creases in paper.

This particular proof is not elegant, but it does show how one might discover the Pythagorean theorem through a natural line of exploration starting with an origami square. Also, it really uses the notion of an origami fold as its key ingredient.

All you have to do is think about what happens when you fold a corner of the origami square to a point along one of the further edges, as illustrated below.

That single fold produces 3 triangles and a quadrilateral.

Notice that the 3 triangles are all similar right triangles. (Here, we must use the fact that a square has 4 right angles.)

Since the entire model is determined by the parameter $x$, a natural question to ask is: What are the lengths of all the other line segments in terms of $x$?

To get the answer, we will repeatedly apply two facts. First, that the 3 triangles are similar to each other. And second, that the edge lengths along a side of the square add up to the side length of the square, which we’ll go ahead and take to be one unit for now.

We’ll start by letting $y$ be the length of the other leg of $T_1$. We’ll then express the other lengths in terms of $x$ and $y$ and find an equation that relates the two. By solving for $y$ in terms of $x$, we will effectively have succeeded in expressing the lengths in terms of $x$.

Here goes!

We add the label $y$ to the diagram:

The hypotenuse of $T_1$ and the leg of length $y$ make up what was the right side of our square, so the hypotenuse of $T_1$ has length $1 - y$:

One of the legs of $T_2$ extends $x$ to the full side of the square:

Using the fact that $T_2$ and $T_1$ are similar, we can find the other side lengths of $T_2$:

One leg of $T_3$ and the hypotenuse of $T_2$ form what was the upper side of the origami square:

Next, we use similarity of $T_3$ with $T_1$ to see that its other sides have length:

$(1 - (1-x)\frac{1-y}{y})\frac{y}{x}$ and $(1 - (1-x)\frac{1-y}{y})\frac{1-y}{x}$.

So far, we’ve used the right, bottom, and top sides of the original square. So now we use the left side. Studying the figure, we see that the original left side of the square has become one leg of $T_2$ and the last two sides we computed of $T_3$. Thus,

$(1-(1-x)\frac{1-y}{y})\frac{y}{x}+(1-(1-x)\frac{1-y}{y})\frac{1-y}{x}+(1-x)\frac{x}{y}=1$.

If you solve for $y$ in terms of $x$ using this equation, you will find that $y = \frac{1-x^2}{2}$.

Thus, the legs of $T_1$ are $x$ and $\frac{1-x^2}{2}$ and its hypotenuse has length $\frac{1+x^2}{2}$.

Notice that as $x$ varies from 0 to 1, the ratio of the leg lengths of $T_1$ covers the entire range from 0 to infinity (i.e. the range of $\frac{1-x^2}{2x}$ is $(0, \infty)$ when $x$ is restricted to the interval $(0, 1)$). This means that a representative of every similarity class of right triangle can be folded by making a suitable choice of $x$.

Therefore, by scaling the origami square if necessary, we can make $T_1$ congruent to any given right triangle.

The form of the expressions for the lengths of the sides of $T_1$ in terms of $x$ suggests the possibility of finding an explicit relationship between these 3 side lengths, especially since one of the side lengths is just $x$, and if we can find such a relationship that is also homogeneous in these lengths, then the relationship will hold for all right triangles since a homogeneous relationship is preserved by scaling. And, indeed, the desired algebraic identity is $x^2 + (\frac{1-x^2}{2})^2 = (\frac{1+x^2}{2})^2$, hence, the Pythagorean theorem. (Making note of this algebraic identity without knowing the Pythagorean theorem might be a bit of a trick, but it doesn’t seem unreasonable to think it possible.)

(Note that $x^2 + \frac{1-x^2}{2} = \frac{1+x^2}{2}$ as well, and this corresponds to the equation $a^2 + b = c$, where $a$ and $b$ are the legs of a right triangle with hypotenuse of length $c$. However, this identity is not homogeneous in the leg lengths, so it is not true that $a^2 + b = c$ for all right triangles. The equation $a^2 + b = c$ holds only for right triangles where $b + c = 1$, i.e. those folded from a unit square.)

In an earlier post, we described David Gale’s method for constructing Pythagorean triples. There, our aim was to find Pythagorean triples, not prove the Pythagorean theorem, so when we spoke of finding the lengths of $T_1$, we had in mind a method that made use of the Pythagorean theorem. One could say that the content of this blog post is that you don’t have to assume the Pythagorean theorem there because you can deduce it instead. In other words, one-fold origami can deliver the Pythagorean theorem from scratch as well as all Pythagorean triples.