TITLE: Which polygons have *simple* periodic billiard paths?
QUESTION [10 upvotes]: 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.

 
 
 


 
 
 
(Image from: W. Patrick Hooper, "Some irrational polygons have many periodic
billiard paths."
PDF download link.)


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?

REPLY [7 votes]: 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.

REPLY [2 votes]: An example that may be difficult for Aaron's line-of-symmetry argument: