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9 points to convex pentagon (Posted on 2009-10-12) Difficulty: 4 of 5
Given a set of 9 points, no 3 collinear, prove there must be a subset of 5 that forms a convex pentagon.

No Solution Yet Submitted by Jer    
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re: Happy Ending | Comment 2 of 4 |
(In reply to Happy Ending by Brian Smith)

Several of my recent problems lately have come directly or indirectly from a really interesting book I'm reading.  I plan to share the book in the forums when I finish the book.

As for this problem...
well I thought I could solve it with careful deduction through methods similar to 5 points to convex quad.  Even though I hadn't actually finished it when I submitted this I thought it would be D4ish. Unfortunately the number of cases to consider becomes huge.  It got too complicated to tease them all out.  This is definitely a d5 or higher problem.  Even if I do manage it it will be way too much to type out as an official solution.

Oh well I guess these things happen.  I apologize for any wasted time.

  Posted by Jer on 2009-10-18 01:22:37

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