# Pick's theorem

For the theorem in complex analysis, see Schwarz lemma#Schwarz–Pick theorem.
i = 7, b = 8,
A = i + b/2 − 1 = 10
The triangle with vertices at the lower left, lower right, and upper right points has i = 12 and b = 14, giving by Pick's theorem A = i + b/2 − 1 = 18; this is confirmed by the triangle area formula 1/2 × base × height = 1/2 × 9 × 4 = 18.

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 lattice points in the interior located in the polygon and the number b of lattice points on the boundary placed on the polygon's perimeter:[1]

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

In the example shown, we have i = 7 interior points and b = 8 boundary points, so the area is A = 7 + 8/2 − 1 = 7 + 4 − 1 = 10 (square units)

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 a polygon that has h holes, with a boundary in the form of h + 1 simple closed curves, the slightly more complicated formula i + b/2 + h − 1 gives the area.

The result was first described by Georg Alexander Pick in 1899.[2] The Reeve tetrahedron shows that there is no analogue of Pick's theorem in three dimensions that expresses the volume of a polytope by counting its interior and boundary points. However, there is a generalization in higher dimensions via Ehrhart polynomials. The formula also generalizes to surfaces of polyhedra.

## Proof

Consider a polygon P and a triangle T, with one edge in common with P. Assume Pick's theorem is true for both P and T separately; 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[3]

$i_{PT} = (i_P + i_T) + (c - 2)\,$

and

$b_{PT} = (b_P + b_T) - 2(c - 2) - 2.\,$

From the above follows

$(i_P + i_T) = i_{PT} - (c - 2)\,$

and

$(b_P + b_T) = b_{PT} + 2(c - 2) + 2.\,$

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

\begin{align} A_{PT} &= A_P + A_T\\ &= (i_P + b_P/2 - 1) + (i_T + b_T/2 - 1)\\ &= (i_P + i_T) + (b_P + b_T)/2 - 2\\ &= i_{PT} - (c - 2) + (b_{PT} + 2(c - 2) + 2)/2 - 2\\ &= i_{PT} + b_{PT}/2 - 1. \end{align}

Therefore, if the theorem is true for polygons constructed from n triangles, the theorem is also true for polygons constructed from n + 1 triangles. For general polytopes, it is well known that they can always be triangulated. That this is true in dimension 2 is an easy fact. 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:

• observe that the formula holds for any unit square (with vertices having integer coordinates);
• deduce from this that the formula is correct for any rectangle with sides parallel to the axes;
• deduce it, now, for right-angled triangles obtained by cutting such rectangles along a diagonal;
• now any triangle can be turned into a rectangle by attaching such right triangles; since the formula is correct for the right triangles and for the rectangle, it also follows for the original triangle.

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.