By: Heather Bridges

We are going to derive Euler's Formula for a lattice polygon which is
in terms of the edges (E), faces (F), and vertices (V). Let's begin by constructing
and triangulating several lattice polygons. Then count the number of interior
points, border points, faces, and vertices.

From this we can deduuce that V = B + I. The triangulated polygons are
all seperated into fundamental triangles ( B=3, I=0) so

We can use this to find the relationship of F to B and I. We know that

Set these formulas equal to each other and solve for F which leaves F
= B + 2I - 2.

Now let's go back to investigate polygons in terms of the edges by susequently
keeping B constant and increasing I by 1.

Therefore we can deduce that E = B + 3 I when B remains constant. Then
keep I constant and subsequently increase B by one.

So E = 2 B - 3 when I remains constant. We can then analyze our formulas
to discover that

E = 2B + 3I - 3.

Now to gather all the formulas we have discovered:

So E = 2 B + 3 I - 3 = B + 2 I - 2 + B + I then we can substitute our
formulas for F & A to get the following:

It is also easy to now work backwards from Euler's Formula to derive
Pick's Formula.

It is also interesting to try to develop Pick's Formula from only using
the characteristics of a fundamental triangle and expanding on that. I encourage
readers to do so.

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