Maybe the best thing about the Pythagorean theorem is how it puts math and non-math people on a pretty equal footing. We all know what it says (right triangle, squares of sides, hypotenuse), we all agree it’s Important Math with a capital M… and most of us don’t have much idea, if any, why it’s true. Seriously. Ask another math person if you don’t believe me. If you’re lucky, they might point you to a picture that looks more or less like this:

This is sometimes called a proof without words, but here are a few words to guide you, just in case. We’ve got the usual notation: *a* and *b* are sides of a right triangle, *c* is the hypotenuse. We build a super-square with side length *a*+*b *and break it up in two ways. On the right, we divide it into one (white) sub-square with side length *a* (area *a*^{2}), another with side *b* (area *b*^{2}), and four (colored) copies of the triangle. On the left, we rearrange the four triangles so that their complement is a new white sub-square with side *c* (area *c*^{2}). White area on the left = white area on the right, so *c*^{2} = *a*^{2} + *b*^{2}.

Lovely as this is, it feels like a nifty conjuring trick. The Pythagorean equation is the most direct thing in the world, *a*^{2} + *b*^{2} = *c*^{2}, and the best we can do is to rearrange triangles inside a big square? Surely there must be a way to cut up *c*^{2} into *a*^{2} and *b*^{2} directly.

Well, there is! I first ran across what I’m about to show you a few weeks ago, loved it, and was surprised that (1) I hadn’t seen it before and (2) it doesn’t seem to be widely known, though the idea actually goes back to Euclid. Perhaps you’ll feel the same way once you see it. Here goes:

*The set-up*. What’s so special about right triangles? Well, one thing is that they have an amazing self similarity property. Draw a line segment out from the vertex with the right angle toward the hypotenuse and perpendicular to it. It divides our original right triangle up into two smaller ones:

Let’s stick to the same notation we had before. Let *c* be the hypotenuse of our original right triangle, running along the bottom. Let *a* and *b* be the sides, *a* the one on the left, *b* the one on the right. Then *a* is the *hypotenuse* of the green right triangle on the left, and *b* is the hypotenuse of the blue right triangle on the right. Obviously (foreshadowing), the areas of the two smaller triangles add up to the area of the original.

The self similarity property is that the two smaller right triangles are both similar to the original one! In other words, all three triangles have the same three interior angles, which means that you can rotate and scale each one into any of the others. Put simply, all three triangles have the same shape. Can you see why? Let’s compare those interior angles: the big triangle has a right angle of 90 degrees, and two others, which we will call (say the one on the left, in the green triangle) and (on the right, in the blue triangle). The key point is that the angles of any triangle have to add up to 180 degrees, so two angles of a triangle always determine the third. The green triangle has an angle of on the left that it inherits from the original big triangle, and a 90 degree angle in the middle, so its third angle, at the top, must be . (Essentially: the green triangle and the big triangle have two angles in common, so they must have all three in common.) Similarly, the blue triangle has an angle of on the right that it inherits from the original triangle, and a 90 degree angle in the middle (again, two angles in common), so its third angle must be . Same angles, same shape.

*The pay-off*. Stare for a minute at those three similar triangles, with the areas of the two smaller ones adding up to the area of the bigger one. Wouldn’t it be great if the triangle with hypotenuse *a* had area *a*^{2}, the one with hypotenuse *b* had area *b*^{2}, and the one with hypotenuse *c* had area *c*^{2}? It’s not true, of course. But it’s almost true! If you read my last post, you know that in fact the area of a right triangle with hypotenuse *c* and interior angle is

.

The green and blue triangle have exactly the same interior angles as the big one. So their areas are given by the same exact formula, with *c* replaced by *a* and *b*, respectively. The areas have to add up, so we have:

.

Now just divide out the common factor of , and you’re left with

.

*What just happened?* The way Pythagoras’s equation fell out of equating areas may seem like a bit of a magic trick too, but it’s actually based on the very fundamental idea of *scale invariance*. To recap: we (1) wrote a completely explicit formula for the area of a right triangle, (2) equated formulas corresponding to equal areas, and (3) found that the bulk of the formulas, everything except the part corresponding to the Pythagorean theorem, went away. The key to understanding all that is this picture:

It shows the area of a right triangle embedded in the area of the corresponding square, with the hypotenuse matching one side of the square. The precise formula for the ratio of the areas, , doesn’t matter so much — what matters is that *when we blow this picture up or down, the ratio of the areas doesn’t change*. That’s scale invariance. If the two smaller triangles add up to the big triangle, then the squares corresponding to the smaller triangles have to add up to the square corresponding to the big triangle. And that’s exactly the Pythagorean theorem.

I like to think of *a*^{2}, *b*^{2}, and *c*^{2} as *units of area* corresponding to each triangle. In a nutshell, the Pythagorean theorem decomposes a right triangle into two smaller, similar ones, and says that if the triangles add up, the units of area have to add up too. It’s deep, it’s direct, and I’ll never forget it. How about you?

Interesting. You had to rely on sin(theta) to convince me of scale invariance. Can you find a way to show scale invariance (i.e. that the all 3 triangles are similar) and that the area of the triangles is the $f* hypotenus^2$ without using Greek letters or transcendental functions (which of course all drop out at the end)? I’d like to show this to Dylan, who is beginning to understand multiplication and geometry, but his eyes will roll over if I mention angles and sinus functions… 🙂

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You can avoid the trig, but you have to talk about angles at least a bit, because the similarity comes down to triangles having the same shape if they have two angles (the corner and one other) in common. Once you have that, you can argue that because of the similarity, each square is a fixed multiple of its triangle, and then the multiple, whatever it is, drops out. I like having the explicit formula for the multiple around, at least for the first pass, because it makes the whole thing seem more concrete, but that might be just me.

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