[b_tterc_p]

> Now place a candle in the middle of the room, one that shines light in every direction. As the light bounces around the different corners, will it always illuminate the whole room? Or will it miss some spots? A side effect of proving the magic wand theorem, Eskin said, is that it conclusively answers this old question.

This is a point light in the middle of a regular polygon right? Why is this noteworthy? Is it that the light settles on all points evenly? Is it in spite of some sort of phase cancellation thing?

[Someone]

To help people understand why this theorem is surprising, first rephrase it from “the candle lights the entire room (bar a finite number of points)” to “the candle can be seen from any position in the room (bar a finite number of positions)”

Next, don’t think of “room”; that puts your mind too much towards simple, almost convex structures. Instead, think of the a floor of a building where all doors are removed.

For example, take the ground floor plan of the Pentagon, with its myriad of rooms and corridors, with all doors removed, and replace all walls by perfect mirrors. Is there a spot to place a candle so that it or it’s reflection, reflection of a reflection, etc. can be seen from all locations in the pentagon, bar a finite number? The theorem says there is.

Now, feel free to make it harder: add back the doors, but don’t completely close them, keeping a rational angle with the walls the door opening is in. Feel free to make the angles as small as you like.

Next, place room dividers wherever you want, as long as they are perfect mirrors, form rational angles with the walls, and don’t completely close of some room or corridor in the Pentagon.

Do you think you’ll be able to completely shield of at least one room, wherever that candle is placed? If so, you’re mistaken.

[NKosmatos]

That would make a great 2D game... You’re given a floor plan at each level (with increased complexity/geometry) and you need to place the candle (light source) in a point where it will illuminate the whole floor.

[crawfordcomeaux]

I want to design floorplans with intentional shadow spaces.

[pbhjpbhj]

There's a good game called Lazer Maze (by ThinkFun) that is related, it concerns shining a directional Lazer though.

[anonytrary]

> bar a finite number? The theorem says there is.

I find that so interesting, as if these positions are an intrinsic mathematical property of the room. I wonder if there's a classification of rooms this way. What do rooms that have the same number of dark points have in common?

This theorem also has implications in the limits of using sound and light for surveillance. The government could ban homes with irrational angles in order to guarantee there are no dark spots for any kind of radiation surveillance.

[popcorncolonel]

Great visualization, thanks!

[ironSkillet]

The only constraint on the polygon is that the angles are rational multiples. There all kinds of crazy non-convex shapes that satisfy this constraint, which is why the result is noteworthy.

[readams]

Rational angles and only a finite number of points are not illuminated. You can still get points that are dark, but not regions.

[anonytrary]

I think this is also interesting when it comes to physics. A dark "point" physically means that the electromagnetic fields permeate the entire room except over a finite set of Dirac delta functions. If you were to try and put something physical in the dark point, you'd quickly turn this into a quantum mechanics problem. I don't think the dark points matter in real life because confining yourself to a single point would require infinite momentum.

This theorem assumes perfect mirrors and Newtonian light. In real life, I predict that the entire room would be illuminated. There's a path of least action for the light to take, but there is a probability of taking other paths. Scattering is not precise. I would think that would add a fudge factor to the paths.

[RcouF1uZ4gsC]

>Imagine a room made out of perfect mirrors, Eskin said. It doesn't have to be a rectangle; any weird polygon will do. (Just make sure the angles of the different walls can be expressed as ratios of whole numbers. For example, 95 degrees or two-thirds of a degree would work, but pi degrees would not.)

My impression was that it did not have to be a regular polygon.

[aidenn0]

It's trivially true for a convex polygon, but non-obvious for many concave polygons.