Friday Puzzler -- Happy Pi Day!
Alex Y
What better day for a puzzle about Pi. This one may appeal mostly to the physics/engineering/math types. It is about a very simple (in concept) machine to calculate Pi to any required number of digits.
Take a cue ball and place it 6" from a bumper on a pool table. Put another ball between the cue ball and the bumper, and roll the cue ball into the object ball. There will be three collisions: The cue ball hits the object ball, which hits the bumper, and then hits the cue ball. Three collisions = Pi to one digit.
To generate more digits of pi, we need to go to the physics department and borrow one of their frictionless rails, and some mass units that ride that rail and that have perfectly elastic collisions with each other and with the immovable wall at the end of the rail. Our billiard ball experiment becomes two equal masses on the rail.
Now if the initial mass is 100 times that of the object mass, it will keep moving toward the wall after striking the object ball, and until it has collided with the object ball a number of times. This system will create 31 collisions, including wall strikes, the first two digits of pi, or int(10*pi) ["int" is used here to mean "take the integral part of" or excel function "rounddown(x,0)"]
If the initial mass is 10,000 times that of the object mass, there will be 314 collisions, int(100*pi)
And in general, if the first mass is 10^(2n) times that of the object mass, there will be int(pi*10^n) collisions, giving us n+1 digits of pi.
Admittedly, this system has some practical and even theoretical limits (you have to ignore the gravitational attraction between the masses, which would affect the results at some point).
This set-up seems to have nothing to do with circles. The puzzle is, why does it generate pi in this way?
Credit: Alon Amit, as related in NYT column by Gary Antonick