# Which polygons have *simple* periodic billiard paths?

I know (or, rather, believe) that it remains unknown whether every polygon has a periodic billiard path. But Howard Masur proved in the 1980's that every rational polygon (vertex angles rational multiples of $\pi$) has (many) periodic billiard paths. My question concerns simple paths: non-self-intersecting, i.e., embedded paths:

Q. Is it known that certain classes of rational polygons have simple periodic billiard paths? If so, which? That certain classes (of rational polygons) have no such simple periodic paths? If so, which?

For example, in an acute triangle, its pedal triangle is a simple periodic path, and the only such.

To be narrowly specific: Might every regular polygon have a simple periodic path?

• It's "Masur", not "Mazur"... Feb 7, 2015 at 1:32
• For starters a rectangle has infinitely many simple periodic paths. Feb 7, 2015 at 1:44
• "Might every regular polygon have a simple periodic path?" is too narrowly specific, with an easy affirmative answer: such a path (also a regular polygon of the same order) is obtained by joining midpoints of consecutive pairs of sides. Feb 7, 2015 at 3:17
• When you say simple path does that include the case where the path can go back over itself? If so, then a right triangle has a simple periodic path. (math.brown.edu/~res/Papers/billiards1.pdf) Feb 7, 2015 at 3:46
• If the answer to my above question is yes, then isosceles triangles also have simple periodic paths. Start perpendicular to one of the two equal sides then go toward the midpoint of the third side. Feb 7, 2015 at 4:13

Consider the simple polygonal billiard path itself. It is a polygon and it is convex, because all interior angles are less than $\pi$. Now start with an arbitrary convex polygon. It is a billiard path on some polygonal table. Indeed, the lines from which the reflections happened are uniquely determined by our billiard path. Continue these lines until they intersect and form a table. Of course this table is not unique, because you can add arbitrarily many sides which lie outside of the polygonal path and do not intersect our polygonal path. Neither the resulting table, nor the path itself has to have any symmetry.

EDIT. If the table is a triangle with acute angles, then a simple billiard path exists, is unique, and it is the triangle whose vertices are the bases of the three heights of the table-triangle. This is due to H. A. Schwarz. (See Courant-Robbins, What's mathematics, Ch. VII, sect. 4). On the other hand, a triangle whose one angle is $\geq\pi/2$ violates the inequality noticed in Will Sawin's remark: the angles of the triangular table must satisfy $\alpha+\beta-\gamma\in(0,\pi)$ for a simple billiard part to exist.

Thus we have a complete description of triangular tables with a simple billiard path, and more generally, of those tables which have a triangular billiard path.

• It is not too hard to see that the interior angles of the polygon you get by doing this are the averages of the adjacent interior angles of the billiard path. From this you get an explicit system of inequalities that the angles satisfy if and only if there is a polygonal billiard path touching every edge. Feb 7, 2015 at 19:59
• Yes, I think these considerations permit to describe explicitly all such polygonal tables. Feb 7, 2015 at 20:50
• The conditions are sufficient for a cyclically ordered list of angles to be realized by some polygons having a polygonal path touching each side. A regular pentagon has such a path but an equiangular pentagon with sides of lengths $1,\epsilon,1.6,\epsilon,1$ seems unlikely to. For a triangle the angles do determine the sides (up to similarity) so acute triangle have paths and obtuse ones do not. Feb 8, 2015 at 4:25

An example that may be difficult for Aaron's line-of-symmetry argument: Here is a rather revised answer. I originally thought that having a line of symmetry (bisecting one or perhaps two edges ) might be a sufficient condition. Joseph showed a convincing counter-example to my reasoning. And Alexandre's observations finished it off. Here is what I can salvage and add.

I'm not so interested in non-polygonal paths but here are a few

1) On the left below is a variation of the right triangle construction showing that a convex (or simply not too badly reentrant) polygonal table with a right angle has a simple periodic billiard path of the form $abcdcba.$ There are arbitrary further sides not shown.

One might require that a path touch all the sides in order to avoid arbitrary modifications which don't affect the path.

2) The top middle diagram is an isosceles triangle table with a path $bacab$ and another $defgfed.$

I'll now restrict to polygons with a simple periodic polygonal billiards path touching each side.

3) A construction given for (acute) isosceles triangle applies to arbitrary acute triangles. Drop the perpendicular from each vertex to the opposite sides. There is a path touching at just those points.

As Alexandre showed, any convex polygon can be a path and the path determines the polygonal table up to similarity (if we require equal number of sides. Will pointed if the path has angles $\alpha_1,\alpha_2,\cdots,\alpha_n$ then the table has angles $\frac{\alpha_i+\alpha_{i+1}}2.$ This provides a necessary condition for a table to have a path. It is a sufficient condition for a circularly ordered lists of angles to be realized by some tables having a path.

4) For a triangular path the table will have angles $\frac{\pi-\alpha_1}2,\frac{\pi-\alpha_2}2,\frac{\pi-\alpha_3}2.$ Hence an obtuse or even a right triangle has no triangular paths. This tells us everything about triangular tables.

5) For quadrilateral tables we see that opposite angles must add to $\pi.$ That is a strong condition (strong enough to demolish my conjecture.) However I have drawn a quadrilateral below with angles $\frac{\pi}4,\frac{\pi}2,\frac{3\pi}4,\frac{\pi}2.$ It is only slightly modified from a right triangle with no path. Since the angles are rational it should be possible to determine if it has a quadrilateral path. But I doubt it does.

6) Regular polygons have polygonal paths. The final table is pentagonal with all angles $\frac{3\pi}5.$ It even has a central line of symmetry. However I again suspect that it does not have a path of the type we seek. • Very nice observations! I like your line-of-symmetry argument, but I am not sure how easy it will be to complete. It seems not so straightforward to connect $a$ to $d$. I'll post a drawing... Feb 7, 2015 at 14:02
• Good idea to restrict to obtuse angles. I'll have to think about this. Feb 7, 2015 at 21:08