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My main point is that I think thought a sufficient condition might be to have an axis of symmetry (bisecting an edge). Joseph showed a convincing counter-example to my reasoning. Now I wonder if it can be saved with the added the condition that no angles are acute.

Before I get to that, a few other remarks.

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

One might require that a spbp touch all the sides in order to avoid arbitrary modifications which don't affect the path, but for some examples below it is convenient not to require this.

2. The top middle diagram is an isosceles triangle with spbp $bacab$ and another $defgfed.$

I'll now mainly restrict to sppbp with the second (or first) p standing for polygonal.

3. A comment gives a construction for an sppbp in an isosceles triangle $PQP'$. I've illustrated it (as I understand it) for the case that the triangle is acute, I'm not sure about the obtuse case. From $P'$ draw the perpendicular to side $PQ$ hitting that side at $b.$ The path starts at the center $a$ of the base and goes to $b.$ The claim is that it continues parallel to the base and creates an sppbp $abb'a.$ (this is easier for me to see when generalized to arbitrary acute triangles, see below)

main point

Perhaps continuity arguments are enough to show that a convex polygon with a line of symmetry has a sppbp (maybe several).

3. (again) We certainly have the degenerate path $aQa.$ For construction 3 can't we consider a path starting from the center $a$ of the base and hitting $PQ$ at a variable point $b?$ There is a choice of $b$ , as in 2. , starting a path $abaca.$ If we move $b$ slightly closer to $Q$ then the path would return to the base a little to the right of $a.$ If we move it close to $Q$ then the next part of the path hits side $P'Q$ even closer to $Q.$ Someplace in between these extremes should be a choice which continues parallel to the base giving $abb'a.$

4. The final picture is a convex hexagon $PQRR'Q'P'$ with line of symmetry $ad$ (so $a$ and $d$ are midpoints of their sides) but no other special structure.

The degenerate path $ada$ is only so interesting. However a simple billiard path starting at $a$ and getting to $d$ without crossing the lie $ad$ will continue on to an sppbp. I've put one in (rather freehand ) which starts $abcd.$ Can we argue by continuity (moving $b$ along edge $PQ$) that there must be such?

If we move $b$ closer to $Q$ then there should be a place where one get a pentagonal sppbp

Remove sides $QR$ and $Q'R'$ and then extend the other sides to create a trapezoid. That should allow a quadrilateral sppbp which would also be valid for the hexagon. Sides $PS$ and $P'Q'$ could be removed instead.

Using every other side and extending gives (im two ways) an isosceles triangle. This allows at least some of the paths from 2. and 3.

This reasoning is a little unsubstantiated, but seems reasonable for any convex $2k$-gon with a line of symmetry. (but see Joseph's example and the comments below which show that it is not that easy.)

The case of an odd number of sides might not be much different.

LATER Two thoughts:

• The construction for an (acute) isosceles triangle generalizes to arbitrary acute triangles: From each vertex drop a perpendicular to the opposite side. The path joining these three points is a sppbp. The (non-simple) periodic paths have sides parallel to this triangle.

• Joseph gives a possible counter-example having a very acute corner, go check his (lovely as always) diagram. I wonder about the added condition that no angles are acute. That would prevent infinite simple paths such as the two he shows. Would a path from the midpoint of the bottom to the right vertex end up stuck there? If not, it would seem to have to go to the midpoint of the top.

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.

My main point is that I think a sufficient condition is to have an axis of symmetry.

3. A comment gives a construction for an sppbp in an isosceles triangle $PQP'$. I've illustrated it (as I understand it) for the case that the triangle is acute, I'm not sure about the obtuse case. From $P'$ draw the perpendicular to side $PQ$ hitting that side at $b.$ The path starts at the center $a$ of the base and goes to $b.$ The claim (which I do not dispute) is that it continues parallel to the base and creates an sppbp $abb'a.$

This reasoning is a little unsubstantiated, but seems reasonable for any convex $2k$-gon with a line of symmetry.

The case of an odd number of sides should not be much different.

My main point is that I think thought a sufficient condition might be to have an axis of symmetry (bisecting an edge). Joseph showed a convincing counter-example to my reasoning. Now I wonder if it can be saved with the added the condition that no angles are acute.

3. A comment gives a construction for an sppbp in an isosceles triangle $PQP'$. I've illustrated it (as I understand it) for the case that the triangle is acute, I'm not sure about the obtuse case. From $P'$ draw the perpendicular to side $PQ$ hitting that side at $b.$ The path starts at the center $a$ of the base and goes to $b.$ The claim is that it continues parallel to the base and creates an sppbp $abb'a.$ (this is easier for me to see when generalized to arbitrary acute triangles, see below)

This reasoning is a little unsubstantiated, but seems reasonable for any convex $2k$-gon with a line of symmetry. (but see Joseph's example and the comments below which show that it is not that easy.)

The case of an odd number of sides might not be much different.

LATER Two thoughts:

• The construction for an (acute) isosceles triangle generalizes to arbitrary acute triangles: From each vertex drop a perpendicular to the opposite side. The path joining these three points is a sppbp. The (non-simple) periodic paths have sides parallel to this triangle.

