# Pick’s Theorem in Two-Dimensional Subspace ofR3

Authors: Lin Si
Journal: The Scientific World Journal
 (2015)

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## Abstract

In the Euclidean spaceR3, denote the set of all points with integer coordinate by Z3. For any two-dimensional simple lattice polygon P, we establish the following analogy version of Pick’s Theorem, kIP+1/2BP-1, where BP is the number of lattice points on the boundary of P in Z3, IP is the number of lattice points in the interior of P in Z3, and k is a constant only related to the two-dimensional subspace including P.


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## Body

## 1. Introduction

In the Euclidean planeR2, a lattice point is one whose coordinates are both integers. A lattice polygon is a polygon with all vertices on integer coordinates. The area A(P) of a simple lattice polygon P can be given by celebrated Pick’s theorem [1]
(1)A(P)=I(P)+12B(P)-1,
where B(P) is the number of lattice points on the boundary of P and I(P) is the number of lattice points in the interior of P.Pick’s formula can be used to compute the area of a lattice polygon conveniently.For example, in Figure1, I(P)=60, B(P)=15. Then, the area of the polygon is A(P)=60+15-1=74.Figure 1There are many papers concerning Pick’s theorem and its generalizations [2–5], which mostly be discussed in two dimensions.Unfortunately, Pick theorem is failed in three dimensions. In 1957, John Reeve found a class of tetrahedra, named as Reeve tetrahedra later, whose vertices are(2)0,0,0T,1,0,0T,0,1,0T,1,1,rT,
where r is a positive integer.All Reeve tetrahedra contain the same number of lattice points, but their volumes are different.In this note, we discussed Pick’s theorem in two-dimensional subspace ofR3. For any (a,b,c)T∈Z3 with (a,b,c)=1, that is, the greatest common factor of a,b,c is one, denote by K, ax+by+cz=0, the two-dimensional subspace of R3. Then we established the following theorem.Theorem 1.
IfP is simple lattice polygon in the K, then the area of P is
(3)k(I(P)+12B(P)-1),
where B(P) is the number of lattice points on the boundary of P in Z3, I(P) is the number of lattice points in the interior of P in Z3, and k is the constant (a3+ab2)a2+b2+c2.Remark 2.
Although the simple lattice polygonP is in the two-dimensional subspace K, the lattice points in P belong to Z3.Let(a,b,c)T=(1,0,0)T in the Theorem; then we can get Pick’s theorem in some coordinate plane of R3.Corollary 3.
IfP is simple lattice polygon in the K, whose normal vector is (1,0,0)T, then the area of P is
(4)I(P)+12B(P)-1.


## 2. Proof of Main Result

For any(a,b,c)T∈Z3 with (a,b,c)=1, there is two-dimensional subspace of R3(5)ax+by+cz=0,
whose normal vector is just (a,b,c)T. We denote this two-dimensional subspace by K.By the theory of linear equations system,(-b,a,0)T and (-c,0,a)T are two linearly independent solutions of (5). We denote (-b,a,0)T by α and (-c,0,a)T by β. Obviously, α and β are also the basis of K.Lemma 4.
For any(a,b,c)T∈Z3 with (a,b,c)=1, there exists the lattice basis with the minimal area in the two-dimensional subspace K.Proof.
The area of parallelogram generated byα and β is
(6)1a2+b2+c2a-b-cba0c0a=a3+ab2+ac2a2+b2+c2=aa2+b2+c2.
Denote(a,b,c)T by n. For any lattice basis in K, k1α+k2β and l1α+l2β, where ki,li∈Z (i=1,2) and k1l1k2l2=0. The area of parallelogram generated by k1α+k2β and l1α+l2β is
(7)1a2+b2+c2n⋮k1α+k2β⋮l1α+l2β=1a2+b2+c2(n,α,β)×1000k1l10k2l2,
where n⋮k1α+k2β⋮l1α+l2β denote the determinant of n,k1α+k2β, and l1α+l2β.
Thus the lattice basisk1α+k2β and l1α+l2β have the minimal area if and only if k1l1k2l2=1.
Letk1=1,k2=0,l1=0, and  l2=1, and α,β are the lattice basis with the minimal area in the two-dimensional subspace K.Lemma 5.
For any(a,b,c)T∈Z3 with (a,b,c)=1, there exists the orthogonal lattice basis in the two-dimensional subspace K.Proof.
By Lemma4, α,β are the lattice basis with the minimal area in the two-dimensional subspace K. By Schmidt orthogonalization, let
(8)γ1=α=-b,a,0T,γ2=β-(β,γ1)(γ1,γ1)=-c,0,aT-bca2+b2γ1=-c,0,aT--b2ca2+b2,abca2+b2,0T=-a2ca2+b2,-abca2+b2,a3+ab2a2+b2T,
where (β,γ1) denote the usual inner product of β,γ1 in R3.
Thus(9)η1=γ1=α=-b,a,0T,η2=(a2+b2)γ2=-a2c,-abc,a3+ab2T
are the orthogonal lattice basis in the two-dimensional subspace K.Proof of Theorem.
By Lemma5, η1,η2 are the orthogonal lattice basis in the two-dimensional subspace K.
The area of parallelogram generated byη1, η2 is
(10)1a2+b2+c2a-b-a2cba-abcc0a3+ab2=a5+a3b2+ab2c2+a3c2+a3b2+ab4a2+b2+c2=a(a4+2a2b2+b4+(a2+b2)c2)a2+b2+c2=a(a2+b2)(a2+b2+c2)a2+b2+c2=(a3+ab2)a2+b2+c2,
which just is the constant k in the theorem.



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