# Does the $n$-gonal billiards conjecture follow from the $m$-gonal conjecture when $m>n$?

For $n\ge 3$, define the $n$-gonal billiards conjecture as the statement

All convex $n$-gons admit periodic billiard trajectories.

To the best of my knowledge this question remains open for all $n$. The case which has received the most attention is the "simplest" case $n=3$. But is this really the simplest case, in a strict sense? I.e., does the $3$-gonal billiards conjecture follow from the $m$-gonal billiards conjecture for any $m>3$? Furthermore, does the $n$-gonal billiards conjecture follow from the $m$-gonal billiards conjecture for any $n$ and $m>n$?

There is one case for which this is easily seen. If $T$ is any triangle then we can let $T'$ be the reflection of $T$ across its longest side and we see that $T\cup T'$ is a convex quadrilateral, thus the $3$-gonal conjecture follows from the $4$-gonal conjecture. However, if $T$ is obtuse then repeating this process gives us a $5$-gon which is not convex. This also breaks down for any $n>3$, as the result can be non-convex.

It is tempting to argue that by shrinking the length of some side to $0$ on an $m$-gon, assuming the $m$-gonal conjecture we obtain periodic trajectories on every $m+1$ gon. However, I see no reason why the length of any periodic trajectories would not diverge as the length of the side approaches $0$, which breaks this line of reasoning.

• Dear Alex, I am curious if there exist non-convex $n$-gonal billiards without periodic trajectories, are you aware of any such example? Jan 9 '13 at 15:36
• @Dmitri None I'm aware of. I can think of a non-convex $5$-gon which I suspect has no periodic trajectories, but I'm not sure how to go about proving it. I mostly avoid non-convex polygons because billiards can behave so strangely there; for example, I seem to recall that if you put a "chimney" on a square properly then all trajectories will eventually stay in one of a finite number of open subsets of measure less than the whole polygon. Jan 10 '13 at 1:29

The shrinking edge strategy may have some hope: It is well known that the number of reflections in a wedge of angle $\theta$ is bounded by $\pi/\theta+1$. Any trajectory entering the region of the corner at a particular angle can only hit the short side once (if oriented so that both angles are obtuse) and the adjacent sides a bounded number of times before exiting. Thus it can exit only in a bounded number of directions, which vary with the direction of the short side. By orienting the short side at will, we put quite a strong condition on the larger billiard. To rephrase: Are there any polygonal billiards to which we can make an infinitesimal and arbitrarily oriented cut of the corner and always find periodic orbits that include the cut? Even if the answer is yes, it may lead to further interesting directions.