There's a spaceperson with a very bouncy ball and a rigid box in the form of a cube with one face missing. One day she throws the ball into the box and notices the ball bounces off each face exactly once before exiting through the missing face.
(The ball travels in a perfectly straight line, being unaffected by air resistance, spin or any other forces other than the reactions with the box. Also the ball bounces symmetrically such that the incoming angle is identical to the outgoing angle and again is unaffected by spin. Also, the box cannot be moved while the ball is in motion.)
How many different combinations are there of the order in which the ball can bounce off all five faces?
On returning to Earth our spaceperson notices that new combinations are possible.
(All conditions are the same except the ball is now affected by gravity.)
How many different combinations are there of the order in which the ball can bounce off all five faces now?
In plane problems of this type, one would create infinitely many reflections of the square in which the ball is bouncing. Then, the trajectory can be shown by a straight line going through the walls, then considering the next, reflected room to interpret the bounced trajectory for as long as necessary.
In the threedimensional instance we can follow such a path with regard to the 4 walls of the box (excluding the face of the cube opposite the open side), then draw any line segment from within the original square (considered as starting at the open face), to a point within some other, reflected (singly, doubly, etc.) version of the square. The midpoint of the line would be where the trajectory hits the face opposite the open face and the endpoint where it reemerges from the open face of the box.
I am assuming that each face is to be hit once only. Otherwise, with sufficiently low component into the box, as many sequences as you'd like could be obtained, with more and more repetitions.
For brevity, even though there is no gravity, nor cardinal directions, lets call the face opposite the open one the bottom. Let's call the other faces N, E, S and W, in that order clockwise as seen from "above" the open face.
If the 2D representation line starts with opposite cardinal directions, say N then S, the trajectory could be made next to go to E or W, but then would have to hit N again before getting to W or E respectively. So NS, SN, EW and WE are not allowed as the first two of the cardinal walls. But the sequences NEWS and NESW are both possible. Similarly for any adjacent pair of cardinal walls, there are two choices for the sequence of the remaining two. So there are 16 sequences of the cardinal walls (we'll insert the bottom later): NEWS, NESW,ENWS, ENSW, SWNE, SWEN, WSNE, WSEN, SEWN, SENW, ESWN, ESNW, NWSE, NWES, WNSE and WNES.
When looking at the tiled diagram of reflected rooms (boxes), we can try various places within the start box and ending box to see how many places the bottom bounce can take place within the sequence, represented by the midpoint of the drawn line segment that represents the total sojourn in the box.
It seems to me that the B can only be between the E and the W of NEWS, making NEBWS. Similarly, for NESW, it looks like the B must be between the E and the S.
That leaves us with the same 16 sequences, just with a B added in the middle:
NEBWS, NEBSW,ENBWS, ENBSW, SWBNE, SWBEN, WSBNE, WSBEN, SEBWN, SEBNW, ESBWN, ESBNW, NWBSE, NWBES, WNBSE and WNBES.
+N+N+N+N+N+N+N+
        Draw any line segment.
W E W E W E W E Its endpoints are where the ball
+S+S+S+S+S+S+S+ enters and leaves the box.
        Its midpoint is where it hits the bottom.
W E W E W E W E Take the sequence of the lines crossed,
+N+N+N+N+N+N+N+ inserting the B at the midpoint.
       
W E W E W E W E
+S+S+S+S+S+S+S+
       
W E W E W E W E
+N+N+N+N+N+N+N+
       
W E W E W E W E
+S+S+S+S+S+S+S+
       
W E W E W E W E
+N+N+N+N+N+N+N+
       
W E W E W E W E
+S+S+S+S+S+S+S+
       
W E W E W E W E
+N+N+N+N+N+N+N+

Posted by Charlie
on 20051117 14:57:22 