# proof of Pick’s theorem

Pick’s theorem:

Let $P\subset\mathbb{R}^{2}$ be a polygon   with all vertices on lattice points on the grid $\mathbb{Z}^{2}$. Let $I$ be the number of lattice points that lie inside $P$, and let $O$ be the number of lattice points that lie on the boundary of $P$. Then the area of $P$ is

 $A(P)=I+\frac{1}{2}O-1.$ To prove, we shall first show that Pick’s theorem has an additive character. Suppose our polygon has more than 3 vertices. Then we can divide the polygon $P$ into 2 polygons $P_{1}$ and $P_{2}$ such that their interiors do not meet. Both have fewer vertices than $P.$ We claim that the validity of Pick’s theorem for $P$ is equivalent      to the validity of Pick’s theorem for $P_{1}$ and $P_{2}.$

Denote the area, number of interior lattice points and number of boundary lattice points for $P_{k}$ by $A_{k},I_{k}$ and $O_{k},$ respectively, for $k=1,2.$

Clearly $A=A_{1}+A_{2}.$

Also, if we denote the number of lattice points on the edges common to $P_{1}$ and $P_{2}$ by $L,$ then

 $I=I_{1}+I_{2}+L-2$

and

 $O=O_{1}+O_{2}-2L+2$

Hence

 $I+\frac{1}{2}O-1=I_{1}+I_{2}+L-2+\frac{1}{2}O_{1}+\frac{1}{2}O_{2}-L+1-1$
 $=I_{1}+\frac{1}{2}O_{1}-1+I_{2}+\frac{1}{2}O_{2}-1$

This proves the claim. Therefore we can triangulate $P$ and it suffices to prove Pick’s theorem for triangles  . Moreover by further triangulations we may assume that there are no lattice points on the boundary of the triangle other than the vertices. To prove pick’s theorem for such triangles, embed them into rectangles   . Again by additivity, it suffices to prove Pick’s theorem for rectangles and rectangular triangles which have no lattice points on the hypotenuse  and whose other two sides are parallel   to the coordinate axes. If these two sides have lengths $a$ and $b,$ respectively, we have

 $A=\frac{1}{2}ab$

and

 $O=a+b+1.$

Furthermore, by thinking of the triangle as half of a rectangle, we get

 $I=\frac{1}{2}(a-1)(b-1).$

(Note that here it is essential that no lattice points are on the hypotenuse) From these equations for $A,I$ and $O,$ Pick’s theorem is satisfied for these triangles.

Finally for a rectangle, whose sides have lengths $a$ and $b,$ we find that

 $A=ab$
 $I=(a-1)(b-1)$

and

 $O=2a+2b.$
Title proof of Pick’s theorem ProofOfPicksTheorem 2013-03-22 13:09:47 2013-03-22 13:09:47 giri (919) giri (919) 5 giri (919) Proof msc 51A99 msc 05B99 msc 68U05