enter image description here Can 3 lights be placed on the outside of any convex N dimensional solid so that all points on its surface are illuminated?

  • $\begingroup$ When you have a circle with a "light" right of it and one left of it, do you count the points at the top and bottom as illuminated? $\endgroup$ – Listing Jan 9 '12 at 9:30
  • $\begingroup$ The top of where the light ray touches the circle is illuminated. The top of the circle is not (the lights must be a finite distance away). $\endgroup$ – Angela Richardson Jan 9 '12 at 9:35
  • $\begingroup$ Can you use mirrors or black holes? $\endgroup$ – Briguy37 Jan 9 '12 at 18:11

Already in three dimensions, the answer is no.

Let the solid be a sphere. Consider two lights in arbitrary positions. Together with the center of the sphere, they define a plane. The two points on the sphere farthest from the plane are antipodal, and are not illuminated by the two lights. The third light cannot illuminate both antipodal points.

The proof generalizes readily to $n$ dimensions. It is insufficient to use $n$ lights, because $n-1$ of them along with the center of the hypersphere define an $(n-1)$-dimensional hyperplane, which yields two dark antipodes, and you know the rest.

Also, $n+1$ lights should be sufficient for any convex body: place them at the vertices of a simplex that strictly contains the body. (A rigorous justification escapes me at the moment, but it intuitively seems very clear.) In particular, this answers @Listing's comment in the affirmative.

  • $\begingroup$ Are $N+1$ lights required? $\endgroup$ – Dan Brumleve Jan 9 '12 at 9:52
  • $\begingroup$ And is the answer in two dimensions yes? $\endgroup$ – Listing Jan 9 '12 at 10:02
  • $\begingroup$ The question about the shortest path joining such points is asked here $\endgroup$ – ARi Sep 18 '13 at 16:38

No - consider a 3D sphere. After placing the first 2 lights, there will still be 2 antipodal points that are unlit. No third light could light up both.


On the other hand, if you are allowed to place lights at infinity, I think you could do it in 2. Consider placing lights at infinity shining from the exact opposite directions. Since a convex body in finite dimensions is a countable intersection of halfspaces, the number of directions in which this procedure can fail to fully light the sphere is countable, whereas the number of possible directions to choose from is uncountable.

Eg, you could fail to light up a cube if you shined from a direction exactly parallel to some of the faces, but if you tilt the angle a little bit it will light it up all the faces.

Edit: To elaborate a little bit, if the directions to shine from are represented by points on the unit sphere, then the set of directions parallel to a given halfspace form a circular arc. The set of all directions parallel to at least one halfspace defining the convex body is then a countable union of circular arcs, which cannot cover the whole sphere.


Well it depends on how you decide to treat asymptotic behavior, i.e. - is the solid illuminated at the tangent of a ray? In a practical sense, because of light diffusion it would, but then your question seems pretty theoretical.

For $2$ dimensions: $$x^2+y^2 = 1.$$

Let light source $1$ be at $(0, \infty)$, and source $2$ at $(0, -\infty)$. This would basically make tangents with the circle at essentially $(1,0)$, and $(-1,0)$.

For $n$ dimensions:

$$x_1^2+x_2^2+ \cdots + x_n^2 =1.$$

It's basically the same, where you take a vector, place it at infinity: $(0,0, \dots, \infty)$, $(0,0, \dots, -\infty)$.

A similar behaviour with tangents occurs, although they'd happen along a region, rather than just a pair of points.

If you must shed light on those tangents, I guess things get a little more complicated, and $4$ sources of light would be your best bet. This should be correct no matter which dimension, because light sources technically should radiate in all directions anyway, and it's really obvious given the fact that you can always break down the problem into 2d, and it'll be exactly the same, no matter where in the many dimensions you take it in.

Edit: This is mainly because if you place a point at $(0,0, \dots, \infty)$, if you look at it from a 2d perspective from any plane that does not include infinity, the light source basically shines directly onto the solid, since the point actually lies within the region.


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