We call a polygon star-shaped if there exists at least one point for which the entire polygon is "visible" from that point. The set of such points we call the kernel of the polygon.
The art-gallery theorem states that $\left\lfloor\frac{n}{3}\right\rfloor$ points are sufficient (and sometimes necessary) to cover an $n$-gon and in particular this shows that polygons for which $n\le 5$ are necessarily star-shaped. However, the point which the art-gallery theorem selects is always a vertex of the polygon. I am interested in points for the interior of the polygon.
The main question I have is: Given an $n$-gon for $n\le5$, we know that the closed $n$-gon is star-shaped. Is the open $n$-gon also necessarily star-shaped? In other words, is the intersection of the kernel with the interior non-empty?
Intuitively, this should be possible by simply taking a point "sufficiently close" to vertex which oversees the polygon, but I am having difficult time finding a rigorous proof.
Also note that this is not true for $n\ge 5$. For example, the following hexagon is star-shaped from the red vertex, but the open hexagon is not star-shaped.
As an off-shoot of the above, I would also like to know whether this can generalize the art-gallery theorem: Is $\left\lfloor \frac{n}{3}\right\rfloor$ points in the interior of the polygon sufficient to cover it?
Best Answer
For every polygon with at most 5 edges there is a triangulation where all diagonals meet in a single point $v$. This point is a visibility center, since every triangle can be seen from it.
In this triangulation every triangle has positive area. Hence we can perturb $v$ in any direction such that we still have a triangulation. If we perturb towards the polygon then the new point lies in the visibility kernel.
We can discuss this right away for the general $n$-gon setting. Also the classical approach goes via a triangulation argument. Here, every guard is responsible for an area formed by a "fan" of triangles. Again we can perturb the triangulation and add the perturb point. Then we add 2 new triangles to the perturbed triangulation and observe, that the perturbed point can see all of the "fan". This is not super-formal. But the idea should be clear.