I am wondering how to show that a polygon P, in the plane such that at most two angles of P exceed $180$ , then P can be guarded by two guards.
I know that by Chavatal we will need $n \le 8$ and I also know that the sum of interior angles of a polygon is $180(n-2)$ but then I don't know how to prove the statement.
I do not know very much about polygon. I know the bound involving floor function of a third of the vertices.
So maybe we can use that we can only have 8 walls or less? If at most two angles are more then 180, then n-2 angles less then 180. I think a region can be guarded if we can draw a straight line from the guard to the point , where the straight line does not leave the polygon
Can anyone let me know any way to do this? What approach should I try? I really have no idea. I have been sitting trying to do anything but I just make zero progress. I really need some advice
As already suggested, let's prove at first that a polygon with at most one angle exceeding $180^\circ$ can be guarded by one guard. Suppose that $P:=A_1...A_k...A_n$ is a polygon. If $\widehat{A_i}<180^\circ$ for all $1\leq i\leq n$, then $P$ is convex and the result is trivial. On the other hand, if only $\widehat{A_k}>180^\circ$, only one guard $G$ at $A_k$ can guard $P$, because he guards each of the triangles $A_{p}A_{k}A_{p+1}$, where $p\in\{1,...,k-2,k+1,...,n\}$.
Now, let $Q$ be a polygon in the plane such that $n\leq2$ angles exceed $180^\circ$. If $n=0$ ou $n=1$, we get the result. If $n=2$ and $\widehat{A}$ and $\widehat{B}$ are the mentioned angles, divide $Q$ into two polygons so that $A$ and $B$ are in distinct polygons and use the previous conclusion.