For the calculations below, B and I indicate boundary and interior points, respectively. Subscripts indicate the attributed polygon. Because 1 Pick's Theorem shows the sum of the areas of the partitions of a polygon equals the area of the entire polygon, 2 any polygon can be partitioned into triangles, and 3 Pick's Theorem is accurate for any triangle, then Pick's Theorem will correctly calculate the area of any polygon constructed on a square lattice.
A modification of Pick's Theorem incorporates the presence of holes in a polygon. For example, consider the figure below. One way to calculate the shaded green area is to determine the area of the polygon P less the area of the hole T.
Alternatively, using Pick's Theorem on the green polygon with an interior triangular and exterior pentagonal border and interior points inside the green shaded produces the following calculations. Notice the discrepancy between these formulas. The Pick's Theorem calculation is one less than the actual value. To further investigate this discrepancy, consider a figure with two holes. We can analyze this case in a similar manner, first by calculating the area of the pentagon and subtracting the area of the holes.
Next, we calculate the area with Pick's Theorem. The figure in question has three boundaries, exterior pentagonal, interior triangular, and interior quadrilateral. The interior points include the interior points of the pentagon less the boundary and interior points of the holes. With two holes, there is a discrepancy of two between the calculations.
The empirical evidence uncovered here leads to a conjecture regarding how to incorporate the number of holes into Pick's Theorem. A modified formula is shown below, where H is the number of holes in the figure. A proof might involve chopping a hole-laden polygon into two separate hole-free polygons by creating a path involving of a chain of segments connecting two arbitrary exterior boundary points and one or more boundary points of each hole. More Extensions: Different Lattice Systems.
What would this investigation look like if it were conducted on a different isometric lattice system? Web design by Measured Designs. Hit enter to search or ESC to close. Close Search. This is an example of what is called proof by contradiction. We assume that we can draw an equilateral triangle on a set of three lattice points.
We then show that this leads to a contradiction — i. The proof involves a simple counting of the interior and boundary points of the polygon with the holes, without the holes and the holes themselves. In Figure 3, we show a simple triangle with one hole. Using 2 , the area is then. To give a flavour of this, we present two separate problems: geometric and algebraic.
Although square integer grids are ubiquitous in our life, it is a fact that one cannot draw one of the simplest figures, namely, an equilateral triangle with its vertices being grid points on such a lattice. Suppose we have drawn an equilateral triangle with sides of length on a square grid. Then, its area is given by the well-known formula.
This contradiction proves the initial claim. The same argument shows that regular hexagons cannot be drawn on a square integer grid either. Another interesting application arises in the so-called Farey sequences,. These are sequences of rational numbers of the form in an increasing order with and and coprime gcd. For example, [4]. One property of such sequences is that if are neighbours of a Farey sequence, then.
We, thus, represent each fraction as an integer point with coordinates on a square grid. Noting that is the slope of a ray that passes through the origin and the point under consideration, we see that all points from the Farey sequence can be obtained in the correct order by sweeping a ray through the origin and noting down each lattice point it hits, provided we always take the closest point to the origin if several lie on the same ray this is because of the requirement that and be coprime.
In Figure 4, we show an example of a Farey sunburst , obtained by mirroring the polygonal curve that we get for in the first octant to obtain a closed shape. Of course, not all do; some will contribute a little less:. The points outside the polygon whose squares are partly inside which count less than they should will tend to balance out the contributions of the points inside whose squares are partly outside which count more than they should.
The squares around a point on the edge but not the vertex of the polygon will always be half inside, half outside, so that such a point will contribute exactly half of a square to the total:. The vertices are a little funny.
I made the diagrams for this article with a picture-drawing tool, originally designed at Bell Labs, called pic. The source code files for the illustrations were all named things like pick.
This was really confusing. Thu, 06 Apr Pick's theorem In a recent article , I discussed Hero's formula for the area of a triangle in terms of its sides, and I said it was an oddity that didn't seem like any other formula in geometry.
We can check the area manually as follows: The entire square has area 9. The proof of Pick's theorem isn't hard.
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