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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(2): Happy Ending | Comment 3 of 4 |
(In reply to re: Happy Ending by Jer)

No apologies needed Jer. I've spent many happy hours in the last week looking up the history of  this problem and trying to untangle this case, N(5) = 9, from the general result, N(n) = 2^(n-2) + 1 hypothesised by Erdos et al, but seemingly not yet proved beyond n=5. 

Without your posting, I might never have known about this bit of mathematics. Thanks.

Just don't ask us to prove N(6)=17.... 
  Posted by Harry on 2009-10-18 14:39:53

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