• Joseph gives a possible counter-example having a very acute corner, go check his (lovely as always) diagram. I wonder about the added condition that no angles are acute. That would prevent infinite simple paths such as the two he shows. Would a path from the midpoint of the bottom to the right vertex end up stuck there? If not, it would seem to have to go to the midpoint of the top.

My main point is that I think a sufficient condition is to have an axis of symmetry.

Before I get to that, a few other remarks.

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

One might require that a spbp touch all the sides in order to avoid arbitrary modifications which don't affect the path, but for some examples below it is convenient not to require this.

2. The top middle diagram is an isosceles triangle with spbp $bacab$ and another $defgfed.$

I'll now mainly restrict to sppbp with the second (or first) p standing for polygonal.

3. A comment gives a construction for an sppbp in an isosceles triangle $PQP'$. I've illustrated it (as I understand it) for the case that the triangle is acute, I'm not sure about the obtuse case. From $P'$ draw the perpendicular to side $PQ$ hitting that side at $b$ The path starts at the center $a$ of the base and goes to $b.$ The claim (which I do not dispute) is that it continues parallel to the base and creates a, sppbp $abb'a.$

main point

Perhaps continuity arguments are enough to show that a convex polygon with a line of symmetry has a sppbp (maybe several).

3. (again) We certainly have the degenerate path $aQa.$ For construction 3 can't we consider a path starting from the center $a$ of the base and hitting $PQ$ at a variable point $b?$ There is a choice of $b$ , as in 2. , starting a path $abaca.$ If we move $b$ slightly closer to $Q$ then the path would return to the base a little to the right od $a.$ If we move it close to $Q$ then the next part of the path hits side $P'Q$ even closer to $Q.$ someplace in between is a choice which continues parallel to the basegiving $abb'a.$

4. The final picture is a convex hexagon $PQRR'Q'P'$ with line of symmetry $ad$ (so $a$ and $d$ are midpoints of their sides) but no other special structure.

The degenerate path $ada$ is only so interesting. However a simple billiard path starting at $a$ and getting to $d$ without crossing the lie $ad$ will continue on to an sppbp. I've put one in (rather freehand ) which starts $abcd.$ Can we argue by continuity (moving $b$ along edge $PQ$) that there must be such?

If we move $b$ closer to $Q$ then there should be a place where one get a pentagonal sppbp

Remove sides $QR$ and $Q'R'$ the extend the other sides to create a trapezoid. That should allow a quadrilateral sppbp which would also be valid for the hexagon.

Using every other side and extending gives an isosceles triangle. This allows at least some of the paths from 2. and 3.

This reasoning is a little unsubstantiated, but seems reasonable for any convex $2k$-gon with a line of symmetry.

The case of an odd number of sides should not be much different.

My main point is that I think a sufficient condition is to have an axis of symmetry.

Before I get to that, a few other remarks.

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

One might require that a spbp touch all the sides in order to avoid arbitrary modifications which don't affect the path, but for some examples below it is convenient not to require this.

2. The top middle diagram is an isosceles triangle with spbp $bacab$ and another $defgfed.$

I'll now mainly restrict to sppbp with the second (or first) p standing for polygonal.

3. A comment gives a construction for an sppbp in an isosceles triangle $PQP'$. I've illustrated it (as I understand it) for the case that the triangle is acute, I'm not sure about the obtuse case. From $P'$ draw the perpendicular to side $PQ$ hitting that side at $b.$ The path starts at the center $a$ of the base and goes to $b.$ The claim (which I do not dispute) is that it continues parallel to the base and creates an sppbp $abb'a.$

main point

Perhaps continuity arguments are enough to show that a convex polygon with a line of symmetry has a sppbp (maybe several).

3. (again) We certainly have the degenerate path $aQa.$ For construction 3 can't we consider a path starting from the center $a$ of the base and hitting $PQ$ at a variable point $b?$ There is a choice of $b$ , as in 2. , starting a path $abaca.$ If we move $b$ slightly closer to $Q$ then the path would return to the base a little to the right of $a.$ If we move it close to $Q$ then the next part of the path hits side $P'Q$ even closer to $Q.$ Someplace in between these extremes should be a choice which continues parallel to the base giving $abb'a.$

4. The final picture is a convex hexagon $PQRR'Q'P'$ with line of symmetry $ad$ (so $a$ and $d$ are midpoints of their sides) but no other special structure.

The degenerate path $ada$ is only so interesting. However a simple billiard path starting at $a$ and getting to $d$ without crossing the lie $ad$ will continue on to an sppbp. I've put one in (rather freehand ) which starts $abcd.$ Can we argue by continuity (moving $b$ along edge $PQ$) that there must be such?

If we move $b$ closer to $Q$ then there should be a place where one get a pentagonal sppbp

Remove sides $QR$ and $Q'R'$ and then extend the other sides to create a trapezoid. That should allow a quadrilateral sppbp which would also be valid for the hexagon. Sides $PS$ and $P'Q'$ could be removed instead.

Using every other side and extending gives (im two ways) an isosceles triangle. This allows at least some of the paths from 2. and 3.

This reasoning is a little unsubstantiated, but seems reasonable for any convex $2k$-gon with a line of symmetry.

The case of an odd number of sides should not be much different.