We have a billiards table with height $h$ and width $w$, as illustrated in the image below. Notice that it has no pockets. Billiards Table

A ball with radius $r\ll w$ (can be regarded as a point) is propelled from the bottom-left corner at an angle $5°\leq\alpha\leq 85°$ with respect the horizontal axis (as depicted in the figure). The movement is constrained to a plane, so there is no depth involved. The velocity of the ball is irrelevant.

There are no dissipative effects, so the ball will keep bouncing forever with no loss of energy. In other words, the collisions (ball-walls) are perfectly elastic (the angles are conserved before and after a collision).

  1. Find the angle $\alpha$ that maximizes the number of bounces until the ball returns to the first collision point. How many bounces $n(h,w,\alpha)$ will it take for this to happen? If you think there is no angle $\alpha$ for which the ball returns to the first collision point, demonstrate it.

NOTE: The first collision point is the point $(x_0,y_0)$ at which the ball first touches a wall.

  • $\begingroup$ I don't quite understand the downvote. $\endgroup$ – Jose Lopez Garcia Jan 15 '16 at 23:43
  • $\begingroup$ May the downvoters at least leave a comment about why they don't like this puzzle? Thanks. $\endgroup$ – Jose Lopez Garcia Jan 16 '16 at 0:20
  • $\begingroup$ How tall are the edges of the table, how high off the table was the ball released, and how fast was the ball thrown? $\endgroup$ – Ian MacDonald Jan 16 '16 at 0:32
  • $\begingroup$ Does the speed matter in this 2D problem? It all happens in the XY plane, so you can say the edges have depth. Is that why you downvoted? $\endgroup$ – Jose Lopez Garcia Jan 16 '16 at 0:39
  • 2
    $\begingroup$ We might as well start at the first time the ball hits the wall. If we launch at an irrational slope, we will never return to the first collision point. This invalidates the whole question. -1 There are certainly some angles for which the ball returns to the first collision point, but they are of measure zero in the arc around the launch point. $\endgroup$ – Ross Millikan Jan 16 '16 at 3:52


as alpha approaches 0, the number of bounces approaches infinity. There is no true solution to this because any alpha you give I can provide a smaller alpha which will have more bounces


any angle not equal to tan(h/w) will cause a rectangle of some size to be drawn, but I think I may be misunderstanding what you mean by "draw a rectangle"

  • $\begingroup$ 1. >! No valid answer. The problem is about maximizing the number of bounces until the first collision point is reached. As your reasoning is logical, you didn't prove there is an $\alpha$ for which the number of bounces is greater than a smaller one. You didn't even prove the ball will ever reach its first colision point, $(x_0, y_0)$ $\endgroup$ – Jose Lopez Garcia Jan 15 '16 at 22:21
  • $\begingroup$ A new restriction has been added. The angle must lie within 5 and 85 degrees. $\endgroup$ – Jose Lopez Garcia Jan 15 '16 at 22:28
  • $\begingroup$ my answer still applies as alpha approaches tan(h/w) $\endgroup$ – Slepz Jan 15 '16 at 22:30
  • $\begingroup$ Your answer for the second question is also wrong. Give it a deep thinking! $\endgroup$ – Jose Lopez Garcia Jan 15 '16 at 22:34
  • $\begingroup$ For a rectangle to be drawn, the ball needs to bounce between 4 stationary points. $\endgroup$ – Jose Lopez Garcia Jan 15 '16 at 22:38

For part 2.

There can be no rectangle. First, we show no rectangle can contain the starting point. Consider this point. This lies in the corner of the table, so in any drawn rectangle must also be a corner. But, this means the ball must return to this point at a 90 degree angle to the direction it left from. Since the whole corner is only 90 degrees, it must either come from off the table, impossible, or the initial angle must be 0. However, an initial angle of 0 would bounce straight back and not form a rectangle.

Now, we show that any rectangle is a closed loop. Consider any corner of the rectangle. This must be on a wall, and must be a 90 degree angle. Thus, both entry and exit angles must be 45 degrees. Since the walls are 45 degrees from each other, this means that once an angle leaves a wall at 45 degree, all subsequent angles will be 45 degree angles. However, this means that when the point returns to the starting corner, it will be at a 45 degree angle and must then repeat the rectangle. By simply reversing the direction, we see that this means all rectangles are closed loops.

If every rectangle is a closed loop, and no rectangle can contain the starting point, then the starting point cannot create a rectangle.

  • $\begingroup$ The rectangle must be made from the first collision point ahead. $\endgroup$ – Jose Lopez Garcia Jan 15 '16 at 22:30
  • $\begingroup$ I would say that there is no rectangle with an edge ray that could point to the origin. As each point of the rectangle lies on the edge, any ray would be off the table. $\endgroup$ – JonTheMon Jan 15 '16 at 22:51
  • $\begingroup$ @JonTheMon: Given the first collision point $(x_0, y_0)$, the rectangule must have coordinates $A=(x_0, y_0)$, $B=(x_1, y_1)$, $C=(x_2, y_2)$, $D=(x_3, y_3)$ $\endgroup$ – Jose Lopez Garcia Jan 15 '16 at 23:01
  • $\begingroup$ Made it clear in the puzzle, check out. $\endgroup$ – Jose Lopez Garcia Jan 15 '16 at 23:09
  • $\begingroup$ Hey, you made your answer visible. Hide it so others don't accidentally look at it! $\endgroup$ – Jose Lopez Garcia Jan 15 '16 at 23:17

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