According to Pick's Theorem, the size of an area $A$ can be calculated by the sum of
the interior lattice points located in the polygon $i$ and the number of lattice points on the boundary placed on the polygons perimeter $b$ divided by two, minus 1.
My question is - can I use this sentence to prove that a polygon with an Area size $5$ has at least $6$ lattice points in its perimeter? (that the shape is actually lying on $6$ lattice points)?
I'm asking this because when I set A = $5$ and b = $6$, I get the result that i= $3$ - but I couldn't draw a polygon with $3$ interior points.
A = i + b/2 -1
than for area size 5:
5 = i + 6/2 -1
i = 3
is it possible to draw a polygon with i=3 anyway ?
This isn't quite a formal proof, but that's not important.
It's easy to see that a shape with an area that is an integer must have an even number of perimeter points because of the $i/2$ in the formula. Likewise, a shape with an area that is not an integer (0.5, 1.5, 2.5, etc) must have an odd number of perimeter points.
Given an area that is a whole number , it is possible to draw a shape with that area with only 4 perimeter lattice points (which is also the minimum).
Given an area that is a whole number + 1/2, it is possible to draw a shape with that area with only 3 perimeter lattice points (which is also the minimum).
Here is an image that shows how to construct a triangle with any area with the minimum number of perimeter points. The triangles on the left have an area of 1, 2, 3, and 4. The triangles on the left have an area of 0.5, 1.5, 2.5, and 3.5. The pattern is very simple.