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.

 
 
 


 
 
 
(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?
 A: 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.
A: 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.

A: An example that may be difficult for Aaron's line-of-symmetry argument:

 
 
 
 
 


