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Need some help with a proof (using the pigeon hole principle)

  1. Nov 7, 2004 #1
    I got 6 problems that I needed to proove using the pigeon hole principle and I was able to solve 5 of them but this last one is giving me some problems.

    In each convex polygon with 2*n vertices there is at least one diagonal that isn't parallel with either one of the sides of the polygon.

    I would appreciate some help to point me in the right direction or maybe an example of a similar proof that uses the pigeon hole principle, thx in advance
     
  2. jcsd
  3. Nov 7, 2004 #2
    isn't this a Descrete Maths problem?
     
  4. Nov 7, 2004 #3
    i'm obligated to use the pigeon hole principle, i can't use anything else (i'm not sure what you mean by 'discrete math', or did you mean that i posted this in the wrong forum?)
     
    Last edited: Nov 7, 2004
  5. Nov 7, 2004 #4

    Hurkyl

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    How many diagonals are there? How many can be parallel to a side?
     
  6. Nov 8, 2004 #5
    I'm trying to find a function that gives the number of diagonals in funtion of the number of vertices but i don't see a connection both of them

    i looked it up and apparantly there is a formula for it, for a polygon with 2n vertices the number of diagonals is 2n*(2n-3)/2, i hope i'll be able to use this
     
    Last edited: Nov 8, 2004
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