2
$\begingroup$

According to Pick's Theorem, the size of an area $A$ can be calculated by the sum of the interior lattice points located in the polygon $i$ and the number of lattice points on the boundary placed on the polygons perimeter $b$ divided by two, minus 1.
My question is - can I use this sentence to prove that a polygon with an Area size $5$ has at least $6$ lattice points in its perimeter? (that the shape is actually lying on $6$ lattice points)? I'm asking this because when I set A = $5$ and b = $6$, I get the result that i= $3$ - but I couldn't draw a polygon with $3$ interior points.

A = i + b/2 -1
than for area size 5:
5 = i + 6/2 -1
i = 3

is it possible to draw a polygon with i=3 anyway ?

$\endgroup$
  • $\begingroup$ You should enclose numbers and other mathematical expressions (including variable names) in dollar signs. Also, you should really add some line breaks. $\endgroup$ – tomasz May 5 '13 at 13:48
  • 2
    $\begingroup$ i.imgur.com/JHWuMx0.jpg Here is an example of a shape with area 5 with 6 perimeter points and 3 interior points. $\endgroup$ – PhiNotPi May 5 '13 at 13:59
  • $\begingroup$ It's possible to draw a polygon with 4 interior and 4 boundary points. Just grab some grid paper and play around for a little while, there is a solution in the shape of a perfect square. $\endgroup$ – Erick Wong May 5 '13 at 14:00
  • 1
    $\begingroup$ You should specify that the corners of your area $A$ are on lattice points. Otherwise, I can easily draw a rectangle with area $5$ that covers no lattice points at all. $\endgroup$ – Ross Millikan May 7 '13 at 14:41
0
$\begingroup$

This isn't quite a formal proof, but that's not important.

It's easy to see that a shape with an area that is an integer must have an even number of perimeter points because of the $i/2$ in the formula. Likewise, a shape with an area that is not an integer (0.5, 1.5, 2.5, etc) must have an odd number of perimeter points.

Given an area that is a whole number , it is possible to draw a shape with that area with only 4 perimeter lattice points (which is also the minimum).

Given an area that is a whole number + 1/2, it is possible to draw a shape with that area with only 3 perimeter lattice points (which is also the minimum).

Here is an image that shows how to construct a triangle with any area with the minimum number of perimeter points. The triangles on the left have an area of 1, 2, 3, and 4. The triangles on the left have an area of 0.5, 1.5, 2.5, and 3.5. The pattern is very simple.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.