Proving Pick's theorem Can pick's theorem be used for a rectangle such that its vertices are not lattice points?
I'm thinking of the rectangle in the picture but I want to shift it half an unit to the right. Thus there would be 6 boundary points and 9 interior points. Pick's theorem would give us an area of 11, but it is a 3 by 4 rectangle. Did I do something wrong in my calculations?
 A: Pick's theorem only applies to simple polygons whose vertices all lie on the intersection points of a square lattice. If this condition is not met, the enclosed area is not given by the theorem.
In the case of the original rectangle as drawn, the area is indeed given correctly:
$14$ boundary points, $6$ interior points, giving an area of $\frac{14}{2} + 6 - 1 = 12$, as expected.
But you cannot expect the theorem to hold if you translate the rectangle so that at least one vertex no longer lies on the lattice. Of course, the area of the shape is unchanged.
(You can, of course, superimpose a new lattice so that the theorem applies. This can be done either by simply also translating the original lattice by half-a-unit in the same direction, or by using a lattice that is twice as "fine" and arranging it so that all the vertices fall on points of this new lattice. In the latter case, the new area will be $4$ times the result you computed before, because each grid square will have only a quarter the area of an original grid square).
