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Jul 17, 2013 at 14:16 vote accept Arnau
Feb 1, 2013 at 13:17 history edited Gil Kalai
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Jan 31, 2013 at 16:56 answer added Gil Kalai timeline score: 4
Jan 31, 2013 at 12:33 comment added Gil Kalai This is an excellent problem. Not much is known but there are few conjectures and results, I will try to answer later.
Feb 18, 2012 at 13:20 comment added Boris Bukh Not only there are examples of sets of $(r-1)(d+1)$ points with no Tverberg partition, but every set in sufficiently general position is such an example (by the dimension-counting). So, every condition that lowers the bound in Tverberg's theorem must necessarily be of the form "there is a particular kind of algebraic relation between the points", which does not sound all that natural geometrically. With that said, I do not know any results of this kind.
Feb 17, 2012 at 13:50 history asked Arnau CC BY-SA 3.0