23 July 2007

proofs and deceptions by dissection

A friend of mine pointed me to this puzzle, which shows a triangle broken up into four pieces, and then the four pieces are rearranged in such a way that it appears that they now have total area one less than they did before. [edit, 5:43 pm: an anonymous commenter said that the link was broken. It's fixed now.]

What's really happening is that the "triangle" on the top and the "triangle" on the bottom are not triangles. However, it's nearly impossible to tell this by eye. The top "triangle" actually has its hypotenuse bent slightly inward; the bottom "triangle" has its hypotenuse bent slightly outward. If you superimposed the bottom "triangle" on the top one, there would be a long, thin sliver of area which would be contained in the bottom triangle but not in the top one; this turns out to have an area of exactly 1, making up for the "missing square".

One naturally expects the large "triangles" -- which appear to be right triangles with leg lengths 13 and 5 -- to be similar to the two small triangles. But the dark blue triangle has legs 8 and 3, and the green one has legs 5 and 2. It seems natural to compare the three right triangles by looking at the size of their smallest angle (since they're right triangles, this determines the shape uniquely); these are arctan(5/13), arctan(3/8), and arctan(2/5), or 21.04, 20.56, and 21.80 degrees, respectively. So the "bending" is on the order of one and a quarter degrees, which is hardly visible to the naked eye, especially since we naturally expect that a line that "looks" straight really is straight.

It's not a coincidence that the various numbers which are involved in this figure are Fibonacci numbers (2, 3, 5, 8, 13). The Fibonacci numbers have the property that the ratio between any two consecutive members of the sequence is very close to φ = (1+√5)/2, or about 1.618; the sequence approaches this limit quite quickly. For example, 13/8 = 1.625. The relevant ratios here are the ratios between Fibonacci numbers which are two positions apart (say, 5 and 13); these very quickly approach φ2 = 2.618...

I hadn't seen this particular example before, but there are other, similar puzzles which rely on these facts about the Fibonacci numbers; see Dissection Fallacy at MathWorld, in which an 8-by-8 checkerboard is broken up into pieces which appear to be rearranged to have area 63 or 65. The underlying idea is very much the same -- that we don't see the slight "imperfections of shape".

This isn't particularly special to the Fibonacci numbers, but can actually occur whenever the slopes of two lines are rational numbers which are close to each other; see, for example, the Curry Triangle at MathWorld, which relies on the fact that 2/5 and 3/7 are approximately equal. A bunch of Java applets illustrating such "dissection fallacies" can be found at Cut The Knot, and another illustration of this particular dissection can be found at Missing square puzzle on Wikipedia.

There's also a classic proof by dissection of the Pythagorean theorem (see about a third of the way down); essentially we can divide the two squares on the legs of a right triangle into pieces and rearrange them to form a square on the hypotenuse. Pedants will complain that you can't do this without proving that when you move regions around and rotate them their areas don't change -- and I suppose this is technically true -- but I think that these sorts of proofs give a much better intuition for why the Pythagorean theorem is true than, say, Euclid's proof.

1 comment:

Anonymous said...

The first link ('this puzzle') is broken