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This question was cross-posted on Math Stack Exchange. Here is a copy of my answer for it there.
This is it. The perfectly centered billiards break. Behold.
Setup
This break was computed in Mathematica using a numerical differential equations model. Here are a few details of the model:
All balls are assumed to be perfectly elastic and ...

61

Here are a couple pictures. If you'd like to do more, I created this with a simple povray script. Feel free to e-mail me and I'll send it to you.
With four reflections.
With 8 levels of reflections.
One with 24 reflections.
And one with 32 reflections, and each mirror having a slightly different tint, a red sphere, and a slightly wider viewing ...

34

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 ...

answered Apr 19 '18 at 13:15

Adam P. Goucher

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27

I am very surprised that nobody mentioned the REAL reason why mathematicians are
so much interested in billiards. It comes from mechanics (celestial mechanics, first of all), and the first question
considered was that of existence of periodic trajectories. Which correspond to periodic orbits in mechanics.
G. Birkhoff in "Dynamical systems" explains: "Before ...

answered Feb 5 '15 at 23:27

Alexandre Eremenko

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26

These so-called "whispering gallery modes" are familiar from studies of microcavity lasers; they can trap the light indefinitely, only limited by diffraction; this web site by Jens Nöckel nicely summarizes the issues; an efficient way to untrap the trajectory is to introduce flattened portions in the boundary (in 2D this would be a stadium rather than a ...

answered Sep 8 '17 at 9:38

Carlo Beenakker

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24

Douglas Zare's shortest path idea seems to me very well-suited for this.
Intuitively, we can view the circles as being rings, and the reflected ray like a rope going through the rings. We pull to obtain the shortest rope (considering the rings fixed, and other suitable idealizations).
The picture below shows how a path connecting $(0,0)$ with $C(m,n)$ may ...

24

There are some potential applications to the design of optical cavities for lasers.
Imagine a region whose sides are mirrored. You might shine a laser in, and have an opening or a partial reflector where the light can come out. Typically, there is some "gain region" in the interior, say a crystal which is excited electrically or by a laser operating at ...

23

The
billiard-ball computer, also known as a conservative logic
circuit, is an idealized model of a reversible mechanical computer
based on Newtonian dynamics, proposed in 1982 by Edward Fredkin and
Tommaso Toffoli. Instead of using electronic signals like a
conventional computer, it relies on the motion of spherical billiard
balls in a friction-...

answered Feb 5 '15 at 19:27

Carlo Beenakker

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23

Here is a surprising application of billiard-related mathematics.
The story starts quite far, but rest assured that it will end up with problems related to billiard trajectories.
Consider the practical problem of finding out the position dependent electrical conductivity of an object by making voltage and current measurements at the boundary.
This is known ...

answered Feb 5 '15 at 19:30

Joonas Ilmavirta

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22

Let $C_r(x,y)$ or $C(x,y)$ be the circle of radius $r$ about the lattice point $(x,y)$.
Suppose we choose a sequence of circles to hit, and ask for the piecewise linear path of shortest length from the origin hitting each of the circles along the path. If this doesn't go inside a circle, then by the least action principle, the angle of incidence will equal ...

22

I'm sorry that my PhD thesis wasn't published indeed and is hard to find (up to my knowledge it is only available at university of Paris 7). As I have a scanned copy of it, I asked people at my software company Riskdata to include it in the list of posted papers, althogh it is not really related to math finance, which I now study. It should be available in ...

18

This is more of a comment inspired by Jim Belk's answer than an answer to the question in itself. It is also more physics than mathematics. However, I hope I can help readers see that this system is related to some very interesting science.
I wanted to say a little bit about the dynamics of the energy transfer in the billiard break. Very naïvely, you might ...

18

Theorem 1.1 in Richard Swartz's paper Obtuse Triangular Billiards I: Near the (2,3,6) triangle rules out easy proofs: He shows that, for any $\epsilon>0$ and any $N>0$, there is a triangle whose angles are within $\epsilon$ of $(\pi/2, \pi/3, \pi/6)$ and for which any closed path involves more than $N$ bounces. So we can't write down some finite list ...

answered Oct 24 '19 at 2:29

David E Speyer

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17

Suppose your polygon has angles $\alpha_i=\frac{\pi}{k_i}$ where $k_i\geq 2$ are integers. Then they must satisfy
$$\frac{1}{k_1}+\frac{1}{k_2}+\cdots+\frac{1}{k_n}=n-2,$$
which means $n-2\leq \frac{n}{2}$, so $n\leq 4$.
When $n=3$, one finds the solutions $(k_1,k_2,k_3)=(2,4,4),(3,3,3),(2,3,6)$ which correspond to the 45-45-90, equilateral, and 30-60-90 ...

