Am 08.08.2014 18:15, schrieb Le_Forgeron:
> If I got your problem correctly:
> 1. you have 4 points: A,B,C,D, not coplanar, and not even aligned by
> group of 3. (A,B,C,D define a non-degenerated tetrahedron)
> 2. you want to find the sphere whose surface includes all the points.
>
> You need additional constraint on D, or the sphere does not exist.

I don't think that's true: Constraint 1. is sufficient.

Every non-degenerate triangle has exactly one circumscribed circle, so 
obviously there exists exactly one circle touching ABC. Let O be that 
circle.

Let N be the rotational axis of O. If D is on N, Let P be an arbitrary 
plane trough N; otherwise, let P be the plane through D and N. Note that 
in either case both D and N are on P.

Note that P intersects O in two points; let these be E and F. Also note 
that D is not on the line EF, otherwise ABCD would be coplanar. Thus, 
DEF is a non-degenerate triangle, and itself has a circumscribed circle. 
Let Q be that circle.

Note that E and F have equal distance to N, thus N runs through the 
center of Q, so rotating Q about N defines a sphere. Let S be that 
sphere. Note that D lies on Q and hence on S; also note that E and F 
both lie on Q, and their rotation about N yields O, thus O lies on S. 
Note that ABC all lie on O, and therefore also on S.

=> ABCD all lie on S.

Finally note that if P if ambiguous, the case is entirely rotationally 
symmetric except for the positions of ABC, and thus S is the same for 
all possible choices of P.

=> S is unambiguous.

=> For any non-degenerate tetrahedron ABCD, there exists exactly one 
circumscribed sphere.

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