Periodic Orbits on a Triangular
Air Hockey Table
Andrew Baxter
Millersville University
April 2005
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.
C
A B
Unfolding
• Drawing from transformational geometry,
we reflect the triangle, keeping the path
straight.
C A
A B
Unfolding
B
• Drawing from transformational geometry,
we reflect the triangle, keeping the path
straight.
C A
A B
A’’ B’’
A periodic orbit exists
when the puck returns
to an image of the
original point at the
B’ C’
original angle.
C A’
i.e. The puck returns to
an image of the original
point on an edge parallel
to the original edge
A B
General Problem
Conjecture: Every triangle admits a periodic orbit.
Acute Right Isosceles
General Problem
Any rational polygon
has infinitely many
periodic orbits
(Masur) ka + b = p/2
(Vorobets, Gal’perin, Stepin)
ma + nb = p
(Halbeisen, Hungerbuhler) ma = nb 1, is there a period
2n orbit?
– We need a pair (x, y) such that
• x + y = n, and
• x ≡ y (mod 3)
– If n is even, use . If n is odd, use .
– Using is a blatant abuse of the simplification that
k-fold duplicates of period n orbits are new period kn
orbits since it is a -fold duplication of (1,1)
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
An Explicit Formula
• For k = 2 and m = 3,
• This P(n) is the number of pairs (x, y) that
represent period 2n orbits
Checkpoint
• We wanted to determine
How to find periodic orbits
How to calculate their periods
Existence of a period 2n orbit for any n
How many period 2n orbits for any n
• We still need to address the simplification
we made earlier that counts k-fold
duplications of period n orbits as period kn
orbits.
Defining Duplicates
• 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
duplicate-free. Otherwise, the orbit
contains d duplicates.
Examples:
(1, 4) is duplicate-free
(4, 10) contains 2 duplicates of (2, 5)
(3, 6) is duplicate-free
(3, 12) contains 3 duplicates of (1, 4)
New Goals
• We now want to determine:
How to determine if a vector (x, y) represents
an orbit that contains duplicates
Is there a period 2n duplicate-free orbit for
any given n?
For a given n, how many duplicate-free orbits
are there?
(We answer this last question by counting the
number of orbits containing duplicates)
Determining Duplicates
• Theorem: A periodic orbit (x, y) is duplicate-free 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 duplicate-free because gcd(1, 4) = 1
(4, 10) contains duplicates because gcd(4, 10) = 2
(3, 6) is duplicate-free because 1≠2 (mod 3) and gcd(1, 2)=1
(3, 12) contains duplicates because 1 ≡ 4 (mod 3)
Existence of Duplicate-Free Orbits
• There exists a duplicate-free period 2n orbit if an
only if n is a natural number such that n ≠ 1, 4, 6,
or 10.
• The duplicate-free orbit has the form:
Counting Orbits with Duplicates
• Orbits containing duplicates are easier to count
than duplicate-free orbits.
• There are D(n) orbits containing duplicates,
where
• m(d) is the Möbius function
Counting Duplicate-Free Orbits
• Every periodic orbit contains duplicates or
is duplicate-free, so there are
F(n) = P(n) – D(n)
duplicate-free orbits.
• More directly,
Derivation of D(n)
(For n = 50 = 2∙52)
(x, y) 2-fold 5-fold 10-fold Dup-free
(1, 49)
(4, 46)
(7, 43)
(10, 40)
(13, 37)
(16, 34)
(19, 31)
(22, 28)
(25, 25)
Total 4
Calculating D(n) and F(n)
• How many period 100 orbits are duplicate-free?
(n = 50 = 2∙52)
Another Example
• How many period 88200 orbits are duplicate-
free? (n = 44100 = 22∙32∙52∙72)
An Interesting Corollary
• F(p) = P(p) if and only if p is prime.
– All period 2p orbits are duplicate-free if and
only if p is prime.
More Sample Values
2n P(n) D(n) F(n) 2n P(n) D(n) F(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 Values
Purple: P(n)
Red: F(n)
Blue: D(n)
2n
Future Directions
• Little remains for the equilateral triangle.
– Numerical approximation for
and
– Is < 1?
• Some orbits on the equilateral triangle carry over
to acute isosceles triangles.
– Perhaps all do under certain conditions