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      Polygonal Bounded Regions Illuminating a Polygon

Given a bounded simple polygonal region in the plane, is it illuminable from every point?

Figure 1: Example of a bounded simple polygonal region

In other words, can a light source be placed so as to illuminate the entire room in figure 1,
assuming each wall is a mirror?  The problem can equivalently be posed in terms of a billiard ball
bouncing around a pool table.  We can therefore ask if there exists a "pool shot" between any
two points on a polygonal pool table.

We make the following conventions:
1.  Light rays issue in all directions from the source. We say that the room is illuminated by
     a given source if each point of the room lies on at least one of the rays.
2. The light source can be placed anywhere in the interior of the room, or on the room's boundary.
3. A light ray or pool ball reflects at the sides of the room in such a way that the angle of
    incidence equals the angle of reflection.
4. A light ray or pool ball that strikes a vertex is considered to end or be absorbed there.

The constructions we present next are due to Tokarsky.  These are examples of pool tables
that do not admit a shot between two particular vertices.  Note that they provide a negative answer
to question 1.
 


Examples of Polygonal Rooms Not Illuminable from Every Point

Example 1: There do not exist any pool shots from A' to A'' on the table shown in figure 2 below.


Figure 2: Example of a polygonal table that does not admit a  pool shot between some particular points.

In order to prove this,  observe that any path in a regular polygon unfolds to a path in another polygon
constructed from mirror images of the first.  The path ABCDEF on the left below corresponds to
the straight line path ABCDEF on the right.


Figure 3: Unfolding a path in a regular polygon.

Similarly, the polygon in figure 2 can be constructed by taking mirror images of a right angled
isosceles triangle ABC with right angle at C, such that any point labeled B or C is a vertex
of this polygon.  Notice that points labeled A can lie either in the interior of the polygon or they
can be vertices of the polygon.  Refer to figure 4 below.  Hence a pool shot on this polygonal table
will fold up to a path in a triangle labeled ABC.


Figure 4: Table constructed by taking mirror images of
               a right angle isosceles triangle ABC with right angle at C

We now claim that a right angle isosceles triangle ABC with right angle at C does not admit
a pool shot of non zero length from A coming back to A. Tokarsky gave and proved this result
in the form of a lemma.

Making use of these results we can now prove by contradiction that the table in figure 4
does not admit a shot form A' to A''.

If there were a pool shot from A' to A'', the path followed by the ball leaving A' must pass
through the interior of one of the eight triangles surrounding A'.  Let this triangle be T.
Hence the pool shot from A' to A'' would fold up to a pool shot from A' to A' in T, which
is impossible by the lemma just given.
Since we reached a contradiction, we conclude that there is no pool shot from A' to A''.
                                                                                                                    QED

Note that this proof also suggests that there does not exist a pool shot between any
two points labeled A on this table.


Example 2: Consider the non convex pool table ABCD in figure 5, which is symmetric
                    with respect to BC and with m(<A) = m(<D) = 10 degrees, and m(<B) = 120 degrees.
                    Does there exist a pool shot from A to D?

Figure 5

Claim
On any symmetric pool table of the type shown in figure 6, where x divides 90
(and B, C are not 180), there does not exit a pool shot from A to D.

Figure 6

By reflecting in BC, a pool shot from A to D would correspond to a path of non zero length
from A back to A in triangle ABC.  We now prove that the latter is impossible.

THEOREM:   Let x divide 90.  Then the triangular pool table ABC as shown in figure 7 below,
where m(<A) = x, and m(<B) = nx, with n a positive integer, does not have a pool shot
of non zero length from A coming back to A.

Figure 7

Proof:  The idea behind the following proof is that knowing the angles of the pool table and the angle y
            at which the initial shot leaves A, we can determine the angles of incidence with the three sides
            AB, BC and AC. In order to return to A the last reflection must be against side BC. We will show
            that every time the shot bounces off side BC it does so at an angle that will prevent it from ever
            returning to A.
           To simplify notation we will measure all angles mod 2x. When we write m(<B) = 0, for example,
            we mean that the measure of the angle B is an even multiple of x.

We can identify two cases based on the possible values for n.
Case (i):  n is even.
                Let 0 < y < x specify the direction of a pool shot leaving A.
                Since n is even, we have that
                        B = nx mod 2x
                           = 0 (mod 2x),  i.e. B is a multiple of 2x
                and  C = (180 - x - nx) mod 2x
                            = x (mod 2x),  since 2x divides 180.

