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This problem is identical to the Art Gallery Theorem. Look it up on Google and you can find a lot of interesting things. The beautiful solution presented below is taken from the Brilliant Wiki and was first given by Steve Fisk in 1979.

Solution:

We will prove this theorem through a sequence of claims. First, a triangulation of a polygon is a decomposition of the polygon into triangles by drawing non-intersecting diagonals between pairs of vertices.

    

Claim 1:  Any polygon $P$ can be triangulated.

Proof:

We prove this claim by induction on the number $n$ of vertices. For $n=3n=3$, the polygon $P$ is a triangle, which is already triangulated. For $n\geq4n\geq4$, we will find a single diagonal (i.e., a line segment that lies within $P$ connecting a pair of vertices), which splits the polygon into two smaller parts. Then the triangulation of the entire polygon can be obtained by pasting together the triangulation of the two different parts. Since the sum of the interior angles of $P$ is $(n-2)180^\circ$, there is a vertex $u$ of $P$ with interior angle less than $180^\circ$. Let $v,w$ be the neighboring vertices of $u$ in the polygon. If the line segment $v,w$ lies within the polygon, then this line segment is the desired diagonal. Otherwise, triangle $\triangle uvw$ contains other vertices. Move the line segment $v,w$ towards $u$ until it hits the final vertex $x$ in triangle $\triangle uvw$. Then line segment $ux$ lies within the polygon and can be taken as the desired diagonal, proving the claim.

  

Our second claim involves coloring the vertices of the triangulated polygon in the following way: given a triangulated polygon, a $3$-coloring is a coloring of the vertices such that every triangle has $3$ different colored vertices.

Claim 2: Any triangulated polygon is $3$-colorable.

Proof:

We will proceed by induction on the number of vertices in the polygon. For $n=3n=3$, the polygon is a triangle and we can choose three different colors for the three vertices. Now, consider any triangulated polygon with $n>3n>3$ vertices. Pick any two vertices $u$ and $v$ that are connected by a diagonal (i.e., connected by an edge in the triangulation but not in the original polygon). We can split the polygon into two triangulated polygons using edge $(u,v)$ and by induction, the two triangulated polygons are $3$-colorable. Let (red, blue, green) be the colors in the first triangulation $T_1$ and let $(1,2,3)$ be the colors for the second triangulation $T_2$. Then identify the color of $u$ in the first triangulation to the number labeling $u$ in the second triangulation, and identify the color of $v$ in the first triangulation to the number labeling $v$ in the second triangulation. Finally, identify the last remaining colors in both triangulations with each other.

Then we obtain a coloring for the entire triangulation by preserving the color of all vertices in the first triangulation and using the colors identified with $(1,2,3)$ for the vertices in the second triangulation. This gives a $3$-coloring of the entire triangulated polygon and proves Claim $2$.

Claim 3: For any $3$-coloring of a triangulation, there exists a color such that the number of vertices of this color is $\leq \lfloor \frac{n}{3}\rfloor$, and placing guards on these vertices will guard the entire museum.

Proof:

Without loss of generality, suppose the colors of the vertices are red, green, blue such that the number $r$ of red vertices is less than or equal to the number $g$ of green vertices, which is less than or equal to the number $b$ of blue vertices. The total number of vertices is $n$, so $r+g+b=n$. If $r > \lfloor \frac{n}{3}\rfloor$, then $g$ and $b$ would also be strictly greater than $\lfloor \frac{n}{3}\rfloor$, and the identity $r+g+b=n$ cannot be satisfied. Therefore, $r \leq \lfloor \frac{n}{3}\rfloor$. Now, by placing a guard at every vertex colored by red, observe that every triangle in the triangulation has exactly one red node and thus, exactly one guard. Also, any point $P$ in the museum is contained in a triangle in the triangulated polygon, and $P$ is visible from the vertex of the triangle colored by red. This gives a placement of at most $\lfloor \frac{n}{3}\rfloor$ guards that guard the entire museum, proving Claim $3$.

Claims $1$, $2$, and $3$ together give a proof of the Art Gallery Theorem.