# Calculus by Apostol Exercise 1.7

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1. Feb 13, 2015

### shinobi20

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Area of an arbitrary polygon P
P = I + B/2 - 1
where I is the number of interior points and B is the number of boundary points

3. The attempt at a solution
I have done part (a) but stuck on part (b) and (c). I have tried to represent the parallelogram as the difference between a rectangle and triangles, but no success as of now. I know I can just assume that a parallelogram has the area same as a rectangle because I have proved that it can be constructed by a rectangle and triangles, but I want to use the equation above in order to actually show that it is equal to the area of a rectangle. Any suggestions?

2. Feb 13, 2015

### DEvens

Can you show that the result applies to any triangle, not just a right triangle?

Can you show by construction that any such polygon can be constructed of triangles with corners that are lattice points?

Don't forget to use the fact the vertices are lattice points.

3. Feb 13, 2015

### RUber

As you said, the difference between a rectangle and right triangles might be a good way to work this.
You know that a rectangle containing the parallelogram and two right triangles has area P = I + B/2 - 1.
You know that a right triangle containg A interior points and C boundary points has area Q = A + C/2 - 1.
So I suggest you just put them together, you should have
Area of parellelogram = $I + B/2 - 1 - Q_1 -Q_2$.
Then by counting the number of boundary points and interior points leftover, you should have a convincing argument.

4. Feb 13, 2015

### shinobi20

I don't get it, what is the relationship of showing that the formula applies to a general triangle and the area of a parallelogram?

5. Feb 16, 2015

### DEvens

It goes back to the fact the vertices are lattice points. If you take any such polygon, you can build it out of triangles that have their vertices as lattice points. A parallelogram is just two such triangles. So the parallelogram will get you thinking in terms of constructing the general polygon.