A gallery floor plan is a concave hexagon. We need to place guards in the room so that every point on each wall is visible from at least one guard's position. What is the maximum number of guards needed?
We will show that at most two guards are needed. One guard is not enough, for example, in the case of the hexagon shown in Figure 1, if the guard stands at a point other than $C$ in the quadrilateral $A B C F$, then they will not see the $D$ vertex; if the guard stands at a point other than $F$ in the quadrilateral $C D E F$, then they will not see the $A$ vertex.

Since the sum of the angles of a hexagon is $4 \cdot 180^{\circ}=720^{\circ}$, a hexagon can have at most three angles greater than $180^{\circ}$. Let's number the concave angles. Draw the segment of the bisector of the first concave angle that extends from the corresponding vertex to the first intersection point of the hexagon's perimeter and the bisector line (Figure 2). This segment divides the original hexagon into two polygons. Together, these two polygons have at most two concave angles, because when the bisector is drawn, one of the original hexagon's concave angles is split into two convex angles, and no new concave angle is obviously created. Then consider the polygon among the two that contains the second concave angle of the original hexagon. (If the bisector of the first angle divided this angle into two parts - Figure 3 -, then consider the part that remains concave.)

Repeat the previous procedure with the concave angle in this polygon. Then we will have a total of three polygons and at most one concave angle. Finally, draw the bisector of this remaining concave angle. Thus, we have divided the original hexagon into at most four convex polygons with at most three segments. (The resulting polygons are convex because the procedure eliminated all concave angles.)

If a guard stands on the boundary of a convex polygon, then they can see every point of the polygon. Divide the at most four parts into two parts, each containing at most two polygons, such that the polygons in one part share a side. If we place one guard on each of these shared sides, then they will see every point of the original hexagon (Figure 4).

Based on the work of Mihály Szalontay (Budapest, ELTE Apáczai Csere J. Gyak. Gimn., II. o.t.)

Note. The solution to the generalization of the problem can be found in our previous issue (N. 116.). The same question was addressed by László Szabó in his article "Classic Art Gallery Problems" published in the POLYGON* journal, 1993/III. 2. issue. The second part of "Classic Art Gallery Problems" appeared in the December 1996 issue (Volume VI, No. 2) of POLYGON*.[^0]

[^0]:    ${ }^{0}$ See the POLYGON competition announcement on page 270.