# Billiard balls and the number theory result

• Dec 3rd 2009, 11:14 AM
Pitagora's
Billiard balls and the number theory result
Hello,

"Given a rectangular billiard table with only corner pockets and sides of integer lengths m and n (with m and n relatively prime), a ball sent at a 45 degree angle from a corner will be pocketed in another corner after m+n-2 bounces ".

This is the Steinhaus and Gardner result.

Can anyone please tell were i can find (on the Internet = for free) the proof of this result.
i could find the books published by this authors on ebay and similar websites but all i need a free version as i can not effort any of them.

Pitagora's
• Dec 6th 2009, 12:07 PM
Hello Pitagora's
Quote:

Originally Posted by Pitagora's
Hello,

"Given a rectangular billiard table with only corner pockets and sides of integer lengths m and n (with m and n relatively prime), a ball sent at a 45 degree angle from a corner will be pocketed in another corner after m+n-2 bounces ".

This is the Steinhaus and Gardner result.

Can anyone please tell were i can find (on the Internet = for free) the proof of this result.
i could find the books published by this authors on ebay and similar websites but all i need a free version as i can not effort any of them.

Pitagora's

Since 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 $\displaystyle m \times n$ rectangles, each one representing an image of the billiard table. The ball is projected from $\displaystyle A$ at $\displaystyle 45^o$ to the bottom edge $\displaystyle AP$. 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 $\displaystyle B$. Then, since $\displaystyle AB$ is at $\displaystyle 45^o$ to the bottom edge of the grid, the distances $\displaystyle AP$ and $\displaystyle BP$ are equal, and each is equal to a multiple of $\displaystyle m$ and a multiple of $\displaystyle n$ respectively. In other words:
$\displaystyle AP = BP\Rightarrow pm = qn$, for some integers $\displaystyle p, q$.
Now $\displaystyle m$ and $\displaystyle n$ are co-prime. Therefore $\displaystyle p$ is a multiple of $\displaystyle n$ and $\displaystyle q$ is a multiple of $\displaystyle m$. Clearly the smallest values that satisfy this equation are:
$\displaystyle p=n$, and $\displaystyle q=m$

$\displaystyle \Rightarrow AP=BP=mn$
So there are $\displaystyle n$ rectangles placed horizontally side by side to make up the line $\displaystyle AP$, the line $\displaystyle AB$ intersecting the (vertical) right-hand edge of $\displaystyle (n-1)$ of these before entering the virtual pocket at B. Similarly AB intersects $\displaystyle (m-1)$ horizontal edges of the $\displaystyle m$ rectangles that make up the line $\displaystyle PB$.

Each of these intersections represents an impact with an edge of the table. There are therefore $\displaystyle n-1+m-1=m+n-2$ impacts in total.