Shooting Pool
Alvaro Francisco Manuel
1. Introduction
I wanted to have a project that people can relate to. I something that I was interested in and
spend lots of time on it. I found it in this project. My project is called ’Many Bounces: 2D
to 3D’. In this project I will be explore two similar ideas. The first part of this project is on a
problem called ’Shooting Pool’. The main purpose of this problem is to number of bounces the
ball will takes as it goes around the pool table. My overall task is to have a formula for any size
pool table. Then I will take a look at a 3D version of ’Shooting Pool’. The problem is called
’Follow the Bouncing Ball’. It is like the game of racquetball. Similar, we want to find a general
formula to see how many bounces it takes before going into a corner.
In this paper you will see my work of the two projects and all the things I have found.
2. Sufficient Background Information
I have some set parameters and definitions:
We will refer our table as a dimension which is (n × m), where n corresponds to the y-axis and
m corresponds to the x-axis.
Each n and m are integers. (i.e n, m ∈ Z)
For our pool table the corners will be refereed to as pockets. Each pocket will have a name:
Lower Left: LL, Lower Right: LR, Upper Left: UL, Upper Right: UR.
Our starting position will be at the Lower Left corner pocket.
The angle we are going to hit the ball will be at a 45◦ .
We are going to assume the angle of incidence equals the angle of reflection and that the ball
will travel freely with no friction.
A bounces is considered to be when coming off a side of the pool table it create a 90◦ .
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3. Problem Statement
Our task is to develop a formula in which for any size pool table we can predict the number of
bounces it will have before going into a corner pocket. We will also develop a formula where
we can predict which corner pocket it will go into. The following diagram demonstrates this idea:
Figure 1: 2×3
As you can see we have the ball starting at the Lower Left corner pocket. It is hit at a 45◦
angle and travels into it hits the top of the table. It bounces off the top at a 45◦ then travels
downward as the arrows indicate. The ball hits the right side of the table and then hits the
bottom. Finally, it goes into a Upper Left (UL) corner pocket. Therefore, there are 3 bounces
in total and goes into UR corner pocket.
4. Notation
I will be using notation to make it clear to understand.
1. T(n × m)is defined to be the size of pool table.
2. B(n × m)is defined to be the number of bounces on a n,m pool table.
3. P(n × m) is defined as the corner pocket the ball lands in.
4. gcd(n,m) is defined to be the Greatest Common Divisor of n and m.
For example, we are going to use Figure ?? with our notation. The starting position is at the
LR pocket.The arrows indicate the path the ball takes as it travels around our pool table. One
can see there are 3 bounces and the ball exits at the UL corner pocket. Using our notation
we have a T(2 × 3), which is our pool table size. As you can see there are three bounces that
take place and we note that as B(2 × 3) = 3. Lastly, P(2 × 3) is going to equal UR (Upper
Right)which informs us about where the ball will exit.
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5. Findings
The way I approached this problem was by doing lots of examples. I used grid paper and a
ruler, and started with small table sizes and followed the path the ball would take. As I did
that, I kept track of the number of bounces and figured out what corner pocket it went into. I
did it systematically, I begin with table sizes with n= 1 and different m’s then by n = 2 and so
on. Here is a sample of my data:
T(n x m) B(n x m) P(n x m)
2x4
1
LR
2x5
5
UL
3x1
2
UR
3x2
3
LR
3x3
0
UR
3x4
5
LR
3x5
6
UR
4x4
0
UR
4x5
7
UL
4x6
3
UL
4x7
9
UL
4x8
1
LR
5x4
7
LR
5x5
0
UR
5x6
9
LR
5x7
10
UR
By working out example problems it helped me figure out a few things. I noticed that any
T(n × m) will have the same number of bounces as the reverse table i.e. T(m × n). For example,
look at T(4 × 5) and T(5 × 4), they have the same number of bounces. That is because the table
is a rectangle and the lengths do not change. But, what does change is the corner pocket the
ball goes into. So, take the same example T(4 × 5) and T(5 × 4), then P(4 × 5) and P(5 × 4)
equal UL and LR, respectively.
