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

4 planes come to mind, initially, and I'm not sure if there are more.

3 points always define a plane.  So let's pick three points, and label the plane that they define, Plane A.  The 4th point will not be in Plane A (as it is given that the 4 points are not coplanar).  Call the perpendicular distance between Plane A and the 4th point, X.

Now create a plane that is parallel to Plane A, halfway between Plane A and the 4th point.  Call this Plane 1.  So Plane 1 is a distance of X/2 away from the 4th point and X/2 away from Plane A.  Since the distance between a point and a plane are always perpendicular, all 4 points are now equidistant to Plane 1.

Notice there are 4 ways to pick 3 points to make a plane with another point not in that plane.  So we can continue and use another 3 points to create Plane B to construct Plane 2, and so on with Plane C & Plane 3, and then Plane D & Plane 4.

So those are the first 4 I see.  I haven't been able to prove if there are no others.  I do know that such a plane cannot exist "outside" of the tetrahedron created by the 4 points.

 

Posted by nikki on 2004-12-17 15:42:53