proof of Pick’s theorem
Let be a polygon with all vertices on lattice points on the grid . Let be the number of lattice points that lie inside , and let be the number of lattice points that lie on the boundary of . Then the area of is
To prove, we shall first show that Pick’s theorem has an additive character. Suppose our polygon has more than 3 vertices. Then we can divide the polygon into 2 polygons and such that their interiors do not meet. Both have fewer vertices than We claim that the validity of Pick’s theorem for is equivalent to the validity of Pick’s theorem for and
Denote the area, number of interior lattice points and number of boundary lattice points for by and respectively, for
Also, if we denote the number of lattice points on the edges common to and by then
This proves the claim. Therefore we can triangulate and it suffices to prove Pick’s theorem for triangles. Moreover by further triangulations we may assume that there are no lattice points on the boundary of the triangle other than the vertices. To prove pick’s theorem for such triangles, embed them into rectangles.
Again by additivity, it suffices to prove Pick’s theorem for rectangles and rectangular triangles which have no lattice points on the hypotenuse and whose other two sides are parallel to the coordinate axes. If these two sides have lengths and respectively, we have
Furthermore, by thinking of the triangle as half of a rectangle, we get
(Note that here it is essential that no lattice points are on the hypotenuse) From these equations for and Pick’s theorem is satisfied for these triangles.
Finally for a rectangle, whose sides have lengths and we find that
From these Pick’s theorem follows for rectangles too. This completes our proof.
|Title||proof of Pick’s theorem|
|Date of creation||2013-03-22 13:09:47|
|Last modified on||2013-03-22 13:09:47|
|Last modified by||giri (919)|