As one can tell, these examples are relatively small. As we reach bigger integers for n and m,
there are more bounces. Thus, I figured out that that we can multiply by the gcd(n,m) and get
the same results. This is shown later in the Theorem section.
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6. Lemmas
Lemma 6.0.1 An T(n × n) pool table will have B(n × n) = 0 and go into the opposite corner.
Proof:
Let n ∈ N such that our dimensions are (n × n). By the definition of a square the
corner angles are 90◦ . So, our hitting angle is 45◦ which is half of 90◦ therefore the ball will
travel into the opposite corner with no bounces. Hence going into a corner pocket.
Lemma 6.0.2 An T (n × 1) will have B(n × 1)=n − 1 and if n is even then P(n × 1) = UL and
if n is odd P(n × 1).
Proof: Let n ∈ N, T (n × 1) be the pool table size. We know that T(1 × 1) has zero bounces and
exits at the UR pocket by Lemma 6.0.1. Since n represents the vertical dimension (y-axis) then
the squares are stacked on top of each other as n increases.So with n = 2 the ball bounces off the
UR corner of the lower square at a 45◦ angle, and travels to the opposite corner which is the UL
corner of the second square which is the exits. Thus this pattern follows for any n we choose.
For any T(n × 1) there will be n number of squares stacked on top of each other, that make up
the table. Since the nth square has the exit corner then that means each previous square has one
bounce. Therefore, there are n − 1 number of bounces and exit corners alternate from UR and
UL exits. Hence it starts off with P(1 × 1)= UR then P(2 × 1)= UL and repeats in this manner
as n increases. Thus, if n is even then P(n × 1)= UL and if n is odd then P(n × 1)= UR.
7. Theorem
Before we begin with our theorems, first we are going to do an example. We start with a
T(2 × 3) table. We can transform this into a T(6 × 9) as follows. Let k = gcd(6, 9) = 3. We then
decompose each of the squares into k 2 = 32 = 9, so each square in the T(2 × 3) table becomes
a k × k = 3 × 3.
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Figure 2: T(1 × 1) and T(2)
Our new table is a T(2 · k × 3 · k) = T(2 · 3 × 3 · 3) = T(6 × 9). As you can see there will be the
same number of bounces as the T(2 × 3), because the perimeter of the table has not changed.
The shapes are congruent. In this way given a T(n × m) table with the gcd(n, m) = k, we can
transform a T( nk ×
m
)
k
to a T( nk · k ×
m
k
· k) table equals T(n × m) table.
Theorem 7.0.1 If the gcd(n,m)=k, then B(n, m) = B( nk , m
) and P(n × m) = P( nk ×
k
m
).
k
Proof: The T(( nk × m
) table can be transformed into a T(n × m) table as explained above. The
k
) and P(n × m) = P( nk × m
).
tables are congruent so B(n, m) = B( nk , m
k
k
Consider the table T(6 × 5). We are going to pay attention to the number of bounces with a
fixed direction. Graph one shows the number of bounces when we fix the n direction which is
the vertical axis(y-axis).In this case n = 6, so every bounces occurs either at the top or the
bottom wall. From the origin there are 6 steps until you get the first bounces(A). 6 more steps
until we get to the second bounce (B). Then 6 more steps to get to bounces(C) and again 6
more steps to get to bounce (D). Finally, we get to a corner with a total of 30 steps. One can
see there is a pattern,and we get a bounce at the top or bottom whenever there is multiple of 6
5
and the bounces alternate from top to bottom.
Now, with a fixed m direction which corresponds to the horizontal axis(x-axis). In this example
m = 5, so the bounces will occur at either the left or right side wall. From our LL position, it
takes 5 steps to get to the right side wall, bounce number one (A’) and another 5 steps to get to
the left side wall, bounce number two. This continues in the same manner, so at every multiple
of 5 we get a bounce. So the third, fourth, and fifth bounce occur at C’, D’, and E’ respectively.
Finally, we land in a corner label F’ at 30 steps in the m direction.
