Given four non-coplanar points, how many different planes exist that are equidistant from all four points?
(In reply to Initial thoughts (spoilers)
I was picturing the same thing, and I agree with nikki. Four planes. But there may be a way to get up to 10 planes, not sure.
I'm also thinking about 2 pairs of points. Take the tetrahedron, points 1 and 2 form a line; points 3 and 4 form a line. The two lines either intesect or they don't. I haven't been able to picture any tetrahedrons where the two lines interesect (if they did, I believe that all 4 points would be in the same plane thus violating the rules of the problem).
So, suppose the two lines never intersect. Each line can determine an infinite number of planes rotating around that line like an axis. If line 1 is in a plane that is parallel to a plane that line 2 is in, then you could have a plane half way between the two planes, and thus add another plane to our list; potentially even adding a combination of 4 things taken 2 at a time, or 6 more planes, for a total of 10 planes. But I'm not sure this can be done with every set of 4 non coplanar points. Haven't proved this yet.
Conjecture: for any two skew lines, it is possible to find one and only one set of 2 planes such that each plane includes one of the lines, and the 2 planes are parallel to each other.
Posted by Larry
on 2004-12-17 16:30:17