Difference between revisions of "Pick's Theorem"
m (→Proof) |
m (→Proof) |
||
| Line 17: | Line 17: | ||
== Proof == | == Proof == | ||
| − | {Outline: show that any triangle on the lattice with no point in its interior has an area of <math>1</math>. Then triangulate the shape and apply Euler's Polyhedron formula for graphs to obtain the desired result.} | + | {Outline: show that any triangle on the lattice with no point in its interior has an area of <math>\frac{1}{2}</math>. Then triangulate the shape and apply Euler's Polyhedron formula for graphs to obtain the desired result.} |
== Usage == | == Usage == | ||
Revision as of 21:46, 4 February 2015
Pick's Theorem expresses the area of a polygon, all of whose vertices are lattice points in a coordinate plane, in terms of the number of lattice points inside the polygon and the number of lattice points on the sides of the polygon. The formula is:
where 
is the number of lattice points in the interior and 
being the number of lattice points on the boundary.
It is similar to the Shoelace Theorem, and although it is less powerful it is a good tool to have in solving problems.
![[asy] size(150); defaultpen(linewidth(0.8)); for (int i = 1; i <= 5; i=i+1) { for (int j = 1; j <= 5; j=j+1) { dot((i,j)); } } draw((1,1)--(1,3)--(3,4)--(2,5)--(5,5)--(2,2)--(4,1)--cycle);[/asy]](http://latex.artofproblemsolving.com/5/5/c/55ced6a62b824d8f7b144136c14f6fe9580d9ad4.png)
Proof
{Outline: show that any triangle on the lattice with no point in its interior has an area of 
. Then triangulate the shape and apply Euler's Polyhedron formula for graphs to obtain the desired result.}
