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CTK Exchange
bryan Huh
guest
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Aug-19-09, 08:20 AM (EST) |
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"wilson's theorem - geometric proof"
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Found this proof on this website: https://www.cut-the-knot.org/blue/GeometricWilson.shtml An important part of this proof is this point: These polygons, too, are congruent in pairs since the vertices of a polygon could be traversed in two directions. IN ANY EVENT, WE SEE THAT THE NUMBER OF SUCH POLYGONS IS DIVISIBLE BY P. This is not clear to me. I see that there are p! permutations using the p points, but if you take out polygons which are congruent by mere direction of traversal, it becomes p!/2. Then if you take out polygons which are congruent by mere rotations (referred to as "shifting of indices" in the proof) we get (p-1)!/2 polygons. Then it looks like we subtract the star polygons which there are (p-1)/2 of to get the total number of "irregular polygons." This is my understanding. But I'm not getting how there are supposed to be "p" of these irregular polygons. Thank you, Bryan |
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alexb
Charter Member
2427 posts |
Aug-19-09, 08:39 AM (EST) |
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1. "RE: wilson's theorem - geometric proof"
In response to message #0
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>Found this proof on this website: >https://www.cut-the-knot.org/blue/GeometricWilson.shtml > >An important part of this proof is this point: > >These polygons, too, are congruent in pairs since the >vertices of a polygon could be traversed in two directions. >IN ANY EVENT, WE SEE THAT THE NUMBER OF SUCH POLYGONS IS >DIVISIBLE BY P. Think of a permutation as a sequence of jumps from a vertex to another vertex so that, for example, 1 3 6 5 ... starts with one and then makes jumps of 2, 3, -1, ... vertices. Now, the same series of jumps can start with any of the p vertices producing (congruent, but distinct for the sake of enumeration) polygons. This is why the number of such polygons is divisible by p. This is truly a simple geometric argument. The sentence about them too coming in pairs is unnecessary and misleading. I am on vacation. Will change the wording when back.
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