Difference between revisions of "Pick's Theorem"
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== Proof == | == Proof == | ||
If a triangle on the lattice points with no point in its interior or on its edges, it has an area of <math>\frac{1}{2}</math>. Such triangle must contain two lattice points distance <math>1</math> from each other and one lattice point on a line parallel to the opposite edge distance <math>1</math> apart. The minimum distance between two distinct lattice points is <math>1</math>. If no two lattice points have distance <math>1</math>, by <math>\frac{1}{2}bh</math> the area is more than 1 and similarly for the height. | If a triangle on the lattice points with no point in its interior or on its edges, it has an area of <math>\frac{1}{2}</math>. Such triangle must contain two lattice points distance <math>1</math> from each other and one lattice point on a line parallel to the opposite edge distance <math>1</math> apart. The minimum distance between two distinct lattice points is <math>1</math>. If no two lattice points have distance <math>1</math>, by <math>\frac{1}{2}bh</math> the area is more than 1 and similarly for the height. | ||
| − | Removing <math>1</math> of the mentioned triangles either removes <math>1</math> boundary point, turns <math>1</math> interior point into a boundary point, accounting for the <math>I+\frac{1}{2}B</math> part. The <math>-1</math> part is accounted for by looking at the area of the unit triangle with <math>3</math> boundary points, <math>0</math> interior points, and <math>\frac{1}{2}</math> area. <math>QED</math> | + | Removing <math>1</math> of the mentioned triangles either removes <math>1</math> boundary point, turns <math>1</math> interior point into a boundary point, accounting for the <math>I+\frac{1}{2}B</math> part. The <math>-1</math> part is accounted for by looking at the area of the unit triangle with <math>3</math> boundary points, <math>0</math> interior points, and <math>\frac{1}{2}</math> area. <math>QED</math> <math>\blacksquare</math> |
| − | |||
| − | <math>\blacksquare</math> | ||
Solution by [[User:a1b2|a1b2]] | Solution by [[User:a1b2|a1b2]] | ||
Revision as of 12:17, 15 November 2019
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 = -2; i <= 2; i=i+1) { for (int j = -2; j <= 2; j=j+1) { dot((i,j)); } } draw((-2,-2)--(-2,0)--(0,1)--(-1,2)--(2,2)--(0,0)--(1,-2)--cycle);[/asy]](http://latex.artofproblemsolving.com/b/4/a/b4acbe6d46fcbed26c9dd24437ae47fcf13f70bc.png)
Proof
If a triangle on the lattice points with no point in its interior or on its edges, it has an area of 
. Such triangle must contain two lattice points distance 
from each other and one lattice point on a line parallel to the opposite edge distance apart. The minimum distance between two distinct lattice points is . If no two lattice points have distance , by 
the area is more than 1 and similarly for the height.
Removing of the mentioned triangles either removes boundary point, turns interior point into a boundary point, accounting for the 
part. The 
part is accounted for by looking at the area of the unit triangle with 
boundary points, 
interior points, and area. 
Solution by a1b2