Putting everything together we see that with a fixed n direction we have a bounce at 6(Top),
12(Bottom), 18(Top),and 24(Bottom). At 30 it goes into a pocket. Similarly, with a fixed m direction we have a bounce 5(Right side), 10(Left side), 15(Right side), 20(Left side), and 25(Right
side). At 30 steps it also goes into the same corner as when we had fixed the m direction. Note
that the Least Common Multiple of 6 and 5 is 30. So, this means the following: n(5)=6(5) =30
AND m(6) = 5(6)= 30.
For the fixed n direction there are (m-1)= 5 − 1 = 4 because m is the multiple to get to 30.
For the fixed m direction there are (n-1)= 6 − 1 = 5 because n is the multiple to get to 30.
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Therefore, the total number of bounce is 9 = 5 + 4 = (6 − 1) + (5 − 1) = 6 + 5 − 2.
Theorem 7.0.2 If the gcd(n,m) = 1 then B(n, m) = n + m − 2.
Proof: Let gcd(n,m)= 1. So, the LCM= n · m. We fix the n direction which tells us that
the bounces occur at the top and bottom and alternate at every multiple such that T \ B := n,
2n, 3n, . . ., (m-1),m·n. Now, we fix the m direction where the bounces alternate from right side
wall to left side wall because m represents the horizontal axis (x-axis). A bounce occurs at every
multiple of m such that R \ L := m, 2m, 3m, . . ., (n-1),n·m.Thus by the definition of a corner
it happens at m · n = n · m = LCM(n,m). So, the number of n bounces is (m − 1) and m
bounces is (n − 1). The total can be found by the sum of (m − 1) and (n − 1) which is equal to
(m + n − 1 − 1) = m + n − 2.
Now, that we can figure the total number of bounces for a generic size pool table we are going
to predict what pocket the ball will go into. Our generic size pool table will be in its simplest
form,(Theorem). We have four option where the ball can land, but we can eliminate one of
them. We will eliminate the Lower Left corner which is explain in corollary ... This leaves us
with three options; Upper Right, Upper Left, and Lower Right. The prediction depends on n,
m and whether or not they are even or odd.
Note that when we fix the ’n’ direction (x direction) the ball hits the top at odd multiples of
’n’ and the ball hits the bottom of the table at even multiples of ’n’. As the ball goes around
the pool table the ball alternates from top to bottom. Similarly, when we fix the ’m’ direction
(y direction) the ball hits the right side at odd multiples of ’n’ and the ball hits the left of the
table at even multiples of ’m’. As the ball goes around the pool table the ball alternates from
right side to left side.
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Theorem 7.0.3 If the gcd(n,m) = 1 then if both n and m are odd then P(n,m)= UR or if n is
odd and m is even then P(n × m) = LR or if n is even and m is odd then P(n × m)= UL.
Proof: Let the gcd(n,m) = 1.
Case 1: Let n and m both be odd. So, there are going to (m − 1) number of top and bottom
bounces. Since n is odd then the ball will end up hitting the top. Now, there are going to be
n − 1 number of right and left bounces and since n is odd, the last bounce is going to hit the
right side. Recall that the lcm(n,m)= n·m, thus the ball is going to land in the top and right
side pocket, which is the P(n × m) = UR.
Case 2:Let n be odd and m be even. So, there are going to be ’m’ bounces but the last one is
going towards the bottom since m is even. Now, there are going to be ’n’ number of bounces
but the last one is going towards the right side because n is odd. Thus the ball is going to land
in the bottom and right side pocket, which is the P(n × m) = LR.
Case 3:Let n be even and m be odd. So, there are going to be ’m’ bounces but the last one is
going towards the top since m is odd. Now, there are going to be ’n’ number of bounces but the
last one is going towards the left side because n is even. Thus the ball is going to land in the
bottom and right side pocket, which is the P(n × m) = UL.
8. Future Work
For my future work, I would like to change up a few things. After studying how a cue ball
travels around a pool table at a specific angle, and starting at a specific corner; some questions
came up. What if I change the angle of my shoot? What if I introduced some side pockets?
What are the differences between a 2-dimension table to a 3-dimension room? The reasons I
would like to explore these questions is that it will get me a better model to the actual game
of Billiards. In the case of change the angle I would be keeping the same starting position, but
would aim at different angles. Recall, we shoot the cue ball at a 45◦ degree angle.
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