The Marble Ring
Solution


Such a lot of questions! I'm not going to spoil your fun by giving you all the answers. But I hope you've thought about the problem a bit already.

We have assumed that no marble has a zero velocity. One implication of this assumption is that if we draw any fixed circle on the table, "eventually" there will be no marbles in the circle, and no marbles will ever enter that circle. This is because any marbles in the circle will have to leave eventually, and any marbles outside of the circle can only pass through the circle at most once. Since there are finitely many marbles, eventually, the circle will be empty with no prospect of any future marbles.

But if every fixed circle eventually has no marbles in it, where are the marbles? Clearly, the position of every marble must be tending to infinity. Does that fact by itself mean that D(t) must also tend to infinity? Well, certainly we could have a group of marbles with the same velocity. Their positions would tend to infinity, but D(t) would be an unchanging constant.

So now let's use the assumption that at least two marbles have different velocities (that is, different speeds or directions). Only then may we conclude that the asymptotic separation of these two marbles becomes infinite, and hence the diameter of the ring which encloses them (and all the other marbles) must grow without bound.

If something (the enclosing circle) starts out of finite size and ends up infinite, then it must have a minimum value. It could simply increase from the instant we begin, a somewhat uninteresting "initial minimum". A more interesting case would occur if the ring size decreases first and then increases. We could actually arrange this by starting the marbles in a circle, and aiming them towards the center (but just off a bit so they don't collide). Then the enclosing circle will decrease in size before increasing. So it's clear that the function D(t) can have a local minimum.

But this can't happen twice. To convince yourself of this, grab a handful of spaghetti loosely, and think of each piece as marking the track of a marble over time. You can always hold the spaghetti in such a way that there is a "neck" or "waist" or narrow part, but you can't have two such waists with a fatter part in between. To convince yourself of this mathematically, realize that if this were so, you could draw a "shadow" of the system, with a short line, a wide line, and a short line, representing the system over time, "viewed from the side". Then we can always arrange the projection so that both ends of the wide line represent marble positions. Then there must be trajectories that pass through the two ends of the wide line, and intersect the shorter lines above and below. But the two lines cannot get closer in both directions. So this is impossible, and the minimization happens only once. Mind you, if all the marbles have the same velocity, then the enclosing ring is of constant size, and the spaghetti is just a cylinder. But we've already said that's a boring case not worth considering further.

As for questions of continuity (easy) and differentiability, I leave those to you.

In 3 dimensions, the corresponding problem might occur when we encounter a region of space containing asteroids. Assume we can ignore gravity, and that the asteroids never collide. Then there will have been some time at which the asteroids were contained in the smallest possible sphere, and after that time, the containing sphere has always increased in size, at a rate which asymptotically tends to a constant velocity, and can be regarded as a linear radial growth around some center that moves with a constant speed.

Back to the marble ring puzzle.


Thanks to the owner of the Ames, Iowa "Pita Pita" for telling me about this problem!

Last revised on 01 November 2003.