Answer to Question 10/99

The answer: Such a path exists. It was originally found by Hugo Steinhaus. (See M. Gardner's Sixth book of mathematical games from Scientific American, Freeman, San Francisco, 1971. It describes several problems of this type.) Below is one possible solution. It is followed by solutions submitted to us, which expand upon the solution and offer an intuitive argument.


One solution:

Say the size of the cube is 3a, and its corners are located at coordinates
(0,0,0,), (3a,0,0), (0,3a,0), (0,0,3a), (3a,3a,0), (0,3a,3a), (3a,0,3a), (3a,3a,3a).
A path satisfying the conditions of the problem connects the points
(0,a,2a) -> (a,2a,3a) -> (2a,3a,2a) -> (3a,2a,a) -> (2a,a,0) -> (a,0,a) -> (0,a,2a).
The length of each straight segment of the path is sqrt{3}a. The projection of this path onto any face of the cube is a rectangle, which proves that at any impact point the angle of impact is equal to the angle of reflection. The following figure depicts the path:





Bill Schwennicke noted that there are infinitely many solutions of this type. Here is his solution of the problem (slightly edited):

Assume a unit cube with opposite corners at (0,0,0) and (1,1,1). An infinitesimal ball heading out from (0,0,0) at an equal angle from each axis would hit the cube at (1,1,1) and then bounce straight back. Offsetting the starting point a little, but heading out at the same angle, the ball would proceed as in the following example: (0,1/8,2/8), (6/8,7/8,1), (7/8,1,7/8), (1,7/8,6/8), (2/8,1/8,0), (1/8,0,1/8), (0,1/8,2/8), etc., thereby making the required circuit. A finite-sized ball with radius r can be substituted for the infinitesimal one by expanding the cube by an amount r in each direction, and having the above coordinates be for the center of the ball.


Andrew Wiggin presented a nice argument:

On every collision, the normal component of the velocity just reverses, and the other two are conserved. Hence the directions are decoupled, and the answer doesn't depend on the number of dimensions. All that matters is that the period of the motion in each direction is the same, i.e., that the absolute value of the velocity is the same in all directions. So the starting velocity should be of the form V = c(±1, ±1, ±1), and the starting point can be anything. The ball will always move parallel to one of the body diagonals of the cube.


P.S. (11/99) Avi Nagar and Heinz Kabelka made the following observation (we present A. Nagar's message, slightly edited): I just wanted to point out the similarity of this problem to a corner-reflector which is desciribed as an experiment in one of the pages of the Melbourne University Physics Department (http://www.ph.unimelb.edu.au/lecdem/oa6.htm). [Before you continue, take a look at this experiment by clicking here.] The problem can be viewed as a combination of two corner reflectors. Since each reflects light (or a billiard ball) parallel to its original direction, we need to find a direction that will ensure that after two corner reflections we'll be back at the exact starting direction/point. The symmetry of the cube ensures that any direction parallel to the big diagonals is a good solution (because not only is the beam reflected from a corner reflector parallel to the incoming beam, but it also has the same offset from the corner).

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