Do a Google for "Soddy Hexlet". The first entry should be "Hexlet -- From MathWorld" which gives a lot of information.

For a beautiful animation see
https://members.ozemail.com.au/~llan/soddy.html

Soddy started with the configuration shown at MathWorld. Your three given pairwise externally tangent spheres configuration can be converted to this by inversion about one of the three spheres.

Let the three fixed spheres A, B, and C be labeled per MathWorld and call this the ORIG configuration. Do an inversion about the point of tangency between spheres A and B and call this the INV configuration.

In INV spheres A and B become parallel planes and sphere C a sphere between them and tangent to them. The hexlet spheres become spheres between planes A and B and tangent to them and tangent to sphere C. They have centers at the vertices of a regular hexagon centered at the center of sphere C. There is another sphere, say D, which is concentric with sphere C and tangent to each of the hexlet spheres.

In ORIG the hexlet spheres are internally tangent to sphere C and externally tangent to sphere D. Let C' and D' and c and d be the centers and radii of C and D. Let S' and s be the center and radius of an arbitrary hexlet sphere S. |C'S'| + |S'D'| = (c - s) + (s + d) = c + d (a constant). Therefore, the centers of the hexlet spheres lie on an ellipsoid of revolution with foci C' and D'. Let A' and B' and a and b be the centers and radii of A and B. ||A'S'| - |S'B'|| = |(a + s) - (s + b)| = |a - b| (a constant). Therefore, the centers of the hexlet spheres lie on some kind of hyperbolic surface. Maybe the intersection of these two three dimensional surfaces is the planar ellipse the I am looking for.
MathWorld says that there are two planes, say P and Q, tangent to the hexlet spheres.

In INV the planes P and Q would be spheres tangent to the hexlet spheres and passing through the center of inversion. I am still having trouble visualizing these spheres.

Hope this has been a help.

-- Bractals

Thanks -- I was googling for "Soddy sphere" and found other links that were much less infomative than this one.
>
>For a beautiful animation see
>https://members.ozemail.com.au/~llan/soddy.html

That is a really nice Java applet, though point and drag controls would have been nice for some of the features. I got a good idea of what is going on just from playing with it.

...

>MathWorld says that there are two planes, say P and Q,
>tangent to the hexlet spheres.
>
>In INV the planes P and Q would be spheres tangent to the
>hexlet spheres and passing through the center of inversion.
>I am still having trouble visualizing these spheres.

Once I started trying to prove that the centers of the hexlet spheres were coplanar, I ended up with these two planes as part of the proof. Here is what I came up with:

Since in the INV configuration, the hexlet forms a perfect hexagon of spheres you can "pull a sphere through the hole" in the hexagon. Start with a plane tangent to the hexlet from above. Think of this plane as a sphere of infinite radius, with its center directly on the axis of symmetry of the hexagon. Then begin to shrink this sphere by moving its center down the axis, while adjusting the radius to keep it tangent to the hexlet. You can always do this because the hexagon is prefectly regular and the hexlet spheres are equal in radius, so if the moving sphere touches one, it touches all. As the center of the moving sphere passes through the plane of the hexlet centers, the sphere will coincide with the image of C. Let the center continue down the axis, and let the moving sphere expand again, eventually becoming the lower tangent plane of the hexlet.

Now since the inversion point ( let's call it I) was inside C in ORIG, it is still inside the image of C in INV. The moving sphere sweeps through every point of the volume inside C, because every point within C is initially outside the moving sphere, then later inside it when it coincides with C, then later still outside it again. Therefore, the moving sphere touches every point within C twice, once when the lower side first reaches that point, and later when the upper surface passes it. The moving sphere also touches every point on the surface C twice, as is easy to visualize. Therefore the moving sphere touches I twice.

Take a snapshot of the sphere when it first touches I and call it P, and another snapshot when it touches I the second time, and call it Q. These two spheres are tangent to all 6 members of the hexlet, and they invert to planes in ORIG, since they pass through I. This proves that the hexlet spheres are sandwiched between two planes P and Q in ORIG, and are tangent to both planes.

Now it is easy to prove that their centers lie on a common plane as well. If P and Q are parallel, take the parallel plane halfway between them, which will obviously contain the centers. If P and Q are not parallel, construct the plane that bisects the angle between them (find their line of intersection L, construct perpendiculars to L in P and Q from a common point, construct the angle bisector K of these perpendiculars, and get the plane containing L and K). Then the centers of the spheres lie on this plane. You could prove this in detail, but I think it is is clear enough from symmetry.

>Hope this has been a help.

Yes, it was very interesting to see the Soddy sphere animations. I hope my post has been a help to you as well.

--Stuart Anderson

Thanks for the visualization of spheres P and Q in the INV configuration (especially for reminding me that the center of inversion is inside sphere C). You are right - with the hexlet spheres between planes P and Q in the ORIG configuration it is obvious that their centers lie in a plane. Thanks again.

-- Bractals

You're welcome. Oh, by the way, I forgot to mention it explicitly, but this also answers your second question. Since the images of A, B and C in INV are fixed, the hexlet can only rotate around its axis of symmetry in INV. The construction of P and Q is the same regardless of the orientation of the hexlet, and P and Q determine the plane in ORIG that contains the centers of the hexlet spheres. Therefore the spheres A B and C determine this plane.

But you probably already saw that.

--Stuart Anderson