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
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