]> Pythagoras's Theorem

Pythagoras's Theorem

One of the most fundamental truths of Euclidean geometry – and, indeed, of the geometry of the real world, for all that its precision here is limited by the scale of the triangle in relation to the local curvature of space-time – describes a relationship among the sides of a right-angled triangle. It has been known since antiquity – it was known in ancient Indian cultures and (probably subsequently) in Mesopotamian cultures. More recently yet, the ancient Greek school of thought known as the Pythagoreans (notable for also sharing, with some Indian cultures, both vegetarianism and a belief in re-incarnation) attributed it to their (possibly mythical) founder, Pythagoras. It asserts that

The sum of the areas of the squares on the two orthogonal sides of a right-angle triangle is the area of the square on the third side.

The third side, facing the right angle, is called the hypotenuse of the right-angle triangle. (The addition of an extra term in the cosine of the angle between two sides generalises this rule to the case of non-right angles.)

Proof

Given a right-angle triangle, I need to show you how you can satisfy yourself that it has the claimed property. I'll explain what to do with the right-angle triangle you've got, illustrating each step with a right-angle triangle of my own. Hopefully, that'll help you see how to apply what I say to yours; and, thereby, lead you to see that the same is true of yours.

So take your right-angle triangle and rotate it through a quarter turn about its right-angle corner (in either sense). You now have two copies of it, with each side of either perpendicular to the corresponding side of the other; and the perpendicular sides all meet in the corner about which you rotated. In particular, for each of the perpendicular sides of your triangle, this gives you two perpendicular copies of that side: complete these to form a square, in each case.

I'm now going to show that, if we translate the combination of those two squares through the displacements along the two copies of our hypotenuse, its images fit around the original – abutting it, with no overlap and no gaps. From this it follows that, using these two translations, this figure tesselates the whole plane. If that's immediately apparent to you, you can skip all the diagrams and reasoning below. Since the two translations used are the same two that tesselate the plane with a square on either hypotenuse, this figure's area must in fact be equal to the area of that square, which is exactly the result to be proven.

To show that the tesselation works, let's first translate our pair of squares through one hypotenuse repeatedly (using alternating colours for each copy, to help tell them apart):

I trust it's clear that you can keep doing this indefinitely, in each direction, with each copy's concave corner at one end of the chosen hypotenuse surrounding the next copy's corner at the other end of that hypotenuse. Now let's take that train of pieces and translate it parallel to the other hypotenuse:

As you can see, and as I'm sure you'll find if you carefully perform the same operations with any right-angle triangle, the copies of the figure never overlap or leave gaps; and we can clearly continue repeating the pattern in each direction indefinitely, thereby tesselating the plane.

Now, indeed, I'm appealing to an intuition that figures tesselated by the same pair of translations must have equal area; but we can equally use our tesselated pair of squares and cut out one square on the hypotenuse, to look at the fragments of each square that it gives us. The tesselation property inescapably implies that these fragments, translated through various combinations of the two hypotenuse-displacements, can be reassembled into the two squares on the perpendicular sides – as can readily enough be verified. (Of course, this in turn relies on a different intuition, that cutting a figure up and rearranging its pieces doesn't change area. The two intuitions are, of course, equivalent.)

There are many other proofs, but this is my favourite – I don't know if anyone else ever thought of it before me, but I dreamed it up for myself (albeit I can't rule out having been prompted thereto by some long-forgotten encounter with it, or something similar), which gives me a natural bias in its favour. I do, that notwithstanding, consider it to be more elegant and to touch more closely the reasons for the truth of the result it proves. Finally – in case you thought I played a trick by my choice of which right-angle triangle to use, or suspect that the variants you've tried only worked because you didn't pick the quirky special case that'd break the reasoning – consider this animation (or a line-only variant), that varies the ratio of the perpendicular sides of the triangle through the full range of relative values possible:

A difference of squares

Pythagoras's theorem is usually stated in terms of adding two squares to make a third; but it can equally be expressed in terms of subtracting the square on one perpendicular side from the square on the hypotenuse, to get a figure with the same area as the square on the other perpendicular side. That's a bit more of a mouthful, but it leads to one interesting result. Embed one square on a perpendicular side in the square on the hypotenuse: we're left with a strip of the latter, whose width is the difference between the lengths of the hypotenuse and the chosen perpendicular side; this strip's area is equal to that of the square on the other perpendicular side.

Now, in general, if we take one square out of another, we'll be left with such a strip: and we can cut it in two pieces that we can reassemble into a rectangle. One side of this is the thickness of the strip, which was the difference between the sides of the two original squares. The diagram beloww shows that the other side of the rectangle is in fact the sum of the two squares' sides: cut along the red line, as indicated, and move one limb of the strip to abut the end of the other.

So, back in our right-angle triangle, we have the square on one perpendicular side equal to a rectangle whose sides are the sum of and difference between the other two sides. The nice thing about this is that we can run it backwards: given a rectangle, we can construct a square with the same area. When we halve the sum and difference of the rectangle, we get back the sides of the two squares whose difference was the rectangle. If we construct a straight line, whose length is the halved difference, and a circle, about one of its ends, whose radius is the halved sum, a second straight line perpendicular to the first, through its other end, will cut the circle at a point that completes a right-angle triangle with the halved sum and difference as hypotenuse and a side; a square on the other side will now have area equal to the rectangle we started with.

This, in turn, lets us construct a square whose area is equal to that of any given triangle. Half turn the triangle about the centre-point of one of its sides. Construct perpendiculars to one of its other sides, at each end; either these are the other side and its half-turned image, or one of them cuts off a triangle that we can translate through the displacement along our edge to rearrange our two copies of our original triangle into a rectangle. Cut this in half with the perpendicular bisector of one of its sides and apply the construction above to make a square with the same area as one of those halves – which is the area of our original triangle.

Now, any polygon can be cut up into triangles, for each of which we can construct a square of equal area. We can add the area of any two squares, by using their sides as perpendiculars of a right-angle triangle, whose hypotenuse square is the sum of the first two; so we can add up the squares from the triangles into which we cut up our polygon. We're thus able to construct a square whose area is equal to that of any polygon. (We can't do the same for a circle, though.) Since comparing the areas of squares is just a matter of comparing the lengths of their sides, this lets us compare the areas of arbitrary polygons.

See also

Discussion of right-angle triangles (and now simplices), whose sides are in whole-number ratios to one another.


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