I'm trying to prove the corollary of the Art Gallery Theorem to include interior guards.
A.G. Theorem: If P is a simple polygon with n vertices, then there is a set S of at most ⌊n/3⌋ vertices of P such that every point of P is visible from S.
Corollary: If P is a simple polygon with n vertices, then there is a set S of at most ⌊n/3⌋ points in the interior of P such that every point of P is visible from S.
I have tried various approaches but they all land up being verbose and seemingly lack rigour.
My attempt: If $n=3$ the result holds trivially. We assume $n\geq 4$. Apply the Art Gallery Theorem and select one vertex, say $v$, in a colour class with ⌊n/3⌋ vertices. Call the subgraph induced by the vertices visible from $v$, $G$.
Since $P$ is a polygon, $v$ has degree 2. Thus $v$ has 2 incident edges. Extend these edges into the interior of $G$ until we reach an exterior wall of $P$. We claim this creates an interior region of $G'$ (proven below). Place $v'$ in this region and add an edge $vv'$ to form a new graph $G'$. We can triangulate the graph $G'$ in such a way that $v'$ is a neighbour of every vertex of $G$ in the graph $G'$. We can thus add edges $v_iv'$ where $v_i$ is a neighbour of $v$. Thus every vertex visible from $v$ is visible from the interior vertex $v'$. Add $v'$ to $S$ and repeat this selection procedure for each of the vertices in the chosen class. We have found a set $S$ of ⌊n/3⌋ interior points such that every vertex in $P$ is visible from the points in S.
I then go on to try and prove the claim. I realised that I would have to consider seperate cases for when a vertex and its incident edges form a convex/concave region.
Any advice? Thanks :) !