Difference between revisions of "Pick's Theorem"

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{{WotWAnnounce|week=September 12- September 18}}
 
 
 
'''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>
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<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 shoestring formula, and though it is less powerful it is a good tool to have in solving problems.
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It is similar to the [[Shoelace Theorem]], and although it is less powerful, it is a good tool to have in solving problems.
  
{{image}}
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<asy>
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size(150);
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defaultpen(linewidth(0.8));
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for (int i = -2; i <= 2; i=i+1) {
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for (int j = -2; j <= 2; j=j+1) {
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dot((i,j));
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}
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}
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draw((-2,-2)--(-2,0)--(0,1)--(-1,2)--(2,2)--(0,0)--(1,-2)--cycle);</asy>
  
 
== Proof ==
 
== Proof ==
{{stub}}
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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.
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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>
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Solution by [[User:a1b2|a1b2]]
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== Discussion ==
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By a lattice polygon, we mean a polygon whose vertices are lattice points. For a lattice triangle, say a triangle if minimal if it has area <math>\frac{1}{2}</math>. In the above proof, it is assumed that any minimal triangle has two vertices with distance <math>1</math>. This is wrong.
  
 
== Usage ==
 
== Usage ==

Latest revision as of 01:37, 13 June 2021

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 no point 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 $1$, by $\frac{1}{2}bh$ the area is more than 1 and similarly for the height. Removing $1$ of the mentioned triangles either removes $1$ boundary point, 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. $QED$ $\blacksquare$ Solution by a1b2

Discussion

By a lattice polygon, we mean a polygon whose vertices are lattice points. For a lattice triangle, say a triangle if minimal if it has area $\frac{1}{2}$. In the above proof, it is assumed that any minimal triangle has two vertices with distance $1$. This is wrong.

Usage

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