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March 14, 2010, 4:15 pm

Square Dancing

By STEVEN STROGATZ

I bet I can guess your favorite math subject in high school.

It was geometry.

So many people I’ve met over the years have expressed affection for that subject. Arithmetic and algebra — not many takers there. But geometry, well, there’s something about it that brings a twinkle to the eye.

Is it because geometry draws on the right side of the brain, and that appeals to visual thinkers who might otherwise cringe at its cold logic? Maybe. But other people tell me they loved geometry precisely because it was so logical. The step-by-step reasoning, with each new theorem resting firmly on those already established — that’s the source of satisfaction for many.

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But my best hunch (and, full disclosure, I personally love geometry) is that people enjoy it because it marries logic and intuition. It feels good to use both halves of our brain.

To illustrate the pleasures of geometry, let’s revisit the Pythagorean theorem, which you probably remember as a2 + b2 = c2. Part of the goal here is to see why it’s true and appreciate why it matters. Beyond that, by proving the theorem in two different ways, we’ll come to see how one proof can be more “elegant” than another, even though both are correct.

The Pythagorean theorem is concerned with “right triangles” — meaning those with a right (90-degree) angle at one of the corners. Right triangles are important because they’re what you get if you cut a rectangle in half along its diagonal:

And since rectangles come up often in all sorts of settings, so do right triangles.

They arise, for instance, in surveying. If you’re measuring a rectangular field, you might want to know how far it is from one corner to the diagonally opposite corner. (By the way, this is where geometry started, historically — in problems of land measurement, or measuring the earth: geo = “earth” + metry = “measurement.”)

The Pythagorean theorem tells you how long the diagonal is, compared to the sides of the rectangle. If one side has length a and the other has length b, the theorem says the diagonal has length c, where

For some reason, the diagonal is traditionally called the “hypotenuse,” though I’ve never met anyone who knows why. (Any Latin or Greek scholars?) It must have something to do with the diagonal “subtending” a right angle, but as jargon goes, “subtending” is about as opaque as “hypotenuse.”

Anyway, here’s how the theorem works. To keep the numbers simple, let’s say a = 3 yards and b = 4 yards. Then to figure out the unknown length c, we don our black hoods and intone that c2 is the sum of 32 plus 42, which is 9 plus 16. (Keep in mind that all of these quantities are now measured in square yards, since we squared the yards as well as the numbers themselves.) Finally, since 9 + 16 = 25, we get c2 = 25 square yards, and then taking square roots of both sides yields c = 5 yards as the length of the hypotenuse.

This way of looking at the Pythagorean theorem makes it seem like a statement about lengths. But traditionally it was viewed as a statement about areas. That becomes clearer when you hear how they used to say it:

“The square on the hypotenuse is the sum of the squares on the other two sides.”

Notice the word “on.” We’re not speaking of the square “of” the hypotenuse — that’s a newfangled algebraic concept about multiplying a number (the length of the hypotenuse) by itself. No, we’re literally referring here to a square sitting on the hypotenuse, like this:

Let’s call this the large square, to distinguish it from the small and medium-sized squares we can build on the other two sides:

Then the theorem says that the large square has the same area as the small and medium squares combined.

For thousands of years, this marvelous fact has been expressed in a diagram, an iconic mnemonic of dancing squares:

Viewing the theorem in terms of areas also makes it a lot more fun to think about. For example, you can test it — and then eat it — by building the squares out of many little crackers. Or you can treat the theorem like a child’s puzzle, with pieces of different shapes and sizes. By rearranging these puzzle pieces, we can prove the theorem very simply, as follows.

Let’s go back to the tilted square sitting on the hypotenuse.

At an instinctive level, this image should make you feel a bit uncomfortable. The square looks potentially unstable, like it might topple or slide down the ramp. And there’s also an unpleasant arbitrariness about which of the four sides of the square gets to touch the triangle.

Guided by these intuitive feelings, let’s add three more copies of the triangle around the square to make a more solid and symmetrical picture:

Now recall what we’re trying to prove: that the tilted white square in the picture above (which is just our earlier “large square”— it’s still sitting right there on the hypotenuse) has the same area as the small and medium squares put together. But where are those other squares? Well, we have to shift some triangles around to find them.

Think of the picture above as literally depicting a puzzle, with four triangular pieces wedged into the corners of a rigid puzzle frame.

