As has been discussed earlier on MO,<sup>1,2</sup> 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](https://mathworld.wolfram.com/IlluminationProblem.html): <img src="https://i.sstatic.net/yxa2r.gif" /> The result is Corollary 3 in this "Everything is illuminated" [paper][1]: > 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. <hr /> - <sup>1</sup>[Current state of Straus's illumination problem](https://mathoverflow.net/a/252441/6094), - <sup>2</sup>[Using mirrors to make a non-convex polygon visible from a fixed interior point](https://mathoverflow.net/a/42007/6094) [1]: http://dx.doi.org/10.2140/gt.2016.20.1737