Calculus Apostol Exercise 1.7 A point (x,y) in the plane is called a lattice point if both coordinates x and y are integers. Let P be a polygon whose vertices are lattice points. The area of P is I + B/2 - 1, where I denotes the number of lattice points inside the polygon and B denotes the number on the boundary.
(a) Prove that the formula is valid for rectangles with sides parallel to the coordinate axes. 
(b) Prove that the formula is valid for right triangles and parallelograms. 
(c) Use induction on the number of edges to construct a proof for general polygons.
I have done part (a) but stuck on part (b) and (c). I have tried to represent the trapezoid and parallelogram as the difference between a rectangle and triangles, but no success as of now. Any suggestions?
 A: This is known as Pick's Theorem. Hint: below, from my sci.math post on 1998/5/3 is a diagram that should help you see how the formula is additive. Use this additivity in your induction step.

To help remember the correct formula, you can check it on easy
cases (unit square, small rectangles, etc) or, better, you can
view how it arises from additivity of area. One can view Pick's
formula as weighting each interior point by $1$, and each boundary 
point by $1/2$, except that two boundary points are omitted. Now
suppose we are adjoining two polygons along an edge as in the
diagram below. Let's check that Pick's formula gives the same
result for the union as it does for the sum of the parts (and
thus it gives an additive formula for area, as required).
        1/2           1/2                        1/2   1/2
   ... - @             @ - ...              ... - @     @ - ...
 /         \         /         \          /         \ /         \
          0 @       @ 0                              @ 0 
            |       |                                
.       1/2 @       @ 1/2       .        .           @ 1         . 
.           |   +   |           .  -->   .                       . 
.       1/2 @       @ 1/2       .        .           @ 1         .
            |       |                                
          0 @       @ 0                              @ 0 
 \         /         \         /          \         / \         /
   ... - @             @ - ...              ... - @     @ - ...
        1/2           1/2                        1/2   1/2

The edge endpoints we choose as the two omitted boundary points.
The inside points on the edge were each weighted $1/2 + 1/2$ on the 
left, but are weighted $1$ on the right since they become interior.
All other points stay interior or stay boundary points, so their
weight remains the same on both sides. So Pick's formula is additive.
