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Difference between revisions of "Pick's Theorem"
(Undo revision 215867 by Marianasinta (talk))
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== Proof ==
 
== Proof ==
−
If a triangle on the lattice [https://artofproblemsolving.com/wiki/index.php/TOTO_SLOT_:_SITUS_TOTO_SLOT_MAXWIN_TERBAIK_DAN_TERPERCAYA TOTO SLOT] points with exactly <math>3</math> points 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>, then 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 exactly <math>3</math> points 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>, then by <math>\frac{1}{2}bh</math> the area is more than 1 and similarly for the height.
 
Removing a triangle either removes <math>1</math> boundary point or 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.
 
Removing a triangle either removes <math>1</math> boundary point or 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.
  

Revision as of 12:08, 20 February 2024

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:

$A = I + \frac{1}{2}B - 1$

where $I$ is the number of lattice points in the interior and $B$ 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]

Proof

If a triangle on the lattice points with exactly $3$ points in its interior or on its edges, it has an area of $\frac{1}{2}$. Such triangle must contain two lattice points distance $1$ from each other and one lattice point on a line parallel to the opposite edge distance $1$ apart. The minimum distance between two distinct lattice points is $1$. If no two lattice points have distance $1$, then by $\frac{1}{2}bh$ the area is more than 1 and similarly for the height. Removing a triangle either removes $1$ boundary point or turns $1$ interior point into a boundary point, accounting for the $I+\frac{1}{2}B$ part. The $-1$ part is accounted for by looking at the area of the unit triangle with $3$ boundary points, $0$ interior points, and $\frac{1}{2}$ area.

Solution by a1b2

Usage

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