(by Polymath, with thanks to several commenters on the original post)

This page is part of the Advanced High School Math Project.

So the CMFE is this: Put *n* equally spaced marks around the
circumference of a unit circle (radius of 1). Then from any one of
those marks, draw the chords that connect it to all the other (*n*-1) marks. The lengths of these chords are then multiplied together, and amazingly, that product is always *n*. So: 532 equally spaced marks, 531 chord lengths, and the product is incomprehensibly 532. Here is the starting diagram for *n*=9:

The proof hinges on considering this
unit circle to be the very important complex unit circle—that is, the
circle of radius 1 centered at the origin of the complex plane. It is
a well-known fact in complex number analysis that the *n*th roots of 1 (or, as mathematicians say, the roots of unity) are *n*
points, equally spaced around a unit circle, starting at the complex
number 1, which appears 1 unit to right of the origin on the real axis
of the complex plane. This is a direct result of DeMoivre's theorem. You'll notice that if you superimpose a set of
axes on the diagram above, the number 1 would fall right at the point
from which the chords emanate. That means that collectively, the *n* marks are indeed the *n*th roots of 1.

If you call one of the other endpoints *r*, then the segment
connecting it to the number 1 is the vector that represents the complex
number 1-r. Which means the length of that segment is:

Okay. So with that in mind, note the following polynomial identity, which holds over the complex numbers:

(*z* is the variable, the *r*'s are the *n*th roots of 1.) This is an identity, because it's always true: note that both polynomials have the same degree (*n*), the same leading coefficent (1), and the same roots (the *r*'s). The only way this can happen is if the polynomials are the same.

Now make the following substitutions: the first root of 1 is
clearly just 1, so r_1 turns into 1. Then also replace all instances
of *z* with *Z*+1 (for reasons that will become clear). We now have the first line below:

The second line above comes from using the binomial expansion on the left and the obvious simplification on the right. Since the 1's cancel on the left, we can divide both sides by Z to get:

Note that this is still an identity, and true for all values of Z (there's a small glitch here, explained below, so check that out before you claim I'm wrong here). So I pick 0 as my value for Z. This gives the way simpler equation:

But if you take the absolute value of both sides, you get simply that the lengths of all the chords have a product of n. Which is JUST ABSURDLY COOL!

Glitch: when you divide both sides by Z, you have to exclude the
case where Z=0, or the division is meaningless. Thus the equation
after the division should only be true if Z is *not* zero. Which means
I ought not be allowed to substitute Z=0 into it in the next step. A commenter pointed out, though, that the left and right hand sides of the equation are both continuous functions of Z that agree everywhere other than Z=0. But this means they also must agree at Z=0, or there would have to be some discontinuity there. So, despite being derived by dividing by Z, they still agree everywhere, even at Z=0.

So. Now you know the CMFE. The main reason it's so cool is this:
the diagonals of a square with a side-length of 1 have lengths
expressible with the square root of 2 (notated SQR(2) hereafter). The
diagonals of a pentagon with a side-length of 1 have lengths
expressible with SQR(5). The diagonals of a hexagon with a side-length
of 1 have lengths expressible with SQR(3). Notice that those are the
kinds of lengths multiplied in the CMFE. Thus it is no surprise that
products of SQR(5), for example, result in the number 5. In other
words, it isn't a coincidence that SQR(2) shows up in 4-sided figures, SQR(5) shows up in 5-sided (and 10-sided) figures, and SQR(3) shows up in 6-sided (and 12-sided) figures. If the number *k* shows up as SQR(*k*) in the expression for the length of a diagonal of a regular *n*-gon, then *k* must be a factor of *n*. (Note that not all regular *n*-gons have diagonals that can be written in that form—some diagonal lengths are not expressible using any combination of roots.) Maybe you have to be a math geek to find these facts cool, but I guess that makes me a math geek.