Re:Post your proofs: Given five points on a sphere, four of ...
by kcl on Tuesday December 04, 2007 @02:18PM
(#86047)
Three non-collinear points determine a plane.
A counterexample to your proof would be to draw a plane through three points, say near the top of the sphere, then distribute the remaining points on the bottom. Then you don't get a proper division by taking your plane as the dividing line, since a working version would require more than a hemisphere of volume.
Thinking in terms of x, y, and z axes is a red herring. The symmetry of the sphere makes them irrelevant.
The farthest any two points can be is on a line passing through the center of the sphere. There are three such point-pairs for the x,y, and z axes. Given at least 5 points, four must lie on the same plane. That plane defines the border between hemispheres. Voila.