Convex Geometry 
The illumination problem
Attention: in this page, whenever we talk of lines we will be refering to directed lines (that is, lines with a fixed chosen direction).
This problem consists in, given a bounded convex body in the plane, finding the least number of directions in the plane of that set such that every point in its boundary is illuminated by one of the chosen directions.
Just for a rough idea: imagine that you have a flat solid body laying on a table and that you wish to illuminate its boundary (the external contour) with the least number of light directions possible, being that all possible directions are on the table surface. A point is only illuminated by a certain direction of light if there is a line with that direction that hits the point directly (and does not simply "graze it") and that is directed from the outsider to the inside of the body (so the point considered is the first point of the body that the light reaches).
To be more rigorous: a boundary point \(A\) of a convex body \(F\) is iluminated by direction \(l\) if the light beam with direction \(l\) illuminates point \(A\) and also some neighborhood of it in the boundary of \(F\).
Or in other words a boundary point \(A\) of a convex body \(F\) is iluminated by direction \(l\) if it satisfies:
- The line \(r\) parallel to \(l\) passing through \(A\) is not a supporting line of the set \(F\);
- \(A\) is the first point of \(F\) that you reach advancing in \(r\) following direction \(l\).
In this figure points \(A\) and \(V\) are not illuminated by direction \(l\) since the lines parallel to \(l\) passing through them are support lines of \(F\).
The directions \(l1\), \(l2\), \(l3\), ... , \(ln\) are enough to illuminate the boundary of \(F\), if every boundary point is illuminated by at least one of these directions.
I will denote by \(c(F)\) the least number of directions (in the plane of \(F\)) enough to illuminate the boundary of \(F\).
To illuminate the boundary of a parallelogram three directions are not enough anymore; we need four directions:
If \(F\) is a parallelogram, then \(c(F)=4\).
This result can be experimented on in the sketch below. The body \(F\) is initially set to be a triangle, being illuminated by three directions. We therefore have that \(c(F)\) is less or equal to three. All the red points can be moved; we can therefore change the light directions, in the upper left corner, to see that the points in the boundary of the triangle are illuminated. We can also move the vertices of the triangle to make \(F\) into any quadrilateral, that (provided it is not a parallelogram) also needs only three directions to be illuminated: one only needs to find appropriate light directions. Note that we are assuming the set to be convex.
(The applet below was produced with the help of JavaSketchpad).
Why is it that, for a parallelogram \(F\), the three directions are not enough to illuminate
the boundary?
Because we have four vertices and two pairs of parallel sides. To illuminate these four vertices with
three directions there would have to be a direction illuminating two vertices and this is not possible.
To illuminate a vertex \(V1\) we need a direction of light such that a parallel line to it passing
through \(V1\) intersects the interior of the parallelogram, so it must be a direction of light with
an angle between the angles of both sides of the parallelogram (as seen in the figures, \(a\) and \(b\)
have to be greater than zero, since if se \(a=0\) or \(b=0\) we get support lines that do not illuminate
the vertices). Trying to illuminate the adjacent vertex \(V2\) (or \(V4\)) with that same direction,
the parallel line passing through that vertex will not contain any point in the interior of the set,
hence it does not illuminate it.