 
      Determine the angles at which the shot bounces off the sides of the triangle.

      side BC: The shot leaving A will hit side BC at point M (figure 8).
                     The angle of incidence AMC can be determined as follows:
                     m(<AMC) = m(<MAB) + m(<ABM),    since in triangle ABM, the measure of the exterior angle
                                                                              at M equals the sum of the measures of the other two
                                                                              interior angles of the triangle (i.e. A and B).
                                     = (y + nx) mod 2x
                                     = y,
                     Hence the shot hits side BC at an angle congruent to y and will therefore reflect at an angle
                     congruent to -y. See figure 8.
      side AB: Suppose that next the shot hits side AB. Let N be the intersection with AB.
                    Then the angle of incidence is MNB and its measure can be determined from triangle MNB
                     as follows:
                     m(<MNB) = 180 - m(<B) - m(<BMN)
                                    = (180 - nx - y) mod 2x
                                    = y,
                     Hence the shot hits side AB at an angle congruent to y and will therefore reflect of AB at
                     an angle congruent to -y. See figure 8.
      side AC: Suppose that the shot now hits side AC and let P be the intersection with AC.
                    Then the angle of incidence is NPC, and its measure is given by:
                     m(<NPC) = m(<ANP) + m(<A),      since in triangle ANP, the measure of the exterior angle P
                                                                          equals the sum of the measures of the interior angles
                                                                          A and N. See figure 8.
                                    = (y + x) mod 2x
                                    = x + y,
                     Hence the shot hits side AC at an angle congruent to x + y and will therefore reflect of AC at
                     an angle congruent to x - y (since - x - y is the same as x - y when computed mod 2x).
 

                In order to return to A, the last reflection has to be against side BC.
                Since every time the shot bounces off side BC it does so at an angle
                equal to +y or -y, the shot must reenter A at an angle equal to +y or -y.
                But -y is impossible since 0 < y < x = m(<A).
                Hence the shot must reenter at the same angle y as it left. This means
                that at some point the shot will hit a side at a 90 degree angle and
                it will reverse its direction by returning on the exact same path on which
                it came.
                If this happens while bouncing off side AB or BC, we have:
                                       y = 90 mod 2x  or  -y = 90 mod 2x,
                which implies that
                                       y = x mod 2x  or y = 0 mod 2x, since x divides 90.
                This is impossible since 0 < y < x.
        If the reversal of the trajectory occurs while bouncing off side AC, then                                        x + y = 90 mod 2x  or  x - y = 90 mod 2x,
                which again implies that
                                       y = x mod 2x  or  y = 0 mod 2x. Contradiction.  
Figure 8
            Case (ii): n is odd.        In this case m(<B) = x, m(<C) = 0.  The analysis is similar to that of Case (i).
       This time a pool shot leaving A at angle 0 < y < x hits side AB at angles
       congruent to +y or -y and hits sides BC, AC at angles congruent to x + y or
        x - y.  If it returns to A, it must return at the same angle y at which it left, and,
       as before this is impossible.

Figure 9

Since we reached a contradiction in all the cases considered, we conclude that
there is no shot of non zero length from A going back to A on the this table.
                                                                                                              QED


These examples can be extended to construct pool tables with any finite number of pool
shots that can not be made.

 Let G be a pool table built from a triangle ABC of the type shown in  example 2, figure 7  and
which is constructed using only successive mirror images of this triangle. If G is constructed
following the rule that every occurrence of B or C is a vertex, then there does not exist a pool shot
between any two points labeled A.

The pool shot is impossible since a path between any two vertices labeled A
corresponds to a pool shot from A to A in triangle ABC, which is impossible by
the  theorem proved in  example 2 .

This general construction result can be used to form various polygonal tables that do not admit
a pool shot between some particular points.


Other problems

While the above examples give a negative answer to question 1, question 2 remains open. That is,
we do not know if for an arbitrary polygonal region P, there exists at least a point (in the interior or on
the boundary of P) such that a light source placed at this point will illuminate the entire polygon P.

Notice also that the polygonal regions discussed were assumed to have mirror walls.  If we remove this
condition then we could ask how many light sources are needed to directly illuminate the entire polygonal
region P. This is the "art gallery" or "watchman" problem for which we have the following result:

For the direct illumination of a polygonal room with n edges, floor[n/3] light sources are occasionally necessary and always sufficient.


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