This question is a rough analog of Kac's "Can One Hear the Shape of a Drum?" A closer analog is the recent "Bounce Theorem" that says, roughly, the shape of a polygon is determined by its billiard-bounce spectrum.1

Suppose there is a polygon $P$ hidden inside a disk $D$. All its edges are mirrors. You shoot in a light ray, and are able to observe the trajectory of the ray's emergence from $D$. If the ray hits a vertex, it dies; otherwise it reflects across edge perpendiculars.

          Left: Polygon $P$ hidden by disk $D$. Right: The ray reflects from $P$'s edges.

Q1. Are there two incongruent polygons $P_1$ and $P_2$ that cannot be distinguished from the bounce behavior of external rays?

Here I want to ignore rigid motions of the polygons. By "bounce behavior" I mean comparing the geometry of the incoming and outgoing trajectories of the ray; what happens inside the disk is not known to you. Imagine all possible incoming rays. Can two incongruent polygons have the same bounce behavior for every possible ray, i.e., be equireflective?

One can think of several variants. Perhaps a bit more information might help prove a negative result:

Q2. Suppose you not only observe the in- and out-trajectories, but also the time it takes for the ray to emerge, effectively yielding the length of the ray path.

Perhaps it is easier to construct equireflective shapes if one could make use of sections of parabolas and ellipses and their special reflection properties:

Q3. Are there two incongruent piecewise-smooth Jordan curves $C_1$ and $C_2$ that cannot be distinguished from the bounce behavior of external rays?

1 Moon Duchin, Viveka Erlandsson, Christopher J. Leininger, Chandrika Sadanand. "You can hear the shape of a billiard table: Symbolic dynamics and rigidity for flat surfaces." 2018. arXiv abs.

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    $\begingroup$ What about light rays that hit corners? $\endgroup$ – Joel David Hamkins Apr 19 '18 at 13:04
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    $\begingroup$ I think it would be also interesting to consider a variant of the problem in which the polygon is assumed to be convex $\endgroup$ – Qfwfq Apr 19 '18 at 19:13
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    $\begingroup$ If I understand what you're talking about, not only is it theoretically possible, it has a practical application. $\endgroup$ – StackOverthrow Apr 19 '18 at 21:56
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    $\begingroup$ @Qfwfq: For a convex polygon, each ray that returns at an angle determines the line containing one edge. Repeating, I believe the entire polygon is determined by intersecting those lines. $\endgroup$ – Joseph O'Rourke Apr 19 '18 at 23:56
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    $\begingroup$ echoing Qfwfq -- the case of a convex polygon seems to lead to a fairly straightforward `yes'. First, you can measure angles at vertices via pencils of parallel lines. Similarly you can measure side lengths by rotating this pencil of rays until you get some total (180) reflection, and measuring the thickness of the fraction of totally reflected rays.... $\endgroup$ – Richard Montgomery Apr 25 '18 at 3:36

For question 3, the answer is yes: take a solid disc and excavate half of the Penrose unilluminable room from it. Then, there are boundary arcs which can never be touched, and you can perturb them without altering the bounce behaviour.

It's possible to make this simple closed curve $C^{\infty}$ if you carefully smooth the corners.

Edit: included a rough sketch.

rough sketch

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    $\begingroup$ I'm probably not thinking enough, but could you describe the excavation in a little more detail (or draw a sketch)? What's confusing me is that the Penrose unilluminable room is usually described as a room which has unilluminable regions when lit with a point source that is inside... $\endgroup$ – j.c. Apr 19 '18 at 13:27
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    $\begingroup$ Are you on an airplane? Perhaps a train...and evidently in some kind of mirror universe. $\endgroup$ – Joel David Hamkins Apr 19 '18 at 13:34
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    $\begingroup$ Thanks! That clears everything up. (I took the liberty of fixing the orientation) $\endgroup$ – j.c. Apr 19 '18 at 13:36
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    $\begingroup$ Beautiful! ${}$ $\endgroup$ – Joseph O'Rourke Apr 19 '18 at 15:02
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    $\begingroup$ I like the suggestion that the infrastructure at the heart of government is some sort of mysterious unilluminable object. $\endgroup$ – Nathaniel Apr 20 '18 at 6:49

my answer for Q1:

"If the ray hits a vertex, it dies". I take this property as the starting point of my solution

Let two neighboring disks D1 and D2 contain two such incongruent equireflective polygons. Let us take a vertex V1 of polygon P1; all rays hitting V1 will die; same behavior for the corresponding point V1' in disk D2, no matter if V1' is a vertex of P2 or not. All rays hitting V1' die, meaning that at some point they hit a vertex of P2, vertex which doesn't necessarily have be in V1' position. But is has to be somewhere on the ray support line, maybe before or after V1'; So this line contains for sure a certain vertex of P2.

This property holds for an arbitrary ray hitting V1'. We can have infinitely many rays hitting V1', each containing a vertex of P2. But P2 has a finite number of vertexes. So we have an infinity of lines - all convergent in V1' , each containing a point from a finite set of points. This leads to V1' itself being a vertex of P2.

Repeating the same procedure for each vertex of P1, we get that the corresponding point in D2 is also a vertex of P2. So the set of P1 's vertexes is included in the set of P2 's vertexes. Now we start from disk D2 with the same procedure, to obtain that the set of P2 's vertexes is included in the set of P1 's vertexes. The two sets are equal. The two polygons have their vertexes in the exact same (geometric) location on their disks - so they are congruent.


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  • $\begingroup$ i don't know why the image is not displayed in the post; please check the link $\endgroup$ – Newton fan 01 Apr 19 '18 at 16:37
  • $\begingroup$ I fixed the image issue---you were missing the exclamation mark. $\endgroup$ – Joel David Hamkins Apr 19 '18 at 16:50
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    $\begingroup$ This assumes that you can hit $V1$ in the first place; I don't see why that's guaranteed on non-convex polygons. Your argument about 'support lines' assumes that the vertex is the first point of contact, which doesn't at all have to be so. $\endgroup$ – Steven Stadnicki Apr 19 '18 at 17:08
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    $\begingroup$ To expand on @StevenStadnicki's comment: In Adam Goucher's construction, there are portions of the boundary of the polygon that are never touched by any rays. This seems impossible to achieve with polygons, but that is exactly what needs proof. $\endgroup$ – Joseph O'Rourke Apr 19 '18 at 23:31
  • $\begingroup$ @JosephO'Rourke "This seems impossible to achieve with polygons" Really? My intuition suggests me that if you take a sequence of polygons which converge to the shape of Adam Goucher, then you'll get the counter example. $\endgroup$ – Surb Apr 20 '18 at 7:52

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