How I Learned to Stop Worrying and Love Pythagoras


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:

Pythagoras-proof-anim.svgThis 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 a2), another with side b (area b2), 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 c2). White area on the left = white area on the right, so c2 = a2 + b2.

Lovely as this is, it feels like a nifty conjuring trick. The Pythagorean equation is the most direct thing in the world, a2 + b2 = c2, and the best we can do is to rearrange triangles inside a big square? Surely there must be a way to cut up c2 into a2 and b2 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 \theta (say the one on the left, in the green triangle) and \phi (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 \theta 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 \phi. (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 \phi 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 \theta. 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 a2, the one with hypotenuse b had area b2, and the one with hypotenuse c had area c2? 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 \theta is

\frac14 \cdot \sin(2 \theta) \cdot c^2.

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:

\frac14 \cdot \sin(2 \theta) \cdot a^2 + \frac14 \cdot \sin(2 \theta) \cdot b^2 = \frac14 \cdot \sin(2 \theta) \cdot c^2.

Now just divide out the common factor of \frac14 \cdot \sin(2 \theta), and you’re left with

a^2 + b^2 = c^2.

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, \frac14 \cdot \sin(2 \theta), 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 a2b2, and c2 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?


For Pi Day 3/14/16: Triangle to Circles, and Back Again

Around this time last year, we heard a lot about 3/14/15 being a once-in-a-century Pi Day, because the date, including the year, matches the first few digits of pi (3.1415…). But the next digit after 3.1415 in the decimal expansion is 9, giving us 3.14159… So pi is a lot closer to 3.1416 than it is to 3.1415, which is why I want to wish you a Happy once-in-a-century Pi Day today, 3/14/16!

Pi fascinates us because it appears in so many unexpected places. Since pi is ultimately about circles (it’s defined, more or less, as the circumference of a circle of diameter 1), what this really means is that circles appear in many unexpected places. I want to show you one.

You need something with a corner (I used a Rubik’s cube), a cable or string of some kind, and two fixed, firm endpoints (I used the ends of a towel rod). Stretch the cable between the two endpoints across the corner. Now move the corner from one side to the other between the two endpoints, keeping the cable stretched across all three, like this:

Question: what’s the shape that the corner traces out? A triangle, a circle, a parabola, an ellipse, something else?

I told you there was a circle coming, so, yes, it turns out to be a circle. (Well, a semicircle, tracing out 180 degrees, or pi!) I find this a little surprising, because this set-up is quite different from a compass, which is based on spinning a fixed length around a center. Here, the length of cable stretched across the corner extends and then contracts as you move the corner from one side to the other. One good reason to try this out physically is that you can really feel the extension and contraction!

Here is an animation that might start to convince you:


The mathematical fact afoot here is Thales’s Theorem, which says that if you put two points on two opposite ends of a circle (so the line between them is a diameter), then the line segments connecting those points with any third point on the circle will meet at a right angle. To give you an idea of why this is true, and play with pi some more, here’s a picture:


The idea here is that if O is the center of the circle, then the line segments OA, OB, and OC all have the same length, because they’re all radii of the circle. This makes the two triangles AOB and BOC isosceles: each triangle has two equal sides, hence two equal angles, as illustrated in the picture. With this notation, the sum of the angles of triangle ABC is 2 \alpha + 2 \beta. But we know that the sum of angles of a triangle is 180 degrees (pi again!). Dividing by 2, we find that \alpha + \beta must be 90 degrees, which is Thales’s theorem.

One last thing: this picture also leads to a formula for the area of the right triangle in terms of the length of the hypotenuse. Let d be this length, i.e., the diameter extending from A to C. Look at triangle BOC again. The sum of its angles must also be 180 degrees, and since we know from above that 2 \alpha + 2 \beta = 180, we find that the missing angle, between segments OB and OC, must be 2 \alpha. This means the height of our triangle is \sin (2 \alpha) times the radius of the circle, or \sin (2 \alpha) \cdot d/2. Since the area of a triangle is half the base times the height, we find

{\rm{Area}} = \frac14 \cdot \sin(2 \alpha) \cdot d^2.

This last formula is actually connected in a very interesting way to the Pythagorean theorem. But that’s another blog post.