Hello Pitagora'sSince no-one has been able to find a proof of this on the internet, I have come up with this.

Look at the attached diagram. It shows a grid made up of rectangles, each one representing an image of the billiard table. The ball is projected from at to the bottom edge . In this diagram, instead of being reflected when it hits a side of the table, the ball continues on its straight-line path. So the new path is, at each impact, the mirror-image of the actual path the real ball would take.

This 'virtual ball' will clearly enter another 'pocket', then, as soon as it strikes two sides simultaneously - in other words, where the diagonal line in our diagram passes through the point where two of the grid-lines meet. Denote this point by . Then, since is at to the bottom edge of the grid, the distances and are equal, and each is equal to a multiple of and a multiple of respectively. In other words:Now and are co-prime. Therefore is a multiple of and is a multiple of . Clearly the smallest values that satisfy this equation are:, for some integers .

, andSo there are rectangles placed horizontally side by side to make up the line , the line intersecting the (vertical) right-hand edge of of these before entering the virtual pocket at B. Similarly AB intersects horizontal edges of the rectangles that make up the line .

Each of these intersections represents an impact with an edge of the table. There are therefore impacts in total.

Grandad