Calculus by Apostol Exercise 1.7

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Homework Help Overview

The problem involves proving a formula for the area of polygons whose vertices are lattice points, specifically focusing on rectangles, right triangles, and parallelograms, with a generalization to all polygons using induction. The formula relates the area to the number of interior and boundary lattice points.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss proving the formula for various shapes and consider the relationship between different types of polygons. There is an exploration of using triangles to construct proofs for parallelograms and general polygons.

Discussion Status

Some participants have made progress on specific parts of the problem, while others are seeking clarification on the connections between the shapes and the application of the formula. Suggestions for combining areas of shapes and counting points are being explored, but no consensus has been reached on the best approach.

Contextual Notes

Participants are navigating the constraints of using lattice points and the implications of constructing polygons from triangles. There is an emphasis on ensuring that all vertices remain lattice points throughout the discussion.

shinobi20
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Homework Statement


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.

Homework 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

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?
 
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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.
 
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.
 
DEvens said:
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.
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?
 
shinobi20 said:
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?

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.
 

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