Combinatorial Properties of Periodic Orbits on an Equilateral

Triangular Billiards Table

Andrew Baxter

i Seminar

October 6th, 2005

A dynamical systems problem involving geometry, analytic geometry, linear algebra, number theory, and combinatorics

but not a bit of functional analysis.

Goals

• Explore the motion of a puck sliding across a frictionless triangular surface bounded by walls.– Billiard ball on a triangular table– Laser in a triangular mirror room

• Specifically, we search for paths that repeat themselves, known as “periodic orbits.”

• Two-fold problem:– Does every triangle admit a periodic orbit?– Count the number of periodic orbits on a given

triangle (e.g. equilateral triangle).

Assumptions

1. A puck bounce follows the same rules as a reflection:

The angle of reflection equals the angle of incidence.

2. A path terminates at a vertex

Definitions• The path a puck follows is called the orbit

• Periodic orbits retrace after a finite number of bounces

• A period n orbit bounces n times before retracing.

Some Periodic Orbits

Unfolding

• Drawing from transformational geometry, we reflect the triangle, keeping the path straight.

A B

C

A B

C A

Unfolding

• Drawing from transformational geometry, we reflect the triangle, keeping the path straight.

A B

C A

B

Unfolding

• Drawing from transformational geometry, we reflect the triangle, keeping the path straight.

A B

C A’

B’ C’

A’’ B’’

A periodic orbit exists when the puck returns to an image of the original point at the original angle.

i.e. The puck returns to an image of the original point on an edge parallel to the original edge (a blue edge).

General ProblemConjecture: Every triangle admits a periodic orbit.

Acute Right Isosceles

General Problem

Any rational polygon has infinitely many periodic orbits

(Masur)

x

k

(Vorobets, Gal’perin, Stepin)

m + n =

(Halbeisen, Hungerbuhler) m = n < /2

(Vorobets, Gal’perin, Stepin)

Equilateral Triangle

• Masur’s result shows there are infinitely many periodic orbits on the equilateral triangle

• We will determine:– How to find periodic orbits– How to calculate their periods– How many orbits of a given period

Odd-Period Orbits• There is a period 3 orbit on the equilateral

triangle.• Start on the midpoint of any side at a 60 angle.

• This is the only periodic orbit with an odd period.• We will treat it as a degenerate period 6.

60

60

60

Equilateral Triangle• We can unfold the triangle infinitely many times

in all directions without overlap

Tessellation

• Unfolding infinitely many times in all directions creates a tessellation (a tiling) with equilateral triangles.

• Orbits appear as vectors

A Coordinate System

• Working in the tessellation is aided by imposing a coordinate system.– Set origin at the initial point– Align the y-axis with the right-leaning diagonals– Leave the x-axis alone– Define the triangles to be unit triangles.

Coordinates System Results

• Length

• Angle

22),( yxyxyx

yx

yyx

2

3arctan),(

Finding Periodic Orbits

Theorem: An orbit (x, y) is periodic if and only if x ≡ y (mod 3) (x and y are integers)

Calculating Period

.0,0if2

,,0if2

,0,0if)(2

),(

xyxx

xyxy

yxyx

yxPeriod

• Here “Period” means the number of lines of the tessellation that the vector crosses, not the minimum number of bounces before the orbit repeats itself.

Calculating Period (proof)

• Period(x, y) = h + r + l

• Overlaying parallelograms over the vector shows l = r + h

• When x and y are integers, r = x and h = y

•

)(2

)(2)(

),(

yx

hrhrrh

lrhyxPeriod

The period 22 orbit (4, 7)

Locating Orbits

• For any given n, the terminal points of the period 2n orbits lie in the same left-leaning channel.

Checkpoint

• We want to determineHow to find periodic orbitsHow to calculate their periodsExistence of a period 2n orbit for any nHow many period 2n orbits for any n

SimplificationsTwo simplifications make our work easier

1. Restrict our attention to the region 0 ≤ x ≤ y.

2. k iterations of a period n orbit are counted as a period kn orbit.• This is called a k-fold iteration or a period kn iterated orbit

containing k iterations.

Existence of Orbits

• For any natural number n > 1, is there a period 2n orbit?– Yes. If n is even, use . If n is odd, use .

– Using is a blatant abuse of the simplification that a k-fold iteration of period n orbits is a new period kn orbits since it is a -fold iteration of (1,1)

2,

2

nn

2

3,

2

3 nn

2,

2

nn

2n

Counting Orbits

• How many period 2n orbits are there?– For example, there are two period 22 orbits (n = 11)

(1,10) (4,7)

Counting Orbits

• We wish to count the number of pairs of integers (x, y) such that

1. x + y = n, and

2. x ≡ y (mod 3)

• This is a special case of a more general combinatorics problem

Adventures in Combinatorics

• How many ways can you partition n into k nonnegative addends a1, a2, …, ak such that

1. a1 + a2 + … + ak = n

2. a1 ≡ a2 ≡ … ≡ ak (mod m) for a given m.

