Assume we have an idealized even- and $n$-sided pool table with no holes, friction and perfectly reflecting walls, and that a ball is set in motion (inside the table) parallel to one of the sides and towards the middle point of the wall of the first contact.

For a 3-sided table the ball would describe an eternal path in another inscribed triangle, for a 4-sided table there are two states of such eternal paths.

I now wonder, for $n$ number of walls is there only one state or path if $n$ is a prime number, and would there be perhaps $m$ number of states if $n$ is a factor of $m$ primes?

  • 3
    $\begingroup$ Are you assuming the table edges are all the same length? i.e. the table is a regular polygon? $\endgroup$ – Ben Millwood Jul 24 '13 at 12:03
  • $\begingroup$ it would be helpful if you attach some images as example. $\endgroup$ – al-Hwarizmi Jul 24 '13 at 12:35
  • $\begingroup$ Yes, I intended the edges to be the same length by using the term "even-sided". $\endgroup$ – Mikael Jensen Jul 24 '13 at 19:47

By symmetry with respct to the mid-perpendicular of the side the motion is parallel to, the ball moves by an even number $2k$ of sides (which may be more than $\frac n2$, as in the case $n=3$) and the (polygonal or star) shape of the path is uniquely determined by the number $2k$, except that $n-2k$ and $2k$ generate the same. So if $n=2m$ is even, we have to check $k$ ranging from $1$ to $\lfloor \frac m2\rfloor =\lfloor \frac n4\rfloor$ inclusive, if $n$ is odd $k$ ranging from $1$ to $\frac{n-1}2$ inclusive. For each such $k$, one closed cycle involves every $d$th edge where $d=\gcd(n,2k)$, so we get $d$ distinct rotational copies of this path and the total number of closed cycles is $$\tag1 \sum_{k=1}^{\lfloor\frac n4\rfloor\text{ or }\frac{n-1}2}\gcd(n,2k).$$ It seems that this sequence $$ 0, 0, 1, 2, 2, 2, 3, 6, 6, 4, 5, 12, 6, 6, 15, 16, 8, 12, 9, 22, 22, 10, 11, 34, 20, 12, 27, 32, 14, 30$$ is not in OEIS (yet). By the way, if $n$ is an odd prime, then $(1)$ equals $\frac{n-1}{2}$.


Let $R$ be a regular $n$-gon, and number the midpoints of its sides $0$ through $n-1$.

Note that if a line passing through the midpoint of a side is parallel to another side, then it passes through the midpoint of another side. For odd $n$ the converse holds; a line passing through the midpoints of two distinct sides is parallel to another side. For even $n$ this only holds if the sides differ by an even number (assuming we have numbered the sides in a sensible way).

Hence we may describe the initial motion as starting from a point $p\in\{0,\ldots,n-1\}$, towards another point $q\in\{0,\ldots,n-1\}$, where $p\neq q$. Denote the path starting from point $p$ moving towards $q$ by $P(p,q)$.

Consider the path $P(0,m)$. After meeting point $m$, the ball is reflected and moves towards point $2m$. By symmetry, the $k$-th point to be met is the point $k\cdot m$, so the points on the path $P(0,m)$ are the integer multiples of $m$, modulo $n$ of course.

Counting orientation, this yields precisely $n-1$ distinct paths meeting $0$ if $n$ is odd, and precisely $\tfrac{n}{2}-1$ distinct paths meething $0$ if $n$ is even. Of course not every path meets the point $0$ necessarily. The length of $P(0,m)$, i.e. the number of distinct points it meets, is $\operatorname{lcm}(m,n)/n$, or equivalently $m/\gcd(m,n)$, whichever you prefer, so there are precisely $\gcd(m,n)$ distinct 'rotations' of this path. The total number of distinct paths, counting orientation, is therefore

$$T(n)=\left\{\begin{array}{cc} \sum_{m=1}^{n-1}\gcd(m,n)&\text{ for } n \text{ odd,}\\ \sum_{m=1}^{\tfrac{n}{2}-1}\gcd(2m,n)&\text{ for } n \text{ even.}\end{array}\right.$$

This is not as bad as it looks; it turns out these sums are multiplicative. For details, see this question. The accepted answer there shows that if $n=\prod_{p\mid n} p^{n_p}$ then $$\sum_{m=1}^n\gcd(m,n)=n\prod_{p\mid n}\big(1+n_p(1-\tfrac1p)\big).$$

So indeed this shows that the number of paths on an $n$-sided table is closely related to the factorization of $n$.

Also, if we don't count orientation, we find the total numbers to be

$$t(n)=\left\{\begin{array}{cc} \frac{1}{2}(T(n)+n)&\text{ if } 4\mid n\\ \frac{1}{2}T(n)&\text{ otherwise}\end{array}\right..$$

  • $\begingroup$ Thank you for the answers. I must say I thought the solution would be much easier than it seems. $\endgroup$ – Mikael Jensen Jul 24 '13 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.