Difference between revisions of "Pick's Theorem"

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'''Pick's Theorem''' expresses the [[area]] of a [[polygon]], all of whose [[vertex | 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:
 
'''Pick's Theorem''' expresses the [[area]] of a [[polygon]], all of whose [[vertex | 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:
  
<math>A = I + \frac{B}{2} - 1</math>
+
<math>A = I + \frac{1}{2}B - 1</math>
  
 
where <math>I</math> is the number of lattice points in the interior and <math>B</math> being the number of lattice points on the boundary.
 
where <math>I</math> is the number of lattice points in the interior and <math>B</math> 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.
+
It is similar to the [[Shoelace Theorem]], and although it is less powerful, it is a good tool to have in solving problems.
  
 
<asy>
 
<asy>
 
size(150);
 
size(150);
 
defaultpen(linewidth(0.8));
 
defaultpen(linewidth(0.8));
for (int i = 1; i <= 5; i=i+1) {
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for (int i = -2; i <= 2; i=i+1) {
for (int j = 1; j <= 5; j=j+1) {
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for (int j = -2; j <= 2; j=j+1) {
 
dot((i,j));
 
dot((i,j));
 
}
 
}
 
}
 
}
draw((1,1)--(1,3)--(3,4)--(2,5)--(5,5)--(2,2)--(4,1)--cycle);</asy>
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draw((-2,-2)--(-2,0)--(0,1)--(-1,2)--(2,2)--(0,0)--(1,-2)--cycle);</asy>
  
 
== Proof ==
 
== Proof ==
{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.}
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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.
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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.
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Solution by [[User:a1b2|a1b2]]
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(Edited by JeffersonJ)
  
 
== Usage ==
 
== Usage ==

Revision as of 15:23, 7 August 2022

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 (Edited by JeffersonJ)

Usage