Given four noncoplanar 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 20041217 17:13:50 