Pick's theorem

In geometry, Pick's theorem provides a formula for the area of a simple polygon with integer vertex coordinates, in terms of the number of integer points within it and on its boundary. The result was first described by Georg Alexander Pick in 1899. It was popularized in English by Hugo Steinhaus in the 1950 edition of his book Mathematical Snapshots. It has multiple proofs, and can be generalized to formulas for certain kinds of non-simple polygons.

Formula

Suppose that a polygon has integer coordinates for all of its vertices. Let i {\displaystyle i} be the number of integer points interior to the polygon, and let b {\displaystyle b} be the number of integer points on its boundary (including both vertices and points along the sides). Then the area A {\displaystyle A} of this polygon is:

A = i + b 2 − 1. {\displaystyle A=i+{\frac {b}{2}}-1.}

The example shown has i = 7 {\displaystyle i=7} interior points and b = 8 {\displaystyle b=8} boundary points, so its area is A = 7 + 8 2 − 1 = 10 {\displaystyle A=7+{\tfrac {8}{2}}-1=10} square units.

Proofs

Via Euler's formula One proof of this theorem involves subdividing the polygon into triangles with three integer vertices and no other integer points. One can then prove that each subdivided triangle has area exactly 1 2 {\displaystyle {\tfrac {1}{2}}} . Therefore, the area of the whole polygon equals half the number of triangles in the subdivision. After relating area to the number of triangles in this way, the proof concludes by using Euler's polyhedral formula to relate the number of triangles to the number of grid points in the polygon.

The first part of this proof shows that a triangle with three integer vertices and no other integer points has area exactly 1 2 {\displaystyle {\tfrac {1}{2}}} , as Pick's formula states. The proof uses the fact that all triangles tile the plane, with adjacent triangles rotated by 180° from each other around their shared edge. For tilings by a triangle with three integer vertices and no other integer points, each point of the integer grid is a vertex of six tiles. Because the number of triangles per grid point (six) is twice the number of grid points per triangle (three), the triangles are twice as dense in the plane as the grid points. Any scaled region of the plane contains twice as many triangles (in the limit as the scale factor goes to infinity) as the number of grid points it contains. Therefore, each triangle has area 1 2 {\displaystyle {\tfrac {1}{2}}} , as needed for the proof.

An alternative proof that these triangles have area 1 2 {\displaystyle {\tfrac {1}{2}}} uses Minkowski's theorem that a symmetric convex set centered at a grid point and with no other interior grid point has area ≤ 4 {\displaystyle \leq 4} . Applying it to a parallelogram constructed from eight copies of a given triangle shows that the triangle's area is at most 1 2 {\displaystyle {\tfrac {1}{2}}} . But by the shoelace formula, the area of such a triangle is a positive half-integer, so the area must equal 1 2 {\displaystyle {\tfrac {1}{2}}} .

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