I am faced with the following question for my undergraduate Graph Theory class:
Suppose a person is standing in a room which has a painting on each of its walls. Prove that if the room has at most five walls, then the person can find a place to stand so as to see all the paintings at once. Prove that if the room has six walls or more, then it is possible that the person cannot find a place to stand so as to see all the paintings at once.
I do not understand the logic behind this at all. I have drawn out cases in which the person can see all walls when there are more than 6 walls. Can someone please explain to me the logic behind this question, or a theorem that may help? (We are currently learning about planar graphs, colorability, etc.) Thanks!
This is related to the Chvátal art gallery problem. The room is a polygon, and a polygon with $n$ sides can always be triangulated into $n-2$ triangles – at most three if $n\le5$. It is easy to see that no matter how these triangles are placed edge-to-edge, there must be a vertex common to all the triangles, so all the paintings can be seen if the person stands at this common vertex.
A counterexample for six walls is a room shaped like the following, but "thinner" so that the centre can't see the two sharp corners: