# Triangulations of lattice polygons

Let L be a 2-dimensional lattice and P- a lattice polygon. Suppose, it is triangulated into lattice tiangles. What are restrictions on their areas? For instance, can a lattice triangle of even area always be divided into lattice triangles of area 1? Is there any general approach to such questions?

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By Pick's Theorem, the area of a lattice polytope is related to the number $i$ of lattice points in the interior and the number $b$ of lattice points on the boundary, according to the formula:

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

Note that if $b$ is odd, then the area must be equivalent to $1/2$ mod $1$. This implies that any lattice triangle with integral area has at least one non-vertex point located on an edge. That extra point can be used to subdivide the triangle into two smaller lattice triangles.

Concerning the splitting of even area triangles into triangles of area 1, I don't know any general results.

Hope this helps!

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Yes, this observation gave rise to this question. – Daniil Rudenko Feb 22 '12 at 13:21

This is not my own formula: A = 2i + b - 1 when the area of one triangle is 1. I haven't tested this, but I think it works on any triangular grid (not just equilateral). I did test A=sqrt(3)*(i+b/2-1)/2 on an equilateral triangle grid. This works because each square can be mapped to a parallelogram comprised of 2 equilateral triangles, and the area of the parallelogram is sqrt(3)/2.

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I think that's $2i+b-2$. For a single triangle, $i=0$, $b=3$, $A=1$. – Gerry Myerson Apr 19 '12 at 12:22
I would think it's -2 except that this can be compared to Euler's polyhedral formula F+V=E+2, flattened to 2D, and F has to include the one face outside the bounded figure. A+1=F – Rick Walcott Apr 19 '12 at 12:27
I can't argue with i=0, b=3, A=1. I should have tested this. – Rick Walcott Apr 19 '12 at 12:35

I tested A = 2i + b - 2 for any triangular lattice where A = 1 for each triangle, and it works. Geometer's Sketchpad implements it very nicely, but it would work in GeoGebra too.

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