# (Non)existence of mirrors with more than two foci

Do there exist any mirrors $M$ in $d$-dimensional Euclidean space $\mathbb{R}^d$ for which there exist three different points $x_1$, $x_2$, $x_3 \in \mathbb{R}^d$ such that if any ray of light passes through one of the points $x_i$ it automatically passes through the other two after a number of reflections on $M$?

The question needs some formalisation for what is meant by $M \subset \mathbb{R}^d$ and “rays of light”, although it is easy to guess what they should be:

• $M$ is a $(d-1)$-dimensional differentiable submanifold of $\mathbb{R}^d$.

• A ray of light is a continuous function $g:\mathbb{R} \rightarrow \mathbb{R}^d$ for which $g^{-1}(g(\mathbb{R}) \cap M)$ is a discrete set. Moreover, $g$ is locally smooth, linear and unit speed at any point $t\in \mathbb{R}$ for which $g(t) \notin M$: $\left\|g'(t)\right\|=1$ and g''(t)=0. At any point $t^* \in \mathbb{R}$ for which $g(t^*)\in M$ we require $g$ to satisfy the reflection law: denoting $a=\lim_{t \to t^* -} g'(t)$ and $b = \lim_{t \to t^* +} g'(t)$ $$b + a \perp n(g(t))$$ and $$b - a \propto n(g(t))$$ (n is a local unit-normal vector field on $M$).

• If you'd be happy to settle not for all rays of light, but for a set of positive measure, then it's easy even in two dimensions: simply arrange a curve into a triangular shape using three pieces of ellipse: one piece with each pair of points as focus. – James Cranch Mar 30 '15 at 11:58
• That's what I had in mind. But I'm concerned by what happens with rays that go from one focus to a point of intersection. Even if we can persuade it to be differentiable, I can't see how to get it to visit the other two foci. Admittedly, this is a bit more than positive measure: it may only fail on a set of zero measure. – James Cranch Mar 30 '15 at 12:58
• That's a nice comment, and applying your construction procedure in a careful (but obvious) way, I think we can assert that all light rays except those hitting an edge (i.e. a point where two elliptic pieces meet) prematurely, will pass over all three foci. – Thibaut Demaerel Mar 30 '15 at 12:58
• Ah, I think your edited comment arrived about the same time as my reply, and we were both thinking the same thing! – James Cranch Mar 30 '15 at 12:58
• Sorry, my original comment was too long. This construction seems also very specific to 2 dimensions (and in that sense I don't see why you say that it's easy even in 2 dimensions. It seems to make it easy only in 2 dimensions. Ok, by glueing together pieces of ellipsoid in higher dimensions in a nice way we can probably produce shapes where the "good" light rays have positive measure, yet because we can not glue those pieces together along the "lines of d-2-dimensional submanifolds" the "bad" light rays seem to have positive measure as well (generically). Right?) – Thibaut Demaerel Mar 30 '15 at 13:05

Here is a way that works for a set of full measure in any dimension, for any number of points, not just $3$. We can construct it for dimension $2$, then rotate it about the $x$-axis.

Let the points be $x_i=(i,0)$ for $i=1,...,n$. Attach semi-ellipses $\gamma_i$ whose major axes and end points are along the $x$-axis whose foci are $x_i,x_{i+1}$ so that for $i$ odd we take the intersection of the ellipse with the upper half plane, and for $i$ even we take the intersection with the lower half plane. Extend this by adding semicircles $\gamma_0$ and $\gamma_n$ about the end points. If we choose the semiellipses so that the total distance to $x_i$ and $x_{i+1}$ is $2$, and use semicircles with radius $1/2$, this is even a simple closed curve, although there are points where it is not smooth.

Light passing through $x_1$ going up bounces off $\gamma_1$ and is reflected to $x_2$ going down, so it reflects off $\gamma_2$ to $x_3$ going up, etc. After it passes through $x_n$ it bounces off $\gamma_n$ and reverses. Light passing through $x_1$ going down bounces off $\gamma_0$ and reverses through $x_1$, going up. The behavior of the horizontal ray through the nonsmooth points and all $x_i$ is not really defined, but you could argue that it passes through all of the points, too.

Mathematica code for drawing this picture:

ell[i_, x_] := (-1)^(i + 1) Sqrt[3/4 (1 - (x - (i + 1/2))^2)]
circ[i_, x_] := Sqrt[1/4 - (x - i)^2]
Plot[{-circ[1, x], ell[1, x], ell[2, x], ell[3, x], ell[4, x], circ[5, x]}, {x, 0, 6},
PlotRange -> {-1.5, 1.5},  AspectRatio -> 1/2]

• Very nice, this is the culmination of the line of thought outlined in the comments above and gives, from an "engineering" point of view, a satisfactory answer to the question (also in higher dimensions, though you must be careful how to rotate about the x-axis when introducing new dimensions ;) ). – Thibaut Demaerel Mar 30 '15 at 16:38
• About the edge points: leave them out, right? (except where the semicircles intersect with the first and last ellipses) Then your shape qualifies as a differentiable submanifold of R^2 and all light rays are "good". In this sense your answer is really the resolution of my original problem in two dimensions. As far as I can see it is not yet a full solution for the problem in higher dimensions since rotating this solution over an "open" angle of 180° about the x-axis (repeatedly) will leave a small chink where a few light rays can escape through. – Thibaut Demaerel Mar 30 '15 at 16:47
• Also a suggestion for an edit for your post: the picture is nice but scaling the x and y-axis consistently may add a lot of value in this case. I'll consider accepting this as an answer soon. – Thibaut Demaerel Mar 30 '15 at 16:51
• It is difficult to have a multipurpose boundary, although I did construct examples of that. Part of the idea here was to make sure each part of the boundary is visible from at most two points at a time. If you shine light from $x_1$ at something directing light from $x_2$ to $x_3$, you typically need another curve to refocus that. – Douglas Zare Mar 30 '15 at 19:30

This is not an answer, but suggests one avenue to explore.

There are "$n$-foci ellipses," but I am not familiar with their reflection properties. This image is suggestive (but certainly not definitive):