Billiard Ball Problem

Billiard ball problem can be seen as one elegant mathematical problem. In this article, we will analyse it mathematically.

Before we begin, let us first see the problem. Given below is the situation. Can you predict, in which pocket will the ball falls in?

Let us now connect this problem with mathematical conditioning. We all know reflection property very well. Now what you have thought is that whenever ball strikes any mirror then it enters into a rectangle which is just a mirror image of it. Now just keep on constructing rectangles with taking common sides CD and BC (initially) then their reflections too.

Remember one thing that the ball will keep on moving in its direction until it reaches one of the vertices of our apparent rectangles. and hence it is obvious that the no. of such apparent rectangles that you will need in the horizontal direction will be 67 and in the vertical direction is 100 (smallest possible). being 67 odd hence, it is obvious that it will fall either in C or D and 100 being even it will eventually fall into D.

Alternatively, we can think, as to travel from the lower-left pocket to any other pocket, the ball must traverse integer multiples of the length and width (that is, the ball must travel so that it is either at the top/bottom and right/left of the pool table to land into a pocket). We also know that since it is launched at a 45 degrees angle its speed is split evenly into the horizontal and vertical component (this means at any point it will have travelled the same horizontal and vertical distance).

If we let the horizontal lengths travelled by “x”, and the vertical widths travelled be “y”, then:

100x = 67y where x,y are integers

And, the first value (s) that this works for is x=67 and y=100. This means the ball will have travelled an odd number of lengths (ends up at the right) and an even number of widths (ends up on the bottom), and we can easily conclude from this that it ends up in the bottom right pocket, giving the answer of D.

Wait think about it again! What all we had done so far is that we have computed the lowest common multiple of the length and width, for a problem that seemed so much more complicated than that. Is not it Mind-blowing!

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