Let us return to the problem. We have 8 points and we want to form two groups of three and one group of two so that the intersection of the convex hulls of the groups is not empty.

We take 7 of the points and apply Tverberg’s theorem we get three groups only one can be only one point so we have the following possibilities (4,2,1), (3,2,2), (3,3,1).

For (3,2,2) we add the remaining point to one of the 2 groups and we get a solution to the problem.

If we have groups of the form (3,3,1) we add a point to the group with one point and we get a solution.

The only set of groups left is the set (4,2,1). Here we use Caratheodory’s theorem. We use it to get three points out of the four containing the single point then we move the remaining point to the group containing the single point and then add the remaining point to one of the groups of two and we have a solution to the final case. In the next of this series of posts I want to look at possible generalizations.’

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August 27, 2009 at 9:24 pm |

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