Explanation and Test Case for Pick's TheoremDate: 06/13/2006 at 02:47:54 From: Anelisa Subject: the use of area theorems to check validity of Pick's theorem I want to be able to test Pick's Theorem for the area of polygons. I want to accompany each polygon with a detailed test of the formula. But I don't really understand Pick's Theorem and its formula. Can you explain the formula and show that it works for a polygon? Date: 06/13/2006 at 12:51:01 From: Doctor Peterson Subject: Re: the use of area theorems to check validity of Pick's theorem Hi, Anelisa. If what you want to do is just to check the formula in specific cases, I would draw a few lattice polygons and use both methods to find the area. For example, consider the quadrilateral ABCD: A---o---B o \ \ o \ o o C \ / o D o o By Pick's rule, A = I + B/2 - 1, where I is the number of Interior lattice points, and B is the number of Boundary lattice points. We have I=2 since there are two dots entirely inside the polygon, and B=5, since besides the four vertices there is one extra dot on the boundary. So A = 2 + 5/2 - 1 = 3 1/2 How can we find the area using ordinary area formulas? We'll have to break it up into simple triangles (or, sometimes, rectangles or parallelograms) whose base and height we know. Having just mentioned parallelograms, I see one neat way to break this up: A---o---B o \\ \ o \ X---o---C \| / o D o o We have one parallelogram and two triangles: ABCX has base 2 and height 1, so A = 2*1 = 2 AXD has base 1 (namely XD) and height 1, so A = 1*1/2 = 1/2 CXD has base 2 (namely XC) and height 1, so A = 2*1/2 = 1 Adding these up, the total area is 2 + 1/2 + 1 = 3 1/2, which agrees with Pick. Demonstrating that it works in a few examples doesn't prove the formula, of course; in our archives we have several discussions of how to prove it. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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