Pick's Theorem

Suppose that a polygon has integer coordinates for all of its vertices. Let $i$ be the number of integer points interior to the polygon, and let $b$ be the number of integer points on its boundary (including both vertices and points along the sides). Then the area $A$ of this polygon is:

$$A = i + \frac{b}{2} - 1$$


Let integer area $A = 5$, these are the possible pairs of boundary $b$ and interior $i$ points that satisfy Pick's


Notice that some pairs $(b,i)$ have multiple unique shapes


For $b=6,i=3$ above, found $7$ unique shapes so far. Are there others that I missed? I'm just visually checking and haven't written any code to determine the exact number, yet.

The Pick's integer solutions are

$$\forall A>0, \quad 1 < n \leq (A+1) : b = 2n, \, i = (A + 1) - n, \quad n,A \in \mathbb{Z}$$

Brute-Force Approach

  1. Find all possible pairs of boundary $b$ and interior $i$ points that satisfy Pick's Theorem.
  2. For each $(b, i)$ pair, attempt to generate all unique simple lattice polygons.
  3. Count the unique shapes for the given area $A$.


Given an integer area $A$, I'm curious about how many unique shapes can be formed on a $2D$ lattice?

  • $\begingroup$ Even counting rotated/reflected figures as distinct, a formula would be of interest. [I would think also difficult to find] $\endgroup$
    – coffeemath
    Sep 18 at 20:16
  • 2
    $\begingroup$ Let $a(n)$ be the number of lattice polygons of area $n$ (modulo some symmetries, of course). How many of the $a(n)$ can you compute? If you can get through $a(5)$ or $a(6)$, you can probably check OEIS for more information $\endgroup$ Sep 18 at 21:26
  • 1
    $\begingroup$ Depends on the definition of a unique shape. Up to isometries? Then even the number of triangles of area $1/2$ is infinite. Up to affine transformations? Then the answer is different. $\endgroup$ Sep 19 at 12:52

2 Answers 2


There are infinitely many unique shapes for any area $A$. Let $A=1$ and we can make triangles with base of 2 and height of 1 and move the top vertex over a lattice point. This can be done for any size base of $2A$ and height 1 for a triangle of area $A$. Add in an interior point and this game can continue for any $i$ you choose also.

enter image description here

And for more interior points, you will move the top green dot along, keeping the rest fixed. Shown are $A = 2,3,$ and $4$, with $i = 1,1,2$ respectively.

enter image description here

  • $\begingroup$ @vengy Yes, I am sure it could be cooked up to do what you want. Of course, that will make an area of 0.5 instead of 1, but some creative arranging will probably solve your problem. $\endgroup$
    – N. Owad
    Sep 19 at 19:57

Complementing N. Owad's answer -- the number of polygons of a given area becomes finite if you consider them up to unimodular equivalence, i.e. up to lattice-preserving affine transformations. The number of convex polygons up to unimodular equivalence with area $1/2,1,3/2,2,5/2,3\ldots$ is $1, 2, 3, 7, 6, 13,\ldots$ -- see https://oeis.org/A187015 and refs. therein, especially Balletti.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .