Index | Problem Statement | Polygonal Regions | Bounded Smooth Regions | The Three Dimensional Case |

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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.

**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.

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).

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.

which again implies that

y = x mod 2x or y = 0 mod 2x. Contradiction.

Figure 8

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

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 problemsWhile 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.

Index | Problem Statement | Polygonal Regions | Bounded Smooth Regions | The Three Dimensional Case |

Links to related topics | Glossary | References