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Lattice points

  1. May 20, 2006 #1
    Let P be a polygon whose vertices are lattice points. The area of P is [tex] Z + \frac{1}{2}B - 1 [/tex]. Z is the number of lattice points inside the polygon, and B is the number on the boundary.

    (a) Prove that the forumula is valid for rectangles with sides parallel to the coordinate axes.
    (b) Use induction on the number of edges to construct a proof for general polygons.

    (a) Would you have to use the exhaustion property? We have to find two regions such that the area of a rectangle with sides parallel to the coordinate axes is a subet and superset of. There can be only one c such that c = [tex] Z + \frac{1}{2}B - 1 [/tex]. The question is, how do we determine the two step regions?
    (b) I dont know that this question is asking me to prove.

  2. jcsd
  3. May 20, 2006 #2


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    For b, what do you mean you don't know what it's asking you to prove. It's entirely clear: prove that for any polygon whose vertices are lattice points that its area is Z + B/2 - 1 where Z is the number of lattice points inside the polygon and B is the number on the boundary. Moreover, it says to use induction to prove it.

    For a, it's really quite straight forward. There's no need for any "exhaustion property"s or anything to do with subsets or supersets. Just imagine a rectangle whose vertices are lattice points and whose sides are parallel to the axes. You should know how to compute the area of an arbitrary rectangle of this form, it's just length x width. But the length and with of such a rectangle uniquely determine the number of Z points and B points, and you can easily just count these numbers, and verify that the formula holds.

    For example, consider the square whose vertices are (0,0), (1,0), (0,1), (1,1). It has 0 interior points, and 4 boundary points. The formula predicts its area to be 0 + 4/2 - 1 = 1, and indeed you know the are of such a square is just 1 x 1 = 1, so the formula does work. Prove this for an arbitrary rectangle.
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