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Pick's theorem says that given a square grid consisting of all points in the plane with integer coordinates, and a polygon without holes and non selt-intersecting whose vertices are grid points, its area is given by:

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

where $i$ is the number of interior lattice points and $b$ is the number of points on its boundary. Theorem and proof may be found on Wikipedia.

Let us suppose that the grid is not square but triangular (or hexagonal). Does a similar theorem hold?

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    $\begingroup$ That is crazy, why have I not heard of this $\endgroup$ Jul 27, 2010 at 8:14
  • $\begingroup$ I found it for the first time in Gardner's Sixth Book of Mathematical Diversions. But it's not a topic explained at school, unfortunately. $\endgroup$
    – mau
    Jul 27, 2010 at 8:22
  • $\begingroup$ Interesting question, for sure. Will have to think about it a bit! $\endgroup$
    – Noldorin
    Jul 27, 2010 at 8:44
  • $\begingroup$ RE: It's not a topic explained at school. I have seen it in several geometry textbooks, but unfortunately it isn't in the "core curriculum" (it isn't necessary to go on to other things) so it is done purely at the teacher discretion. $\endgroup$
    – Jason Dyer
    Jul 30, 2010 at 16:36

5 Answers 5

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The short answer is that, no, there can be no formula for polygons with vertices in the hexagonal lattice in terms of just boundary and interior points. This is based on the fact that primitive triangles on this lattice--ones with no lattice points on their boundary (besides the vertices) or in the interior--can have different areas, whereas for the square lattice all primitive triangles have area $\frac{1}{2}$.

However, as Casebash has partly gotten at in his answer, you can approximate things well if you compute what, in the below paper, is called the "boundary characteristic" of the polygon, a number that is somewhat complicated to think to compute, but which gives a decent proxy for how many of each type of primitive triangle the polygon contains.

Kolodziejczyk has been the main one doing work on hexagonal lattice results of this type that I know; he's worth looking up for similar results. Ding Ren is another, and the older work of Grunbaum, etc., still bears on the problem.

"A Fast Pick-Type Approximation for the Area of H-Polygons," Ren, Kolodziejczyk, et al., American Mathematical Monthly, 1993.

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    $\begingroup$ You mean there are two triangles with $b=3, i=0$ of different area? Can you draw a picture? $\endgroup$ Sep 1, 2014 at 10:54
  • $\begingroup$ The linear map that maps $1\mapsto 1$ and $\frac{1+\sqrt 3 i}2\mapsto i$ maps the hexagonal lattice to the square lattice, and thereby a primitive triangle of the hexagonal lattice becomes a primitive triangle of the square latice. $\endgroup$ Aug 21, 2015 at 16:04
  • $\begingroup$ Note that your comment/paper refer to the vertices of hexagonal tilings (i.e. missing the center points of hexagons). A hexagonal/triangular lattice is an affine transformation of a rectangular lattice (as others have noted) and does have a Pick like formula. $\endgroup$
    – Eric
    Mar 24 at 5:19
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Find the area of the polygon on the triangular grid in the usual way using Pick's formula, but multiply by $\sqrt{3}$ and divide by 2. This works because each square can be mapped to a parallelogram comprised of two equilateral triangles, and the area of the parallelogram is $\frac{\sqrt{3}}{2}$.

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This is a very interesting question. I don't have a complete solution yet, but I did get some results. Consider an arbitrary distribution of points. Let $P(i,b)=i+\frac{b}{2}-1$ where $i$ is the number of internal points and $P$ is the number of boundary points. Let $P(A)=P(i_A,b_A)$, where $A$ is a simple polygon with all its vertices on the points. Wikipedia shows that $P(C)=P(A)+P(T)$ where $T$ is a triangle that shares a single edge with $A$ and $C$ a simple polygon formed by the union of $A$ and $T$. Since all simple polygons can be triangulated, $P(C)=sum P(t)$ for all $t$ in the triangulation of $C$.

The proof then has a second part that shows $P(t)$ equals the area for any triangle. So if we want to generalize it for other grids, we have to find a property equal to $P(T)$ for any triangle.

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  • $\begingroup$ Oh, it must also satisfy P(C)=P(A)+P(B) for all shapes A and B $\endgroup$
    – Casebash
    Jul 27, 2010 at 9:24
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(For an integer λ and a lattice polygon P) denote by λP polygon P stretched by λ times. Then number N(λP) of points inside the polygon λ is a quadratic polynomial in λ with leading coefficient S(P). This is the form of Pick's theorem that holds for any lattice (and obvious analogue works in any dimension — unlike usual Pick's formula that has no analogue in 3d even for the cubic lattice).

For the square lattice it yields ordinary Pick's theorem since for a parallelogram P on the square lattice $N(\lambda P)\approx\lambda^2(i+\frac{b}{2}-1)S(P)$ as λ→∞ and general theorem follows from the theorem for parallelograms by additivity.

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A downloadable Word document called Pick A Shape, at http://www.1000problems.org/ claims that if the units are measured in triangles instead of squares, the formula is A=2i+b-1

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  • $\begingroup$ The formula is actually A=2i+b-2, and it works for any triangular lattice, not just equilateral. A=1 for each triangle in the lattice. $\endgroup$ Apr 19, 2012 at 19:26
  • $\begingroup$ 1000problems.org/algebra/pickashape.htm the full link until it is moved or removed. $\endgroup$
    – JasoonS
    Jun 10, 2016 at 21:39

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