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Algebraic expression for triangles within polygons

  1. Jul 8, 2011 #1
    1. The problem statement, all variables and given/known data If a polygon has n>=4 sides, what is the probability, in terms of n, that a triangle made up of vertices of the polygon shares at least one side with the polygon.



    2. Relevant equations



    3. The attempt at a solution Treating vertices of polygons as possible outcomes from which 3 are chosen: Sample space S = nC3. I can't figure out how to express the number of outcomes within event E={number of triangles sharing at least one side with polygon}. My way of thinking about it is that values in E will be all combinations that contain at least a pair of consecutive numbers (which, since they represent vertices of the polgon, constitute a side of both the polygon and the triangle). Unfortunately, I can't come up with a way to express this algebraically. It may not even be the right approach.

    Thanks for any help/advice.
     
  2. jcsd
  3. Jul 9, 2011 #2

    NascentOxygen

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    Staff: Mentor

    I think you are on the right track. nC3 appears to the number of unique triangles that can be constructed.

    When it comes to determining those which share no common side ....

    consider any particular side, and you can immediately rule out two of the remaining points for vertices of your triangle because a line drawn to those would be common to the polygon. Sketching a few polygons might help with generalising the formula.
     
    Last edited: Jul 9, 2011
  4. Jul 9, 2011 #3
    Here are my thoughts using your suggested strategy:

    The number of unique triangles sharing no common edge with polygon = n(n-3)(n-?). This is because there are n possible polygon vertices to choose for the first triangle vertex. The next triangle vertex cannot be the first vertex (-1) and cannot be an adjacent vertex (-2). The final vertex cannot be the first or second vertex (-2) and cannot be adjacent to either. At first glance I thought this would rule out 4 additional vertices, but that is wrong because vertices that are adjacent to two vertices selected for a triangle should only be counted once. So I can't see how to finish counting unique triangles using this strategy.

    Another way of doing it is to take the union of the following events: (1) the triangle shares exactly 1 common side with the polygon and (2) the triangle shares exactly 2 common sides with the polygon. (1) can be expressed as n(n-4) because there are n sides that could be common, but 2 vertices have already been used for that side and the two polygon vertices adjacent to the selected side will be counted by event (2). Event (2) can be expressed as 2n because there are n sides of the polygon that can be used for the common side, and then there are 2 remaining vertices of the polygon that can complete the triangle. So you get 2n+n(n-4) = n(n-2). I guess you also get the same result by just taking n(n-2) to reflect using a side of the polygon for the first two vertices and then any vertex of the polygon except those contained in the selected side to complete the triangle. Either way, that results in n(n-2) / nC3
     
  5. Jul 9, 2011 #4
    For everyone's information, the answer is 3!(n-3) / (n-2)(n-1). I just don't know how to get there.
     
  6. Jul 11, 2011 #5

    NascentOxygen

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    Staff: Mentor

    Taking a closer look, by sketching 4,5,6,7 and 8-sided polygons I arrive at the total number of possible triangles = (n-1)(n-2)-2

    A pattern for the number of triangles which don't share the polygons side is a little more taxing.
     
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