• We need k = 2, m = 3 for our purposes.

A Bijection

• There is a bijection between the set of these k-part modulo m partitions of n and the number of partitions of n using only the addends k, m, 2m, …, (k-1)m.

A Generating Function

• The number of partitions of n using only k, m, 2m, …, (k-1)m as parts is known to have the following generating function

1

1

)1()1(

1k

i

imk xx

An Explicit Formula

• For k = 2 and m = 3,

• This O(n) is the number of pairs (x, y), 0 ≤ x ≤ y, that represent period 2n orbits

3

2

2

2)(

nnnO

Checkpoint

• We wanted to determineHow to find periodic orbitsHow to calculate their periodsExistence of a period 2n orbit for any nHow many period 2n orbits for any n

• We still need to address the simplification we made earlier that counts k-fold iterations of period n orbits as period kn orbits.

Iterated Orbits

• Definition: Given periodic orbit (x, y), let d be the largest value such that (x/d, y/d) is a periodic orbit. If d=1, then the orbit is prime. Otherwise, the orbit contains d iterations.

Examples:

(1, 4) is prime

(4, 10) contains 2 iterations of (2, 5)

(3, 6) is prime

(3, 12) contains 3 iterations of (1, 4)

New Goals

• We now want to determine:How to determine if a vector (x, y) represents a

prime orbitIs there a prime period 2n orbit for any given n?For a given n, how many prime orbits are

there?

Proving Primality

• Theorem: A periodic orbit (x, y) is prime if and only if one of the following is true:1. gcd(x,y)=1, or

2. If (x, y) = (3a,3b), then a≠b (mod 3) and gcd(a,b)=1

Examples:

(1, 4) is prime because gcd(1, 4) = 1

(4, 10) contains iterates because gcd(4, 10) = 2

(3, 6) is prime because 1≠2 (mod 3) and gcd(1, 2)=1

(3, 12) contains iterates because 1 ≡ 4 (mod 3)

Existence of Prime Orbits

• There exists a prime period 2n orbit if an only if n is a natural number such that n ≠ 1, 4, 6, or 10.

• The prime orbit has the form:

4). (mod 26,6

4), (mod 03,3

odd, is ,

,2)1,1(

),(

22n

22n

23

23

n

n

n

n

yx

n

n

nn

Counting Iterated Orbits

• Orbits containing iterates are easier to count than prime orbits.

• There are I(n) iterated orbits , where

• (d) is the Möbius function

.)()()(|

nd d

nOdnOnI

.0

primes,distinct s'with )1(

,11

)( 21

otherwise

ppppd

d

d irr

Counting Prime Orbits

• Every periodic orbit contains iterates or is prime, so there are

P(n) = O(n) – I(n)

prime orbits.

• More directly,

.)()(|

nd d

nOdnP

Derivation of D(n)

(x, y) 2-fold 5-fold 10-fold Prime

(1, 49)

(4, 46)

(7, 43)

(10, 40)

(13, 37)

(16, 34)

(19, 31)

(22, 28)

(25, 25)

Total 4 425 P 210 P 15 P

(For n = 50 = 2∙52)

.)()()(|

nd d

nOdnOnI

Calculating I(n) and P(n)

• How many period 100 orbits are prime? (n = 50 = 2∙52)

9)50( 3250

2250 O

5124)50( 5250

550

250 OOOI

459)50()50()50( IOP

Another Example

• How many period 88200 orbits are prime?

(n = 44100 = 22∙32∙52∙72)

7351)44100( O

7532

44100

75344100

75244100

73244100

53244100

7544100

7344100

5344100

7244100

5244100

3244100

744100

544100

344100

244100

144100)44100(

O

OOOO

OOOOOO

OOOOOP

1680

567116807351)44100( I

An Interesting Corollary

• P(p) = O(p) if and only if p is prime.– All period 2p orbits are prime if and only if p is

prime.

More Sample Values

2n O(n) I(n) P(n) 2n O(n) I(n) P(n)

4 1 0 1 20 2 2 0

6 1 0 1 22 2 0 2

8 1 1 0 24 3 2 1

10 1 0 1 26 2 0 2

12 2 2 0 28 3 2 1

14 1 0 1 30 3 2 1

16 2 1 1 32 3 2 1

18 2 1 1 34 3 0 3

Graph of Sample ValuesPurple: O(n)

Red: P(n)

Blue: I(n)

2n

Cumulative Functions

n

i

iOnO1

)()(ˆ

n

i

iPnP1

)()(ˆ

)(ˆ)(ˆ

)(ˆnO

nPnQ

The total number of orbits of period 2n or less.

The number of prime orbits of period 2n or less.

The proportion of orbits of period 2n or less that are prime

Consider the following three functions

Analytic Number Theory

20 40 60 80 100

200

400

600

800

100 200 300 400 500

0.56

0.58

0.62

and are both approximately quadratic

)(ˆ nO )(ˆ nP