Can one use Pick's theorm to prove that area size 5 covers at least 6 grid points?

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 ?

• You should enclose numbers and other mathematical expressions (including variable names) in dollar signs. Also, you should really add some line breaks. – tomasz May 5 '13 at 13:48
• i.imgur.com/JHWuMx0.jpg Here is an example of a shape with area 5 with 6 perimeter points and 3 interior points. – PhiNotPi May 5 '13 at 13:59
• It's possible to draw a polygon with 4 interior and 4 boundary points. Just grab some grid paper and play around for a little while, there is a solution in the shape of a perfect square. – Erick Wong May 5 '13 at 14:00
• You should specify that the corners of your area $A$ are on lattice points. Otherwise, I can easily draw a rectangle with area $5$ that covers no lattice points at all. – Ross Millikan May 7 '13 at 14:41

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.