-----Original Message----- From: math-fun [mailto:math-fun- bounces+davidwwilson=comcast.net@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Friday, October 17, 2014 9:39 PM To: math-fun Subject: Re: [math-fun] Marble problem
Each point on the surface of the marble corresponds to some ordering, the ordering that prevails when that point is in contact with the floor. We can call two points equivalent if they induce the same ordering, and draw a map on the marble's surface whose regions, edges, and vertices represent the equivalence classes. (The edges and vertices correspond to degenerate orderings where two or more points are at equal heights.)
I think the edges are all pieces of great circles; if the points are in general position there are k = n(n-1)/2 such circles. Now, k circles can cut a sphere into at most k^2 - k + 2 regions [A014206], so one might think that with three points you could get 8 orderings! Well, obviously something is wrong here: the problem is that because three points can be equidistant from the floor, the great circles sometimes meet in threes, not in pairs. For 4 points, the number of circles is 6; they meet in threes in 4 pairs of antipodal points, and there are also 3 pairwise meetings. I am running out of steam here because I don't have a scribbling surface handy.
Consider three points (bubbles). There are three great circles associated with pairs of these points. The planes of these great circles must pass through the center of the sphere (marble), they must also be perpendicular to the plane of the points. This implies that the planes of the great circle pass through the same axial line of the sphere, and cut the sphere surface into only 6 regions instead of the possible 8 by great circles in general position.