<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>mathematics on oddComputer.com</title><link>https://oddcomputer.com/math/</link><description>Recent content in mathematics on oddComputer.com</description><generator>Hugo</generator><language>en</language><atom:link href="https://oddcomputer.com/math/index.xml" rel="self" type="application/rss+xml"/><item><title>Lattice Polygons</title><link>https://oddcomputer.com/math/lattice_polygons/</link><pubDate>Mon, 01 Jan 0001 00:00:00 +0000</pubDate><guid>https://oddcomputer.com/math/lattice_polygons/</guid><description>&lt;h3 id="figure-1-an-example-lattice-polygon"&gt;Figure 1: An example lattice polygon&lt;/h3&gt;
&lt;p&gt;&lt;img src="https://oddcomputer.com/images/LatticePolygons/01_PicksTheorem.png" alt="A quadrilateral lattice polygon on a coordinate grid with four lattice vertices enclosing one interior lattice point in a mathematical analytic geometry diagram" title="test"&gt;
A sample of a lattice polygon. The formula is called &amp;ldquo;Pick&amp;rsquo;s Theorem&amp;rdquo; and in this example B=8, and I=1. This paper is not about Pick&amp;rsquo;s Theorem, but it uses the same rules for counting border points. They do not have to be verticies of the polygon, they can also be points along an edge. For fun notice that A=(8/2)+1-1=4. Try this out on several shapes if you like, but be informed that the rest of this paper will ignore the results of Pick&amp;rsquo;s and do something way simpler.&lt;/p&gt;</description></item></channel></rss>