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Pick's theorem
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Pick's theorem

Given a simple polygon constructed on a grid of equal-distanced points (i.e., points with integer coordinates) such that all the polygon's vertices are grid points, Pick's theorem provides a simple formula for calculating the area A of this polygon in terms of the number i of interior points of the polygon and the number b of boundary points on the polygon:
A = i + ½b − 1.

In the example shown, we have i = 9 and b = 14, so the area is A = 9 + ½(14) − 1 = 9 + 7 − 1 = 15 (square units).

This formula so simple that it has been correctly used by first-grade children, drawing figures on square tiles on floor or wall, or stretching strings from pegs in pegboard. They learn how to add, along with subtraction as "take away". They learn to "halve" by a one-to-one correspondence between counters.

Note that the theorem as stated above is only valid for simple polygons, i.e. ones that consist of a single piece and do not contain "holes". For more general polygons, the "− 1" of the formula has to be replaced with "− χ(P)", where χ(P) is the Euler characteristic of P.

The result was first described by George Pick in 1899. It can be generalized to three dimensions and higher by Ehrhart polynomials. The formula also generalizes to surfaces of polyhedra.


Consider a polygon P and a triangle T, with one edge in common with P. Assume Pick's theorem is true for P; we want to show that it is also true to the polygon PT obtained by adding T to P. Since P and T share an edge, all the boundary points along the edge in common are merged to interior points, except for the two endpoints of the edge, which are merged to boundary points. So, calling the number of boundary points in common c, we have iPT = (iP + iT) + (c − 2) and bPT = (bP + bT) − 2(c − 2) − 2.

From the above follows (iP + iT) = iPT − (c − 2) and (bP + bT) = bPT + 2(c − 2) + 2.

Since we are assuming the theorem for P and for T separately,

       = iP + ½bP − 1 + iT + ½bT − 1'''
       = (iP + iT) + ½(bP + bT) − 2
       = iPT − (c − 2) + ½(bPT + 2(c − 2) + 2) − 2
       = iPT + ½bPT − 1.

Therefore, if the theorem is true for polygons constructed from n triangles, the theorem is also true for polygons constructed from n+1 triangles. To finish the proof by mathematical induction, it remains to show that the theorem is true for triangles. The verification for this case can be done in these short steps:

The last step uses the fact that if the theorem is true for the polygon PT and for the triangle T, then it's also true for P; this can be seen by a calculation very much similar to the one shown above.