Impossible Bounce Problem

Geometry Level 2

Consider a polygonal billiard table with two marked points, one for a "ball" and one for a "hole". Suppose the ball is "hit" once to go off in any direction and the reflects off the billiard table an unlimited number of times. Is it possible to configure this such that the ball will never reach the hole, no matter which direction the ball is aimed at?

(This assumes geometric purity: the "ball" has no mass and the "ball" and "hole" are considered single points. Also, if the ball hits a vertex point as opposed to a side then the reflections are absorbed.)

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2 solutions

David Vreken
Jan 17, 2018

It is impossible for a ball placed on the red dot on the right isosceles triangle billiard table pictured below to be reflected back to itself, given any number of reflections:

(See here for a proof.)

Therefore, using this right isosceles triangle as a building block, a polygonal pool table can be constructed (through a series of reflections of the building block) such that a ball can never reach the hole, as long as the hole and the starting position of the ball are on reflected red dots, and as long as all the reflected blue dots are on a corner or outside the polygonal pool table. (If a reflected blue dot is on the inside or edge of the polygonal pool table, it may be possible to make a path from the ball to the hole through that dot. See here for more details.)

Using these guidelines, there are several polygonal pool tables that can be constructed. Here is one with 24 sides (which is really just a variation of the Castro solution):

Here is one with 25 sides:

And here is one with 27 sides:


Note: While solving this I noticed that a right isosceles triangle is not the only shape that has the property that it is impossible for a ball to be placed on a certain point and reflect back on itself. A square, a rectangle, a 30-60-90 triangle, and a 60-90-120-90 kite also have this property. These are pictured below, where the red dot would be the starting point of a ball:

I am not sure if there are any other shapes with this property. Also, I was unable to make a polygonal pool table using any of these shapes as building blocks (other than the right isosceles triangle) with the given property that the ball will never reach the hole. Perhaps someone else can do this or prove that it is not possible.

While this obviously doesn't prove anything in general, is there some reason why this strategy doesn't might not allow less than 24 sides?

Jason Dyer Staff - 3 years, 4 months ago

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When I tried to make a polygonal billiard table with less than 24 sides, I found that there was not enough room to create one with adjacent reflected red dots (2 units away either vertically, horizontally, or diagonally) for the starting position of the ball and the hole and still have room for the walls of the billiard table to come in between. The next best thing was to have two red reflected dots 4 units away horizontally. Now, the starting position for the ball must have the 8 points around it covered by the walls, and since these points must be corners out of the way of the starting point this necessitates that another 4 points are used (2 away from the starting point left, right, up, and down) for a total of 12 points. The same argument can be made for the point at the hole, for another 12 points. One of those points in between the two reflected dots can be shared, but an additional point must be used for the bridge between the starting point and the hole. Therefore, there are at least 12 + 12 - 1 + 1 = 24 points needed in the polygonal billiard table, for at least 24 sides.

David Vreken - 3 years, 4 months ago

For the square/kite: what do you envision happening when you shoot the ball directly to the opposite vertex? I can't see anything besides it going back to the red dot.

John Gallagher - 3 years, 4 months ago

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According to the given rules, if the ball hits a vertex point the reflections are absorbed

David Vreken - 3 years, 4 months ago

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Yep totally missed that, thanks.

John Gallagher - 3 years, 4 months ago
Jason Dyer Staff
Jan 8, 2018

These are the solutions given in the setup for Open Problem #3. You can assume this is drawn on unit graph paper such that all polygon vertices fall on integer coordinates. The proof is at the same post.

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