My proof of the coolest math fact ever (CMFE) is below (I will explain after the proof why I think it's so cool). I don't claim that this is the easiest proof, but shorter proofs probably use math that I don't entirely understand. The CMFE was stated in a previous post, but I will restate it here. Coincidentally, earlier today, someone commented on that post with a hint for the proof, and I'm going to assume that person knows what he/she is talking about, because that's exactly the right first step.

Unfortunately, there's a little glitch in the proof that I've never quite understood, and maybe someone could clear it up for me. It doesn't sink the proof, but I'm not quite sure I understand why not.

Okay. 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 an amazing 532. Here is the starting diagram for *n*=9:

The proof hinges (as the commenter suggests) 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. 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 (but this is the glitch, so see below 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 is only true if Z is *not* zero. Which means I ought not be allowed to substitute Z=0 into it in the next step. The fix, I think, is to take the limit as Z approaches 0 on both sides of the equation, leaving you with the same result, but never requiring Z to actually equal 0. Note that if Z=0, then *z*=1, which is the crucial point from which all the chords emanate. So is this limit process still legal because excluding one endpoint of a segment doesn't change its length? Or am I confusing a general identity with the application to a specific case? A clear explanation of this would be greatly appreciated.

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 expressable with the square root of 2 (notated SQR(2) hereafter). The diagonals of a pentagon with a side-length of 1 have lengths expressable with SQR(5). The diagonals of a hexagon with a side-length of 1 have lengths expressable 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 transcendental~~ UPDATE: cat lover points out in the comments that those diagonal lengths, while not expressible with roots, are probably not transcendental; I agree completely.) I don't know why I find that sooooo cool, but I'm a math geek, and I'm not going to try to make excuses for my opinions of coolness.

QED.

Wait, you're a math geek? #282?

Posted by: Your Wife | July 05, 2006 at 10:23 PM

Hmmm wha? Oh... oh yeah, yeah, I totally understood all of that....... Yeah.

:P

Posted by: Le Bohemian | July 05, 2006 at 11:42 PM

OK, let's finish up that last little issue. You had z^n - 1 = (z-1)(z-r_2)...(z-r_n). Divide through by z-1:

(z^n-1)/(z-1) = (z-r_2)...(z-r_n). But it's a basic theorem about partial sums of geometric series that the left side is 1 + z + z^2 + ... + z^{n-1}. Now plug in 1 on both sides: you get 1 + 1 + ... + 1 = n on the left side, and (1-r_2)...(1-r_n) on the right side, which as you mentioned earlier must have a magnitude equal to the product of the chords. QED!

Posted by: dave glasser | July 06, 2006 at 12:27 AM

Am I studpid or doesn't CMFE work for n=2?

It only works if the two marks are respectively on the exact opposite side of the circle.

Posted by: Ringelnatz | July 06, 2006 at 06:02 AM

~~~ Me stupid ~~~~

I missed the words "equally spaced"...

~~~ Me stupid ~~~~

Posted by: Ringelnatz | July 06, 2006 at 06:05 AM

Dave,

I like your use of the formula for a partial geometric sum...it simplifies the proof a lot. But I think it still has the same problem: dividing by z-1 is only valid if z is never 1, and then later you use z=1. Same issue in a different outfit, I think.

Posted by: Polymath | July 06, 2006 at 07:08 AM

Hmm. So, I guess, you're happy to believe that 1 + z + ... + z^{n-1} = (z-r_2)...(z-r_n) for all z other than 1, right? (Because the only thing I would have invalidated by the division would have been z=1.) And both sides are degree n-1 polynomials, so like you said in a different part of the proof they have to be the *identical*, because two polynomials are just not going to disagree in precisely one point.

Another way to look at it is that we're just manipulating objects in the ring of polynomials C[z], so that the statement that (z^n - 1)/(z-1) = 1 + z + ... z^{n-1} is not a statement about values that functions take at various values of z, but of two polynomials being the same object.

(I guess in retrospect you can apply that exact argument to your original proof too!)

Posted by: dave glasser | July 06, 2006 at 08:12 AM

Now, I always thought the coolest math fact ever was e^(pi*i) + 1 = 0.

This is cool, too, though.

Posted by: Master_Bratac | July 06, 2006 at 03:00 PM

"...they have to be the *identical*, because two polynomials are just not going to disagree in precisely one point."

An excellent point. I'm convinced!

Posted by: Polymath | July 06, 2006 at 10:07 PM

It does work for n = 2. Remember that the unit circle is of radius 1. The length of its diametre, the chord between two equally spaced marks, is 2, of course.

Posted by: Kristjan Kannike | July 12, 2006 at 11:20 AM

The problem reduced, in geometric analysis, to one of the following forms.

I'm using Prod[f(i),a, b] to represent the product of f(a)*f(a+1)*f(a+2)..*f(b)

If n is even,

productOfSegments = (1/2)*Prod[ 2*Sin[i*pi/n], 1, n/2]^2

If n is odd,

productOfSegments = Prod[2*Sin[i*pi/n], 1, (n-2)/2]^2

I think I typed that in correctly. Hmm. I've not done much with products and the like before, so I'm not sure where to go from there. Interesting, though.

Posted by: Foxy | July 12, 2006 at 11:21 PM

Are you sure the diagonal lengths of an n-gon can be *transcendental*? If so, can I have a reference? Certainly they aren't necessarily constructible (in the Euclidean sense) or even expressable in radicals, but my intuition says that they should at least be algebraic. Oh, and, very cool math fact, btw :-) Slick proof, too!

Posted by: Cat lover | July 14, 2006 at 04:40 PM

Thanks for the correction, cat lover. I agree that those numbers are probably algebraic. My HEAD knows that there are algebraic numbers that can't be written with roots, but I think I don't know it yet in my heart of hearts.

Posted by: polymath | July 15, 2006 at 11:31 PM

The diagonal lengths are definitely algebraic, since the end points are algebraic numbers, and the distance function is an algebraic function.

Posted by: kevin cline | August 01, 2006 at 04:17 PM

Ran across your site by looking for interesting math problems. For the above problem you can avoid dividing by zero if you first move RHS to LHS and then factor out the Z. Then equate other factor to zero and solve. QED

Posted by: clkirksey | March 02, 2007 at 07:21 PM

Generalize and prove results for z^n=a+bi, where |a+bi|=1.

What happens when |a+bi|≠1

Posted by: subbu | November 01, 2011 at 07:43 PM

Can you explain why you substituted z with Z+1 at that one part in your proof? Why can you do so, and why do you do so?

Posted by: Jsong | March 15, 2012 at 12:15 PM