Bert practises potting snooker balls on an elliptical table. He places a ball at one focus, strikes it with his cue in a random direction so that it travels towards the edge of the table, and when it bounces back in from the edge it always goes down the hole positioned at the other focus. Ideally this will only work when he puts negligible spin on the ball.

At night he dreams of a three-dimensional shape in which there's no gravity and he can do the same. What such shapes exist?


A proper ellipse, namely one that isn't a circle, has exactly two distinct foci. These lie on its long diameter, equidistant from the diameter's midpoint. One of their interesting properties is that the sum of the distances from any point on the ellipse to each of the two foci is constant. More information can be found here and here.


Here's how to find all such shapes:

The question requires the ball to start from one point, bounce off a wall, then go to another specific point (no matter which direction it was originally hit in). Now, we can assume (due to the movement in 3D space with no gravity or spin) that the ball will be moving in a straight line before the bounce, and moving in a straight line after the bounce. Two intersecting lines define a plane (they intersect at the point at which the ball bounces), so any given trajectory of the ball will happen entirely within a single plane.

Now, this requires us to find a 3D shape for which every 2D plane taken through the start and end points has the property we want; in other words, every cross section taken through the line from the start to end point is an ellipse, with its foci at the points in question. This already severely limits the possible shapes we can have.

Finally, we need the bounce off the wall to actually stick to the plane in question; if we choose coordinates so that the ball is hit "horizontally", we don't want any "vertical" component to the resulting motion. This means that for any plane through our start and end points, the tangent at the side of our 3D object is perpendicular to that plane. Or in other words, if we take the axis from the start to the end point, and take any plane at right angles to that axis (rather than along that axis, like we've previously been discussing), it cuts our 3D object in a circle (the only shape for which a tangent at a point is always perpendicular to a line from that point to the centre).

Given that we need the object to have only circular cross-sections perpendicular to the axis, it must be a solid of revolution; and given that we need it to have only elliptical cross-sections along the axis, it must be a solid of revolution of an ellipse, specifically (where the foci of the ellipse are the start and end points, and the revolution is around the line between them). Assuming a continuous surface, there are no other possibilities. (This is the same solution that was mentioned in the other answer, just explained in slightly different terms.)

We can also easily see that the surface can't be discontinuous or concave; if it were, it would be possible to hit the ball into its edge (or an internal edge/corner), and the exact position of the ball would then have a large effect on the angle at which it bounced off. It's impossible for the entirety of the large range to reach the same point with no additional bounces.

Note that I'm assuming here that the size of the ball is insignificant; without that assumption, there might be additional (fairly uninteresting) possibilities involving holes in the wall that are too small for the ball to fit through.

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  • $\begingroup$ This is a very nice answer. (And I had to think a bit before I understood the third paragraph.) $\endgroup$ – user36946 May 24 '17 at 1:49

Isn't it just

an ovoid?

Am I missing something obvious here? Are you looking for an object in real life that matches the description?

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    $\begingroup$ ... more specifically a (rot13) cebyngr fcurebvq. $\endgroup$ – 2012rcampion May 23 '17 at 22:19
  • $\begingroup$ @2012rcampion precisely :) $\endgroup$ – CactusCake May 23 '17 at 22:25
  • $\begingroup$ CactusCake - no, I'm not looking for a real life object. I hadn't encountered "ovoid" in the meaning of (rot13) cebyngr fcurebvq before, but, @2012rcampion, yes, that will work! And it's quite easy to prove too. But is that the only solution? $\endgroup$ – user36946 May 23 '17 at 23:57

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