Given a set of 9 points, no 3
collinear, prove there must be a subset of 5 that forms a convex pentagon.
(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....
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Posted by Harry
on 2009-10-18 14:39:53 |
re(2): Happy Ending