In this interpretation, the tilted square is the empty space in the middle of the puzzle. The rest of the area inside the frame is occupied by the puzzle pieces.

Now let’s try moving the pieces around in various ways. Of course, nothing we do can ever change the total amount of empty space inside the frame — it’s always whatever area lies outside the pieces.

The brainstorm, then, is to rearrange the pieces like this:

All of a sudden the empty space has changed into the two shapes we’re looking for — the small square and the medium square. And since the total area of empty space always stays the same, we’ve just proven the Pythagorean theorem!

This proof does far more than convince; it illuminates. That’s what makes it “elegant.”

For comparison, here’s another proof. It’s equally famous, and it’s perhaps the simplest proof that avoids using areas.

As before, consider a right triangle with sides of length a and b and hypotenuse of length c, as shown below on the left.

Now, by divine inspiration or a stroke of genius, something tells us to draw a line segment perpendicular to the hypotenuse and down to the opposite corner, as shown above on the right.

This clever little construction creates two smaller triangles inside the original one. It’s easy to prove that all these triangles are “similar” — which means they have identical shapes but different sizes. That in turn implies that the lengths of their corresponding parts have the same proportions, which translates into the following set of equations:

We also know that

because our construction merely split the original hypotenuse of length c into two smaller sides of length d and e.

At this point you might be feeling a bit lost, or at least unsure of what to do next. There’s a morass of five equations above, and we’re trying to whittle them down to deduce that

Try it for a few minutes. You’ll discover that two of the equations are irrelevant. That’s ugly; an elegant proof should involve nothing superfluous. With hindsight, of course, you wouldn’t have listed those equations to begin with. But that would just be putting lipstick on a p… (the missing word here is “proof”).

Nevertheless, by manipulating the right three equations, you can get the theorem to pop out. See the notes below for the missing steps.

Would you agree with me that, on aesthetic grounds, this proof is inferior to the first one? For one thing, it drags near the end. And who invited all that algebra to the party? This is supposed to be a geometry event.

But a more serious defect is the proof’s murkiness. By the time you’re done slogging through it, you might believe the theorem (grudgingly), but you still might not see why it’s true.

Leaving proofs aside, why does the Pythagorean theorem even matter? Because it reveals a fundamental truth about the nature of space. It implies that space is flat, as opposed to curved. On the surface of a globe or a bagel, for example, the theorem needs to be modified. Einstein confronted this challenge in his general theory of relativity (where gravity is no longer viewed as a force, but rather as a manifestation of the curvature of space), and so did Riemann and others before him, when laying the foundations of non-Euclidean geometry.

It’s a long road from Pythagoras to Einstein. But at least it’s a straight line… for most of the way.

NOTES:

• The ancient Babylonians, Indians and Chinese appear to have been aware of the Pythagorean theorem several centuries before Pythagoras and the Greeks. For more about the history and significance of the theorem, as well as a survey of the many ingenious ways to prove it, see:

E. Maor, The Pythagorean Theorem: A 4,000-Year History (Princeton University Press, 2007).

• Incidentally, on p. xiii of his book, Maor explains that the word “hypotenuse” means “stretched beneath,” and points out that this makes sense if the right triangle is viewed with its hypotenuse at the bottom, as depicted in Euclid’s Elements. He also notes that this interpretation fits well with the Chinese word for hypotenuse: “hsien, a string stretched between two points (as in a lute).”

• If you enjoy seeing different proofs, a nicely annotated and extensive collection of 84 of them — with creators ranging from Euclid to Leonardo da Vinci to President James Garfield — is available here.

• With any luck, the first proof above should have given you an Aha! sensation. But to make the argument completely airtight, we also need to prove that the pictures aren’t deceiving us — in other words, they truly have the properties they appear to have. A more rigorous proof would establish, for example, that the outer frame is truly a square, and that the medium and small squares meet at a single point, as shown. Checking these details is fun and not too difficult.

• Here are the missing steps in the second proof above. Take this equation:

and multiply it by a on both sides to get

Similarly massaging another of the equations yields

Finally, substituting the expressions above for d and e into the equation c = d + e yields

Then multiplying both sides by c gives the desired formula:

• Thanks to George Hart of the Museum of Mathematics for sharing his hands-on demo using Pythagorean crackers, Carole Schiffman for her comments and suggestions, and Margaret Nelson for preparing the illustrations.

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