Pick's theorem says that given a square grid consisting of all points in the plane with integer coordinates, and a polygon without holes and non selt-intersecting whose vertices are grid points, its area is given by:
$$i + \frac{b}{2} – 1$$
where $i$ is the number of interior lattice points and $b$ is the number of points on its boundary. Theorem and proof may be found on Wikipedia.
Let us suppose that the grid is not square but triangular (or hexagonal). Does a similar theorem hold?
Best Answer
The short answer is that, no, there can be no formula for polygons with vertices in the hexagonal lattice in terms of just boundary and interior points. This is based on the fact that primitive triangles on this lattice--ones with no lattice points on their boundary (besides the vertices) or in the interior--can have different areas, whereas for the square lattice all primitive triangles have area $\frac{1}{2}$.
However, as Casebash has partly gotten at in his answer, you can approximate things well if you compute what, in the below paper, is called the "boundary characteristic" of the polygon, a number that is somewhat complicated to think to compute, but which gives a decent proxy for how many of each type of primitive triangle the polygon contains.
Kolodziejczyk has been the main one doing work on hexagonal lattice results of this type that I know; he's worth looking up for similar results. Ding Ren is another, and the older work of Grunbaum, etc., still bears on the problem.
"A Fast Pick-Type Approximation for the Area of H-Polygons," Ren, Kolodziejczyk, et al., American Mathematical Monthly, 1993.