# Thread: the crazy billiard table

1. ## the crazy billiard table

A man designs a 2m by 2m billiard table with only 4 pockets, one in each corner, marked A, B, C, D. A ball is shot from directly in front of one pocket at a 45 degree angle. Obviously, because the billiard table is square, it will go into the diagonally opposite hole.

What happens now if the table is increased to 2 * 3, 2 * 4 , 2 * 5 etc. and the ball is again hit at a 45 degree angle. Is there a way to predict what pocket the ball will go into in each trial?

What about for a table 3 * 3, 3 * 4 3 * 5 etc. Can you predict what pocket it will go in?

Is there a way to write a generalised formula for this, to find out which pocket it will go into for a very large table, 342 * 275 for example?

2. Originally Posted by bruxism
A man designs a 2m by 2m billiard table with only 4 pockets, one in each corner, marked A, B, C, D. A ball is shot from directly in front of one pocket at a 45 degree angle. Obviously, because the billiard table is square, it will go into the diagonally opposite hole.

What happens now if the table is increased to 2 * 3, 2 * 4 , 2 * 5 etc. and the ball is again hit at a 45 degree angle. Is there a way to predict what pocket the ball will go into in each trial?

What about for a table 3 * 3, 3 * 4 3 * 5 etc. Can you predict what pocket it will go in?

Is there a way to write a generalised formula for this, to find out which pocket it will go into for a very large table, 342 * 275 for example?
use pythagoross theorem I attach some pictures hope that will help you
begin from the two green arrow blue numbers for the segments on the sides which the ball did

for the 342 * 275 table I draw I complex pic here I use orange numbers to determine the new segments after the ball hit the four sides

Is there any questions ?

3. Code:
D 2 G    8       C
2
F                8

H
4

E

10

A                B
Actually quite "easy" to draw the paths; above is a 10 by 22 table ABCD; cue ball shot from A:
1: hits BC at E (10 above B)
2: hits AD at F (2 below D)
3: hits CD at G (2 from D)
4: hits BC at H (8 from C)

And you're basically finished: you have 4 lines (AE, EF, FG, GH):
ALL other paths will form lines parallel to these lines; continuing:
from H simply draw a line to side AD parallel to line FG.

And carry on similarly until a path line hits one of corners.
In this example, corner C will be hit after 14 bounces off the sides.

If you really want to have fun, get a sheet of graph paper and draw
a table 27 by 34 (easy enough using the little squares as units).
Go as I explained above: this time you'll hit corner D, after 59 bounces!

Haven't tried for a general case formula (probably won't either!), but
obviously has something to do with LCM(long side, short side).