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        riddles >> putnam exam (pure math) >> Chords on the Unit Circle
        
(Message started by: william wu on Jan 22^nd, 2003, 1:20pm)

Title: Chords on the Unit Circle
Post by william wu on Jan 22^nd, 2003, 1:20pm

Pick K equidistant points on the unit circle. Choose one of the points and call it P. Draw line segments connecting P to all the other points on the circle. What is the surprising product of the lengths of these line segments? Prove it.

Title: Re: Chords on the Unit Circle
Post by Icarus on Jan 30^th, 2003, 7:27pm

The product is [hide]K[/hide]

Proof: [hide]

Let r be a primative K^th root of unity, so r^K - 1 = 0. All the K^th roots of unity are rⁿ for n=0...K-1. On the unit circle in the complex plane, these roots are K equidistant points. Choose 1 for the point P.
The distance from 1 to rⁿ is |rⁿ-1|. So the product is |(1-r)(1-r²)...(1-r^K-1)|

now x^K-1 = (x-1)(x-r)(x-r²)...(x-r^K-1) = (x-1)(x^K-1+x^K-2+...+x+1)

Divide both sides by x-1, then set x=1 in the remaining equation, and you get
(1-r)(1-r²)...(1-r^K-1) = K
[/hide]
QED

Title: Re: Chords on the Unit Circle
Post by wowbagger on Jan 31^st, 2003, 2:37am

on 01/22/03 at 13:20:59, william wu wrote:

Pick K equidistant points on the unit circle. Choose one of the points and call it P.

If you weren't allowed to choose your points, but to prove the result holds for any K equidistant points (any one of which may be P) - which is quite obviously true -, I would have argued that one should point to choosing the K^th roots of unity (and 1 as P) can be done without loss of generality.

Title: Re: Chords on the Unit Circle
Post by Icarus on Feb 3^rd, 2003, 8:19pm

I actually started to include that point (map the unit circle of the complex plane with 1 mapping to P ...), but left it out because I thought it detracted from the main argument, and should be obvious enough to anyone with sufficient math background to snooping around this forum.

Title: Re: Chords on the Unit Circle
Post by harpanet on Mar 25^th, 2003, 1:42pm

Quote:

Divide both sides by x-1, then set x=1 in the remaining equation, and you get...

Just idly browsing before bed-time and came across this one. Now, please correct me if I am wrong, but if you divide by x-1 and x equals 1 then you are dividing by 0. I was always taught to look out for these when doing algebraic proofs as they can easily catch you out (or 'prove' non-sensical things :))

Title: Re: Chords on the Unit Circle
Post by Icarus on Mar 25^th, 2003, 5:05pm

That is often a problem, but it does not occur here. Dividing by x-1 proves the equality

(x-r)(x-r²)...(x-r^K-1) = (x^K-1+x^K-2+...+x+1)

for all x except 1. Extending the equality to x=1 is simply a matter of noting that both sides are continuous functions that are defined at 1 as well. Taking the limits as x-->1 shows equality at x = 1.

This is so familiar a fact to those experienced in higher mathematics, that we usually take it for granted, just as one might go from (x-1)²=0 to x=1 without showing any intervening steps. Thus it did not occur to me to explain it at the time.

Title: Re: Chords on the Unit Circle
Post by harpanet on Mar 26^th, 2003, 7:06am

Thanks for the info Icarus.