Suppose we have a polygonal path $P$ on the plane resulting from removal of an one of a convex polygon's edges and a ray of light "coming from infinity" (that is, if we were to trace the path backwards in time, we would end up with an infinite ray) and reflecting from the sides of $P$ according to standard law of reflection (see below for discussion of interaction with vertices). Is it possible that $P$ will stay inside $P$ forever?

More generally, we could ask this for more general polygonal paths and instead of talking about "inside of $P$" we might talk about for example its convex hull. I know that it is possible with more general curves, even with "infinitely-sided" polygons, which can be shown with idea of this answer (illustrated here). The linked question was also a partial motivation to this one.

About corners: the easiest (and preferable for me) way to deal with them is to require that the light should never hit it (the light then disappears or whatever, I leave it up to your interpretation). Another idea I had is to make the light reflect so that it makes equal angles with bisector of the internal angle at the vertex. However, if it helps to construct an example, feel free to suggest another rule.

At first I thought that an example most likely exists, but I couldn't find one (I had to work with pen and paper though, I don't have software with such functionality). I know that the path cannot be eventually periodic, because the process of light traveling is reversible. I see no obvious obstruction which would disallow more complicated infinite paths.

Thanks in advance for all the feedback.

Edit: in the view of Joseph's example, I would like to clarify that I want the endpoints of the polygonal path to also be contained in it and the light is supposed to vanish on them.

Edit2: I don't know how much easier the problem will become, but seeing the paper Joseph refered to in his first answer I thought that the following weakening might have ability to trap a ray of light: instead of polygonal chain, let's just consider a set of closed line segments. The line segments are allowed to intersect, and the light is supposed to never hit an endpoint of an line segment or an intersection point of two segmentss. Let's see if anyone can figure out such a trap...

Edit3: I forgot to add this to the description of the bounty, I plan to award either (these are the weakest possibilities in both directions):

  • Anything that implies possibility of trapping light with a set of mirrors as described in edit2, or
  • Anything that implies impossibility of trapping light in a convex polygon as described in the first paragraph.

If neither of the above is achieved by anyone, I am willing to award the bounty to any noteworthy attempt in proving either.

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    $\begingroup$ Here is a suggestion on why it is likely impossible to trap a ray with segment mirrors. The ray cannot be trapped into a periodic path. So it must be trapped aperiodically. There must be some "transparent" finite-length entry window through which the ray arrives from infinity (say, on the hull of the mirrors). After entrance, that entry window must be avoided. But billiard flow in this circumstance is volume-preserving, so Poincare's recurrence theorem says that entry window will be revisited. This is not a proof, just a suggestion why the answer is likely No. $\endgroup$ May 16, 2016 at 13:46

2 Answers 2


Update. I answered too quickly. The construction I describe traps a ray whose source is inside.

Mitchell, Zachary, Gregory Simon, and Xueying Zhao. "Trapping light rays aperiodically with mirrors." Involve, a Journal of Mathematics 5.1 (2012): 9-14. (Journal link.)

Abstract. We construct a configuration of disjoint segment mirrors in the plane that traps a single light ray aperiodically, providing a negative solution to a conjecture of O’Rourke and Petrovici. We expand this to show that any finite number of rays from a source can be trapped aperiodically.


To obtain a polygon, one would have to connect their disjoint segments into a path, but I think this would not be difficult. Update. Apologies. Now that I found their construction, which mimics an irrational sloped billiard path reflecting inside a square, it is not immediately evident how to inject the ray from outside the construction...

Incidentally, it is not possible to trap light rays from a continuum of directions, even with curved mirrors:

Dawson, RJ MacG, B. E. McDonald, J. Mycielski, and L. Pachter. "Light Traps." (1996). (PDF download.)

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    $\begingroup$ Unfortunately I don't have access to this paper, hence I can't confirm that this is exactly what I'm looking for. I will take a look at them whenever I have an opportunity though, so thanks. $\endgroup$
    – Wojowu
    Apr 30, 2016 at 15:51
  • $\begingroup$ @Wojowu: Sorry for the lack of details. I cannot access the Involve paper now either. Otherwise I'd post a drawing of their construction. $\endgroup$ Apr 30, 2016 at 15:57
  • $\begingroup$ @Wojowu: Now it is clear I have not answered your question. Sorry! $\endgroup$ Apr 30, 2016 at 16:21
  • $\begingroup$ Yes, I've found a description of this paper somewhere and I've figured it probably shows some result (related to, but still) different from my question. Feel free to keep this answer though, since it's content is interesting by itself. $\endgroup$
    – Wojowu
    Apr 30, 2016 at 16:24
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    $\begingroup$ @Wojowu: I guess if you make a point-hole in the polygon connecting the segments in their construction, to let in the single ray, then it is possible (since the ray will never revisit the hole). $\endgroup$ Apr 30, 2016 at 16:29

I post this in response to Wojowu's request in a comment; I don't really consider this usefully addressing the question raised.

I thought of a simpler construction. Take a square with mirrored sides, and poke a point-hole in the interior of the top edge. When a ray from outside with an irrational slope (w.r.t. the box sides) enters through the hole, it is trapped. For if it escaped through the hole after $k$ reflections, then it would contradict the irrationality of the slope. It is known that the path of this "billiard" is aperiodic (a result due to G.A.Galperin, I believe).

And of course almost all rays have irrational slope.

But as soon as the point-hole is enlarged to a positive-width hole, the ray eventually escapes (Poincaré recurrence).

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    $\begingroup$ I agree that this doesn't really help with answering my question, precisely due to the last remark. Nevertheless thanks for your time. $\endgroup$
    – Wojowu
    May 2, 2016 at 5:52

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