

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 r^{n} for n=0...K1. 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 r^{n} is r^{n}1. So the product is (1r)(1r^{2})...(1r^{K1}) now x^{K}1 = (x1)(xr)(xr^{2})...(xr^{K1}) = (x1)(x^{K1}+x^{K2}+...+x+1) Divide both sides by x1, then set x=1 in the remaining equation, and you get (1r)(1r^{2})...(1r^{K1}) = 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:
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:
Just idly browsing before bedtime and came across this one. Now, please correct me if I am wrong, but if you divide by x1 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' nonsensical 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 x1 proves the equality (xr)(xr^{2})...(xr^{K1}) = (x^{K1}+x^{K2}+...+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 (x1)^{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. 

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