I am interested in Pick Theorem.

Pick Theorem Assume $P$ is a convex lattice point polygon. If $B$ is the number of vertexes of $P$ and $I$ is the number of lattice points which in the interior of $P$. Then the area of $P$ is $I+\frac{B}{2}-1$.

I find this theorem in a magazine. It provides a proof based on Minkowski Theorem. At first, I think the method may be too difficult and I could find an easy one. But I failed. So I want to ask is there any other proof here?

Also, this problem may not be easy as it looks like. I think it may be related to the Euler number of graph, because Pick Theorem is similar to the two dimensional surface version of Euler character. So there is another question what is the relationship between Pick Theorem and Euler character.

In addition, is there an approach to make it into higher dimension? I have know idea because if you imagine a convex lattice point polyhedron, you will find that its area is not a rational number like the version above. So maybe we should generalize it from another way.

  • 1
    $\begingroup$ See pages 735-736 here: mast.queensu.ca/~murty/murty-thain.pdf $\endgroup$
    – tc1729
    Jan 13, 2014 at 5:43
  • $\begingroup$ @SiddharthPrasad So what is the relationship between it and Euler Character? $\endgroup$
    – gaoxinge
    Jan 13, 2014 at 5:47
  • 2
    $\begingroup$ Have you bothered to look at the wikipedia page? "The Reeve tetrahedron shows that there is no analogue of Pick's theorem in three dimensions that expresses the volume of a polytope by counting its interior and boundary points. However, there is a generalization in higher dimensions via Ehrhart polynomials. The formula also generalizes to surfaces of polyhedra." en.wikipedia.org/wiki/Pick's_theorem $\endgroup$ Jan 13, 2014 at 5:54
  • 1
    $\begingroup$ A nice discussion on Pick's Theorem can be found in Chapter 5 ("How Many Goats in the Orchard?") of Ian Stewart's book Another Fine Math You’ve Got Me Into... (New York, W.H. Freeman and Company, 1992). Another explanation can be found here: www.math.utah.edu/mathcircle/notes/pick.pdf‎ $\endgroup$
    – JRN
    Jan 13, 2014 at 5:55

1 Answer 1


I published a paper on generalizing Pick's theorem in Advances in Mathematics in 1993, (Pick's Theorem and the Todd Class of a Toric Variety). There, I interpret $I + B/2 + 1$ to be the sum of angles subtended by a polygon $P$ at all the lattice points. In other words, take a tiny circle around a lattice point $v$. Let $\angle_P(v)$ be the fraction of the circle that is inside $P$. If $v$ is interior, $\angle_P(v) = 1$. If $v$ is interior to an edge of $P$, $\angle_P(v) = 1/2$. The sum of the angles at the vertices of $P$ is $(n - 2)/2$, where $n$ is the number of vertices. These sum up to $\sum_{v\in Z^2} \angle_P(v) = I + B/2 + 1$.

That paper has what I consider a very easy proof of Pick's theorem that doesn't even depend on triangulating the polygon. Take a lattice polygon $P$ and let $1_P(x)$ be its characteristic function: $1_P(x) = 1$ if $x\in P$ and $1_P(x) = 0$ otherwise. Translate $1_P$ by all elements of $Z^2$ and add them up: $T_P(x) = \sum_{v\in Z^2} 1_P(x - v)$. Then let $\tilde{P}$ be the rotation of $P$ by 180 degrees. Then, it's easy to see $T_P(x) + T_{\tilde{P}}(x)$ is a constant function, apart from a set of measure 0, with constant value $2 \text{Area}(P)$. On the other hand, you can also easily see the constant value is $2 \sum_{v\in Z^2} \angle_P(v)$.

As far as generalizing, $\angle_P(v)$ still makes sense in higher dimensions. However, $\sum_{v\in Z^d} \angle_P(v)$ doesn't usually equal $\text{vol}(P)$ of $d$ dimensional polytope $P$. The key is to replace $\angle_P(v)$ by new angle measures that do work in the formula. One way of doing this based on rational functions is given in the paper. In subsequent work, another way of doing it is found using a generalization of the Riemann Zeta function.


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