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Topic: Chords on the Unit Circle (Read 2366 times) 

william wu
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Chords on the Unit Circle
« on: Jan 22^{nd}, 2003, 1:20pm » 
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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.


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Icarus
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Re: Chords on the Unit Circle
« Reply #1 on: Jan 30^{th}, 2003, 7:27pm » 
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The product is K Proof: 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 QED


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wowbagger
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Re: Chords on the Unit Circle
« Reply #2 on: Jan 31^{st}, 2003, 2:37am » 
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on Jan 22^{nd}, 2003, 1:20pm, 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.

« Last Edit: Jan 31^{st}, 2003, 3:29am by wowbagger » 
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Icarus
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Re: Chords on the Unit Circle
« Reply #3 on: Feb 3^{rd}, 2003, 8:19pm » 
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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.


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harpanet
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Re: Chords on the Unit Circle
« Reply #4 on: Mar 25^{th}, 2003, 1:42pm » 
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Quote:Divide both sides by x1, then set x=1 in the remaining equation, and you get... 
 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 )


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Icarus
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Re: Chords on the Unit Circle
« Reply #5 on: Mar 25^{th}, 2003, 5:05pm » 
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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.


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harpanet
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Re: Chords on the Unit Circle
« Reply #6 on: Mar 26^{th}, 2003, 7:06am » 
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Thanks for the info Icarus.


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