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

I'm pretty sure there are 7.

Call the points A, B, C, and D

Its pretty clear that taking one of these points it is possible to find a plane parallel to the plane containing the other three and equidistant to the first point.

So this is 4 planes

There are also planes that separate the points by pairs.  To see them take the line determined by the first pair A and B.  There are an infinity of planes equidistant from this line and point C (they all contain the line which is the midpoint of the perpendicular from C to line AB.)   Only one of these is also the same distance from D (and from the line CD.)

There are 3 of these planes:  separating

AB from CD,

AC from BD, and

AD from BC. 

Grand total = 7 planes. 

-Jer

Posted by Jer on 2004-12-17 17:13:50