16

As to the fact that this improbable scenario arose from chaos, it seems that numerical error really was at fault (my bisection code was faulty).
This behavior is more to be expected.
EDIT: According to the literature, as Carlo Beenakker pointed out, such "whispering gallery modes" wherein trajectories can stay arbitrarily close to the surface can only ...

15

Start from the simplest path, a triangle with angles $\alpha, \beta, \gamma$, and build the unique triangle for which this path is a billiard path. It's easy to see that the latter triangle has angles $\frac{\alpha+\beta}{2}, \frac{\gamma+\beta}{2}, \frac{\alpha+\gamma}{2}$ and is therefore acute. Any acute triangle can be obtained in such a way.
The next ...

15

I asked Rich Schwartz, who is one of the experts in this area (as noted by the OP). Here, with Rich's permission, is his response:
I am not sure why it is so hard. All I can really say is that, after
a lot of experimentation, I can't really see any pattern to it. It
might be hard in the same way that building the fountain of youth is
hard: nobody ...

14

The problem is essentially the Sinai billiard. That takes place on a finite square table with a circular hole removed (=peg added). There are standard bounces off the straight edges as well as off the peg.
There is a standard procedure of "unfolding" across flat edges: you just take a reflected copy of the table across any flat edge. There is a ...

14

If a ray of light at angle $\alpha$ above the horizontal hits your curve $y = f(x)$ from below at a point where the tangent to the curve has angle $\beta$ below the horizontal, it will reflect at angle $\alpha + 2 \beta$ below the horizontal, and then come back up at $\alpha + 2 \beta$ above the horizontal.
In particular, if $\alpha + 2 \beta = \pi/2$ it ...

13

The keyword you are looking for is " Microorganism Billiards".
Very recent topic, but now seems to be catching up in fluids/bio community.

11

The answer is given by the general theory of rational billiard flows (i.e., those on polygons whose angles are rational multiples of $\pi$). On such a polygon $Q$ the tangent vectors to any given orbit are parallel to a finite set of unit vectors, so that the orbits with initial direction $\theta$ lie on an invariant surface $M_\theta$ which consists of a ...

10

As far as I know, the specific question you ask (entirely illuminable from at least one point) remains open.
(Tokarsky showed that placing a light in some spots can leave some points dark.)
But you may be interested to know that an old conjecture has been settled,
as I reported in
this earlier question:
Lelièvre, Monteil and Weiss have shown that ...

answered Oct 18 '16 at 13:11

Joseph O'Rourke

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9

Gregory Galperin invented billiard method of computing $\pi$, see Playing Pool With $\pi$ (The Number $\pi$ From A Billiard Point Of View)
To calculate $\pi$, take two identical balls. Put one near a wall and roll the other ball toward it. The first ball will hit the second, which will bounce off the wall and come back to hit the first ball. Click click ...

8

There is a software by Jeff Weeks, Curved Spaces which allows for flying through various such spaces. It includes a mirrored tetrahedron but with $\pi/2$ dihedral angles. One can construct own spaces though, by providing a list of 4$\times$4 matrices generating the (linearized version of the) transformation group.
Snapshot:
(obviously not that inspiring, ...

answered Nov 25 '14 at 8:18

მამუკა ჯიბლაძე

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8

Schwartz, Richard Evan. Mostly surfaces. Vol. 60. American Mathematical Society, 2011.
On p.219ff of Schwartz's book, he sketches "an elementary proof, due to Boshernitsyn, that every rational polygon has at least one periodic billiard path."
(He doesn't cite an explicit reference for Boshernitsyn's proof.)
&...

answered Oct 3 '16 at 23:02

Joseph O'Rourke

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7

Poncelet published his theorem ("Poncelet's porism) in 1822, after he returned to France following his captivity as war prisoner in Russia:
J.V. Poncelet, Traité des propriétés projectives des figures (Paris, 1822).
The book has been scanned and can be read here.

7

For a recent progress see http://arxiv.org/pdf/1412.2853.pdf
a local version of this conjecture is proven.

7

The short answer, for arbitrary dimensions is this: You are asking for what polytopes can serve as the fundamental domain of an affine Coxeter group (represented as a group of Euclidean isometries of R^n). These are classified, and in particular, you are right about Question 1. I'll try to elaborate. A good reference for this is Humphreys "Reflection ...

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