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For my research I am interested in the transmission characteristics between a transmitter (Tx) and a receiver (Rx) situated in a circular room. In particular, it is important for me to know the number of paths a ray can take such that it reflects exactly once off the walls of the room.

Reflections from a transmitter to a receiver in a room

Since reflections occur such that the incident ray has the same angle relative to the normal as the reflected ray, I tried to use vectors to attack the problem but the math became very unwieldy.

Empirically, I have found that depending on the situation of the transmitter and receiver, there could be 2, 3, or 4 paths—no more, no less. There is an exceptional case where the transmitter and receiver are co-located at the centre, in which case there are infinitely many paths.

Can my experimental result be validated (or denied) analytically?

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  • $\begingroup$ I imagine at best your result will only be true "generically". But like your central configuration there is probably a relatively low-dimensional algebraic variety of configurations that similarly break with your observation. This is a nice problem. Have to run now but will come back to it tonight if it's not already answered. $\endgroup$ – Ryan Budney Apr 8 at 20:18
  • $\begingroup$ You seem right with $2,3,4,\infty$, that's very intuitive, by drawing a few pics. Also, you're right, vector calculus is probably not adapted to this. All this depends a bit on how much math/what type or math must your research paper contain. Guess you can (1) either take this for granted, (2) or draw a few pics explaining the possible cases, (3) or include some complete advanced math - which might be actually quite complicated - once you have answers here on MO. In any case, nice question! $\endgroup$ – Teo Banica Apr 8 at 22:37
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    $\begingroup$ This problem was solved in 1992 (the keyword is "billard") and is well documented, see e.g. unige.ch/~gander/Preprints/Billiard.pdf (there is a nice optical experiment at the end of the paper with pictures of caustics). $\endgroup$ – Luc Guyot Apr 9 at 16:41
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We use complex numbers to prove that there are at most $4$ such points unless both transmitter and receiver are at the center.

Identify the circular room with the unit circle $|z|=1$ in the complex plane, and let $r$ and $t$ be the complex numbers corresponding to the receiver and transmitter, with $|r|<1$ and $|t|<1$. If $z$ is a point of reflection then the condition $\theta_r = \theta_t$ comes down to $(z-r)(z-t)$ being a real multiple of $z^2$; that is, to the ratio $(z-r)(z-t)/z^2$ being a real number. Write $$ (z-r)(z-t) / z^2 = 1 - (r+t) z^{-1} + rt z^{-2}, $$ and note that a complex number $w$ is real if and only if it equals its own complex conjugate $\overline w$. Since $z$ is on the unit circle, $\overline z = z^{-1}$, so our condition is $$ \overline{rt} z^2 - (\overline r + \overline t) z + (r+t) z^{-1} - rt z^{-2} = 0. $$ Multiplying by $z^2$ yields a polynomial of degree $4$ in $z$. Thus there are at most $4$ solutions, even without the condition $|z|=1$, unless the polynomial vanishes identically, in which case every $z$ is a solution. But the polynomial vanishes identically if and only if $r+t = rt = 0$, which is to say $r=t=0$, so we recover the degenerate case where receiver and transmitter are both in the center of the circular room.

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Here is a plot of a grid of lines coming out of the transmitter centred at $(0,-0.5)$ in green, together with the first reflection lines, in yellow. There is 300 emission lines. From the picture you can see a big region where there appears to be two yellow lines through each point, then a significantly smaller region where there are four. The boundary of this region presumably is where all the triple intersections are, although I'm not certain what's happening at those three cusp points.

emitter at (0,-0.5)

And here is a more dramatic image, with the emitter at $(0,-0.8)$.

emitter at (0,-0.8)

On closer inspection I think that boundary curve consists of points with three yellow lines intersecting. i.e. there's nothing unexpected going on here.

emitter at (0,-0.94)

emitter at (0,-0.3)

Regarding giving a detailed proof, I think there is a reasonable way to go about this. If we call the emission point $p$, and the first impact-point on the boundary circle $q$, then the 2nd impact-point on the boundary circle we will call $f(p,q)$.

If $q$ has angle $\theta$ then $f(p,q)$ has angle $\theta + \pi + \delta_p(\theta)$ The nice thing about the function $\delta_p : \mathbb R \to \mathbb R$ is it is $2\pi$-periodic and continuous. When $p=0$ $\delta_p$ is the zero function. When $p$ approaches a point on the boundary of the circle, $\delta_p$ is approximating a sawtooth function -- a sawtooth function with period $2\pi$. Here are a few $\delta_p$ plots, below.

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  • $\begingroup$ It seems like displaying the green lines doesn't convey any additional information, and for someone slightly colorblind like me having the lines be green and yellow makes it tough to see what's going on. $\endgroup$ – Daniel McLaury Apr 9 at 13:11

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