Pick's Theorem is a well-known way to calculate the area of a plane figure whose vertices all lie on grid points.

Usually the grid is orthogonal - but that is far too square for us here in the Hall of Hexagons. So we'll look at an isometric grid, which is (naturally) much better for drawing hexagons.

Pick's Theorem states that the area of a simple plane figure whose vertices all lie on grid points is

where i is the number of grid points that lie inside the figure, and e is the number of grid points that lie on the egdes of the figure. The area is calculated in units of the smallest parallelogram on grid points (see right). Here, "simple" just means having no holes.

The proof of Pick's Theorem on the isometric grid is rather easier than on the orthogonal grid, and even involves hexagons in a minor role.

[1] If it works for area A and area B then it works for the combined area C = A+B (and vice versa)
Proof: Let eA, eB, eC be the number of points on the edges of A, B, C, and let i_A, i_B, i_C be the number of points inside each. Let p be the number of points on the dividing line between A and B.

Now:
i_C = i_A + i_B + (p - 2) and e_C = e_A + e_B - 2p + 2
C = A + B = (i_A + e_A/2 - 1) + (i_B + e_B/2 - 1)
= i_A + i_B + p - 2 + e_A/2 +e_B/2 - p + 1 - 1
= i_C + e_C/2 - 1 QED

[2] The theorem is true for the smallest triangle (shown right) (i + e/2 - 1 = 0 + 3/2 -1 = 1/2), and hence (from [1]) is true for any figure made of these triangles - in particular for any figure whose sides on all on axes, including parallelograms on the axes. The theorem is also true for a half-parallelogram (again from [1]).

[3] The theorem is true for any triangle with vertices on grid points (shown red at right), because we can enlarge the triangle so that it becomes a hexagon whose sides on all on axes, by adding half-parallelograms (shown blue at right). (We know the theorem works for the hexagon and for all the half-parallelograms - so from [1] it works for the red triangle).

[4] Any figure with vertices on grid points can be divided into triangles with vertices on grid points (choose a convex vertex B; join the adjacent vertices A,C if possible, or join B to a vertex inside ABC - there must be one visible to B; continue until you have all triangles). The theorem works for all these triangles, and hence for the original figure. QED.

Having proved Pick's theorem on an isometric grid, it is easy to extend it to an orthogonal grid. We observe that shearing the grid relative to a horizontal axis will not change the area of any figure on it, nor will it change the number of vertices inside the figure or on its edges. Furthermore stretching the grid vertically will not change the number of vertices inside the figure or on its edges, but will change the area of the figure, and of the smallest parallegram (which is our unit of measurement) by the same fraction.

So, with a shear and a stretch of the vertical axis we have Pick's Theorem on an orthogonal grid.

On an orthogonal grid, the small parallelogram has become a square; if the grid points are spaced one unit apart, the area of this square is one. Thus Pick's Theorem is:

On an isometric grid where the grid points are spaced one unit apart, the area of the small parallelogram is sqrt(3)/2. Pick's Theorem expressed numerically is therefore:

It is impossible to draw an equilateral triangle with its vertices on orthogonal grid points. Pick's Theorem tells us why: any figure with vertices on orthogonal grid points will have a rational area, whereas an equilateral triangle has an irrational area.

Of course the same applies to a regular hexagon. But both equilateral triangles and hexagons can be drawn in abundance on an isometric gird.

page date: 7May07.      I enjoy correspondence stimulated by this site. You can contact me here.