Given four non-coplanar points, how many different planes exist that are equidistant from all four points?

(In reply to re: Initial thoughts + conjecture (spoilers) by Larry)

Nice work, nikki and larry, but I suggest one modification to Larry's solution. I agree that each pair of lines also determines an additional solution, however I differ on the count. It seems to me that each pair of lines determines the new plane (not each pair of points). There are three ways to make two pair of lines, not six. This would give 7 solutions not 10, when added to nikki's.

I think that Larry's conjecture is true. Consider the lines AB and CD. While there are an infinite number of planes determined by AB, only one of these is parallel to CD (It exists as long as the 4 points aren't coplanar). Similarly there is one plane thru CD parallel to AB. By their selection, these two planes are parallel to each other.

 

Posted by SteveH on 2004-12-17 17:17:43