Five Noncollinear Points
Date: 2 Apr 1995 18:05:43 -0400 From: Daniel Gomez Subject: Combinations/permutations I am a ninth grader and am stuck on the following question from my math homework. Can you help? In general, how many a. undirected lines b. circles are determined by five noncollinear points?
Date: 2 Apr 1995 19:30:20 -0400 From: Dr. Ken Subject: Re: Combinations/permutations Hello there! Well, the answer is that it depends on what you mean by "noncollinear," a word that different people use to mean different things (when talking about more than 3 points). If you mean that all five points don't lie on the same line, then you'll get one answer. If you mean that no three of them lie on the same line, then that's another thing. Here are a couple of configurations of five points that satisfy the first definition, but not the second: _____________________________________________ 1) o o o o o _____________________________________________ 2) o o o o o _____________________________________________ Now here are five points that do satisfy the second definition: _____________________________________________ o o o o o ______________________________________________ Now, there's an axiom in geometry that any two points determine a unique line. If no three points are collinear, that means that no two points will determine the same line as any other two points, so there will be no overlap. So the number of lines determined will be the number of two-element subsets of this five-element set. As for question B, there could again be several answers, no matter which definition you use for noncollinear. For instance, draw a circle, and then pick five noncollinear points on it. You can see that these five points all determine the same circle. But if you play around with a compass, you'll also be able to convince yourself that you can make five points in the plane, each three of which determine a distinct circle. Do you know how you could prove that? It's a neat problem. Anyway, I hope this helps you! -Ken "Dr." Math
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