I can't believe I haven't blogged about the coolest math fact ever yet. It is not easy to prove—which is another way of saying that I couldn't prove it. I solicited proofs on an internet board, though, and I got one that I could understand. Which I might post in a couple of weeks when I have time over my spring break.

So.

Divide the circumference of a unit circle into *n* equal arcs with points P_0, P_1, ...P_(*n*-1). From P_0, draw the *n-1* chords that connect it to the other points. Since it's a unit circle and the points are equally spaced, the lengths of these *n-1* chords depend only on *n*. Now find the product of the lengths of the *n-1* chords. Try to guess at the punchline, which I'll type between the asterisks below in white type...you can highlight it to see the theorem when you're ready.

*The product of the lengths of the *n-1* chords always equals *n*.*

I know...for you non-math people, that doesn't sound amazing. But...it's soooOOOOooooOOO hard to describe how improbable that result is. I'm sure the mathies in your life will back me up on that.

You are such a nerd.

Posted by: caitmcq | March 07, 2006 at 07:20 PM

a geek, cait. i'm a geek. i used to be very secretive about it and everything, but i'm quite out of the closet now.

we're here, we're geeky, get used to it. there would be some big demonstration about it on the mall in washington except all the geeks are busy with math problems.

Posted by: Polymath | March 07, 2006 at 10:38 PM

More power to ya! I might have said "geek," but feared you might be offended by its original meaning. (I admire you as a word-geek also.)

Posted by: caitmcq | March 09, 2006 at 07:33 PM

Hot holy damn, that's awesome. I love all those little coincidences in mathematics. Freakin' crazy.

Posted by: Xanthir | June 23, 2006 at 11:34 AM

So what was the proof on this, anyways? I worked through the trig to get it as a product of sines but had no idea how to solve that product. Thanks.

Posted by: skywalkthisway | July 01, 2006 at 01:05 AM

hint for a proof: complex roots of unity.

Posted by: Oliver | July 05, 2006 at 02:35 PM

Setting up a problem with so much as this example, it's not too surprising that you get a result that's so closely linked to the setting up of the problem.

Posted by: Jim | July 22, 2006 at 08:17 AM

Setting up a problem with so much symmetry as this example, it's not too surprising that you get a result that's so closely linked to the setting up of the problem. (Heh, I missed out the important word)

Posted by: Jim | July 22, 2006 at 08:17 AM

huh?

Posted by: Caity | January 17, 2007 at 05:57 PM

Not to sound dumb, but when will I use Commutative, Associative, or Identity properties in real life?

Posted by: Nici | February 06, 2007 at 08:10 PM

put the unit circle on the complex plane with p_0=1. then p_0, p_1,...,p_(n-1) are roots of x^n -1=0. for the moment we will calculate the distance between a point x and all those point, not from p_0. this is abs((x-p_1)(x-p_2)(x-p_3)...(x-p_(n-1)). by fundamental theorem of algebra, this factors to abs((x^n -1)/(x-p_0)) or (x^n -1)/(x-1)). when divided out this becomes abs(x^(n-1)+x^(n-2)+x^(n-3)+...+x+1). now since p_0=1 we can plug in x=1 to find the distance we want. this will be abs(n)=n.

QED

Posted by: lolol | February 06, 2010 at 04:57 PM