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I'm working on a project to do with bouncing rays inside polygons and now I've reached a crucial stage of this project with which I need help, related to the problem stated below. Your help will be highly appreciated.

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Problem

Let $Q$ be a polygon. Suppose that if I shoot a ray at a specific angle from a specific point in the interior of $Q$, and I let the ray bounce off the edges continuously, I'll have at least one point inside the polygon with exactly $n$ lines (with different slopes) passing through it. In this case, we will refer to the polygon $Q$ as a $P_n$-polygon.

  • Does a $P_n$-polygon exist for every $n\geq 3?$
  • Is every $P_n$-polygon necessarily also a $P_m$-polygon for all $m < n$?

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(A square is an example of a $P_0,P_1$ and $P_2$-polygon depending on the position and the angle I shoot the ray from, inside the square.)

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mjqxxxx
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JCr
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  • How is a square an example of a $P_0$? What is your initial starting point, and which point can't you reach directly? In fact, I argue that given any starting point. every point can be reached via infinitely many reflections (this is obvious using the typical reflection grid setup), so a square is not a $P_n$ polygon for any $n$. – Calvin Lin Oct 07 '20 at 23:24
  • Different starting points and angles will give you different results. Suppose you shoot a ray from the centre of the square so that it's parallel to two of its edges. Then if you choose a point $p$ on the ray, then this ray will be the only one going through that point $p$. Since there is at least one point $p$ inside the square with exactly one ray (and no more) going through such point, the square is by definition a $P_1$-polygon. – JCr Oct 08 '20 at 01:17
  • Also in this scenario, any point not on this ray, will never be reached by this ray, since the ray will just keep on bouncing / reflecting perpendicularly between the two eges. So the square also contains at least one point $q$ with no rays going through such point. So the square is also a $P_0$-polygon by definition. – JCr Oct 08 '20 at 01:19
  • By choosing a suitable starting point and angle to shoot the ray from - inside the square - you'll also have at least one point $r$ inside the square with exactly two rays (and no more) going through such point (after some reflections from the egdes of the square). Another way to put it is the ray will intersect itself at point $r$ exactly once inside the square. So again by definition the square is a $P_2$-polygon. – JCr Oct 08 '20 at 01:36
  • Finally notice that it doesn't matter how you choose your starting point and angle to shoot the ray from - inside the square - you can never get a point $s$ with exactly 3 (or more) rays going through $s$ (that is, two or more self-intersections at point $s$) . – JCr Oct 08 '20 at 01:43
  • Just consider the angles of the lines themselves. Since each reflection flips one of the coordinates of the direction of the line, there are only two different possible slopes; the odd-reflected one and the even-reflected one. So any intersection must be between two segments of these two different slopes, and there's no 'third angle' a line could come in from. Hence the square is not a $P_n$-polygon for any $n > 2$ – JCr Oct 08 '20 at 01:46
  • The main thing to remember here is that when we talk about rays going through a point $p$, we mean rays with different slopes going through $p$. – JCr Oct 08 '20 at 02:16
  • In a square why not for each point there are infinite rays that can go through? – Moti Oct 08 '20 at 05:15
  • In a square for each point there are at least four rays with different slopes that go through... or may be I am not following what you are attempting. Two perpendicular to the sides and two at $45^0$ – Moti Oct 08 '20 at 05:18
  • I think there's a misunderstanding here. But I'll clarify it further. – JCr Oct 08 '20 at 14:24
  • STEP 1: You choose 'one' starting point and 'one' angle inside the square to shoot the ray from. – JCr Oct 08 '20 at 14:24
  • STEP 2: You let this same 'one' ray reflect off the edges of the square continuously (infinitely many times). – JCr Oct 08 '20 at 14:26
  • STEP 3: Check if there is a point inside the square with no ray going through it, and a point inside the square with the ray (with one slope only) going through it, and a point inside the square with the ray going through it (with two different slopes) and so on... – JCr Oct 08 '20 at 14:26
  • STEP 4: Choose another starting point and/or angle, repeat steps 2 and 3 and see if you can get any different behaviour from before. – JCr Oct 08 '20 at 14:27
  • Each different behaviour from these reflections inside the square determines exactly one type of polygon. – JCr Oct 08 '20 at 14:36
  • Taking a quick look at the comments, I believe the confusion arises from "ray" having double usage. As such, to clarify, I believe what you want is "Choose an initial starting point. Chooset 1 ray. Trace out the infinite path it takes. Choose a given point. If there are $n$ distinct lines through it, then this is a $P_n$." – Calvin Lin Oct 08 '20 at 16:30
  • Yes. That's the idea. – JCr Oct 08 '20 at 17:13
  • For a similar reason of angle chasing, with an equilateral triangle, initial point the center, given point the center, it is a $P_0$, $P_1$ (if we allow for the ray to pass through a vertex with the "natural" reflection across the angle bisector), and a $P_3$. However, it cannot be a $P_2$, so the answer to the latter question is "no". – Calvin Lin Oct 09 '20 at 16:56
  • That's an interesting result. Thanks for the update. It would be interesting to see the picture for $P_3$ – JCr Oct 09 '20 at 17:37
  • Could you tell me what initial angle you used at the centre (as the initial point) to get $P_3$ at the centre (as the given point)? Also could you tell me why $P_2$ doesn't work? – JCr Oct 09 '20 at 18:32
  • Your notation is confusing, since the set you name $P_n$ is not independent of starting point $x$ and ray angle $\alpha$. I suggest a change of notation, e.g. $(P,,x,,\alpha) \in E_n$ means $\exists y \in P$, such that ray starting at $x$ with angle $\alpha$ goes through $y$ with $n$ different angles. – Joce Oct 10 '20 at 14:53
  • I edited the question, hopefully making the notation more clear and the definition more precise. @JCr feel free to correct it if I misunderstood the intent. – mjqxxxx Sep 12 '25 at 18:54

1 Answers1

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The first question has an affirmative answer: a $P_n$-polygon exists for every $n \ge 0$. (Every polygon is both $P_0$ and $P_1$, I believe, but it's not necessary to prove that.) Constructing a $P_n$-polygon is straightforward; just work backward from a single path that demonstrates the property. More specifically:

First draw a piecewise linear path (closed or not) that passes through the origin $n$ times and is otherwise non-self-intersecting. At each vertex of the path, place a small "bumper" perpendicular to the angle bisector at that vertex. These bumpers will form part of the polygon. (If the path is not closed, then the bumpers at the two endpoints should be perpendicular to the path at those points.) Finally, complete the polygon by connecting the bumpers (using piecewise linear paths) into a single loop enclosing the initial path, being careful not to cross any existing segments.

It's not necessarily obvious that the final step is possible; indeed, you can draw the initial path in such a way that it's not. But a little thought shows that as long as the "petals" (minimal portions of the initial path leaving from and returning to the origin) are convex and not nested, none of the bumpers will lie inside a petal, and so they can be safely connected to form the polygon. The figure below shows an example of this construction for $n=3$. Shooting a ray along any of the red segments (in either direction, starting at any point) will trace out the entire red path, which crosses the central point three times.

construction of a P3-polygon

The second question seems much more difficult. I strongly suspect the answer is no (e.g., that there are $P_8$-polygons that aren't also $P_5$); but proving the negative requires exhibiting a $P_n$-polygon and understanding its billiard dynamics starting with every point and direction. The best approach is probably to use a highly symmetric example, in the hope that the dynamics become more tractable.

mjqxxxx
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