As has been discussed earlier on MO,^1,2 recently an impressive advance was proved concerning internally illuminating a mirrored polygon. Here is the result:

Corollary 3. Let $P$ be a rational polygon. Then for any $x \in P$ there are at most finitely many points $y$ for which there is no geodesic trajectory between $x$ and $y$.

Here "no geodesic trajectory" means no illumination path, using angle-of-incidence equals angle-of-reflection from the polygon edges. A rational polygon is one each of whose angles is a rational multiple of $\pi$. So, a match lit at $x$ will leave at most a finite number of points in $P$ dark. Examples were known:

The theorem is proved in this "Everything is illuminated" paper:

Lelievre, Samuel, Thierry Monteil, and Barak Weiss. "Everything is illuminated." Geometry & Topology 20, no. 3 (2016): 1737-1762.

The authors say they depend on "recent breakthrough results of Eskin-Mirzakhani [EM] and Eskin-Mirzakhani-Mohammadi [EMM],"

Here finally is my question:

Q. Suppose $P$ has all rational vertex angles, except for two irrational vertex angles. (A polygon cannot have just one irrational angle, because the angle sum is $(n-2)\pi$.) Is it now true that, for any $x$, the entirety of $P$ is illuminated, i.e., no point is dark?

I am wondering if this is implied by their results. I do not have sufficient grasp of their use of the "Magic Wand Theorem" and other translation-surface technology to understand if the answer to Q is just a corollary, or a possibly open question. It makes some sense to me that one irrational angle will suffice to fill $P$ with light, because the lightray reflections become ergodic.

• ¹Current state of Straus's illumination problem,

• ²Using mirrors to make a non-convex polygon visible from a fixed interior point

As has been discussed earlier on MO,^1,2 recently an impressive advance was proved concerning internally illuminating a mirrored polygon. Here is the result:

Let $P$ be a rational polygon. Then for any $x \in P$ there are at most finitely many points $y$ for which there is no geodesic trajectory between $x$ and $y$.

Here "no geodesic trajectory" means no illumination path, using angle-of-incidence equals angle-of-reflection from the polygon edges. A rational polygon is one each of whose angles is a rational multiple of $\pi$. So, a match lit at $x$ will leave at most a finite number of points in $P$ dark. Examples were known:

The result is Corollary 3 in this "Everything is illuminated" paper:

Lelievre, Samuel, Thierry Monteil, and Barak Weiss. "Everything is illuminated." Geometry & Topology 20, no. 3 (2016): 1737-1762.

Here finally is my question:

Q. Suppose $P$ has all rational vertex angles, except for two irrational vertex angles. (A polygon cannot have just one irrational angle, because the angle sum is $(n-2)\pi$.) Is it now true that, for any $x$, the entirety of $P$ is illuminated, i.e., no point is dark?

I am wondering if this is implied by their results. I do not have sufficient grasp of their use of the "Magic Wand Theorem" and other translation-surface technology to understand if the answer to Q is just a corollary, or a possibly open question. It makes some sense to me that one irrational angle will suffice to fill $P$ with light, because the lightray reflections become ergodic.

• ¹Current state of Straus's illumination problem,

• ²Using mirrors to make a non-convex polygon visible from a fixed interior point