arXiv:math/0509292v3 [math.DS] 28 Jul 2006
PERIODIC ORBITS OF BILLIARDS ON AN
EQUILATERAL TRIANGLE
ANDREW M. BAXTER1 AND RON UMBLE
Abstract. Using elementary methods, we find, classify and count
the periodic orbits on an equilateral triangle. A periodic orbit
is either prime or some d-fold iterate thereof. Let µ denote the
Möbius function. For each n ∈ N, the number
of (classes of) prime
P
n/d+2
n/d+2
orbits of period 2n is exactly d|n µ(d) ⌊ 2 ⌋ − ⌊ 3 ⌋ .
1. Introduction
A billiard ball in motion on a frictionless triangular table bounces
about along a path completely determined by its initial position, angle
and speed. When the ball strikes a bumper at any point other than
a vertex, we assume that the angle of incidence equals the angle of
reflection. An orbit is any path of the ball; an orbit that strikes a
vertex is terminal and terminates at that vertex; an orbit that retraces
itself after n bounces is periodic with period n; all other orbits are
infinite. When considering periodic orbits over a long period of time
(as we do here), we may ignore the speed and begin our observations
at the point of the first bounce.
In this paper we give a complete answer to the following “billiards
problem:” Find, classify and count the periodic orbits of a given period
on an equilateral triangle. Tabachnikov [9] compiled results regarding
the existence of periodic orbits on polygons with various boundary
curves. In fact, all non-obtuse and certain classes of obtuse triangles
are known to admit a periodic orbit [3], [5], [7], [10]. The first examples
of periodic orbits were discovered by Fagnano in 1745; his period 3
orbit, known as the “Fagnano orbit,” is the only orbit with odd period
on an equilateral triangle (see Theorem 1 below). Much later in 1986,
Masur [5] proved that every rational polygon (whose interior angles
Date: July 11, 2006.
1991 Mathematics Subject Classification. Primary 37E15;
Secondary
05A15,05A17,51F15.
Key words and phrases. Billiards, periodic orbit .
1
The results in this paper appeared in the first author’s undergraduate thesis
supervised by the second author.
1
2
ANDREW M. BAXTER1 AND RON UMBLE
are rational multiples of π) admits infinitely many periodic orbits with
distinct periods, but he neither constructs nor classifies them. Katok [4]
showed that the number of orbits of period ≤ n grows subexponentially
as a function of n. Here we use elementary methods to derive our
explicit counting formula.
The paper is organized as follows: Section 2 reviews the geometric
techniques we need to study the problem. In Section 3 we use analytic
geometry to identify orbits. In Section 4 we construct a bijection between periodic orbits of period 2n and the partitions of n with 2 or 3
as parts; we show that there are O(n) = ⌊ n+2
⌋ − ⌊ n+2
⌋ period 2n or2
3
bits. Section 5 concludes the paper and contains the proof of our main
theorem: Define the p-fold iterate of a period k orbit to be its p-fold
retracing. An orbit that is not the p-fold iterate of any orbit for any
p > 1 isP
prime. Let µ denote the Möbius function; there are exactly
P(n) = d|n µ(d)O (n/d) prime orbits on an equilateral triangle. For
example, P(4) = 0, whereas O(4) = 1. Indeed, the single period 8 orbit
counted by O(4) is in fact a 2-fold iterate of the period 4; there are no
prime period 8 orbits.
This project grew out of an undergraduate research seminar directed
by Zhoude Shao and the second author during the spring of 2003. Participants included John Gemmer, Sean Laverty, Ryan Shenck, Stephen
Weaver and the first author. To assist in our study of the billiards problem, Stephen Weaver created his “Orbit Tracer” software [11], which
generated copious amounts of experimental data and produced several
of the diagrams below. Dennis DeTurck suggested the general billiards
problem and consulted with us on several occasions. We thank these
individuals for their significant contributions to this work.
2. Orbits and Tessellations
To begin, let us find and classify the periodic orbits in an equilateral
triangle. Consider a tessellation of the plane by equilateral triangles positioned so that its three families of parallels are horizontal, left-leaning
and right-leaning. Since any two adjacent triangles in the tessellation
are reflections of each other in their common edge, orbits are represented by lines in the plane. To see this, let △RST be a triangle of
the tessellation and let ℓ = ℓ0 be a line passing through it. Assuming
that the point A0 = ℓ ∩ RS is not an endpoint, let ℓ1 be the reflection
←
→
of ℓ in RS and let A1 = ℓ1 ∩ △RST − A0 . Assuming A1 6= T, let ℓ2
be the reflection of ℓ1 in the line of the tessellation containing A1 and
let A2 = ℓ2 ∩ △RST − A1 . Continuing this process indefinitely, let X
PERIODIC ORBITS OF BILLIARDS ON AN EQUILATERAL TRIANGLE
3
S
denote the closure of set X; the orbit i≥0 ℓi ∩ int (△RST ) is called the
folding of ℓ in △RST (see Figure 1).
Figure 1. Folding line ℓ into △RST.
An orbit in △RST can be unfolded by reversing this process. Begin
at any point P and move along the orbit until it strikes some side.
Reflect the orbit in that side and continue along the reflection, entering
the interior of some triangle adjacent to △RST. Continue along the
reflection until it strikes some side and reflect again. Continuing this
process indefinitely produces a ray with initial point P .
Counting periodic orbits amounts to counting equivalence classes.
Since we are concerned with periodic orbits, we only consider line segments in the tessellation that represent one period.
Definition 1. Given a periodic orbit P Q with P and Q on distinct
horizontals of the tessellation, let
U = {all reflections of P Q in lines of the tessellation}.
A periodic orbit α has type P Q if either α ∈ U or α is the translation
of some orbit β ∈ U.
Since P and Q each lie some positive distance from the nearest vertex,
the equivalence class of P Q is uncountable with one exception. The
Fagnano period 3 orbit, with initial point at the midpoint of a side
and angle of incidence 60◦ , represents the only singleton class. If we
translate the Fagnano orbit in either horizontal direction and extend,
we obtain a period 6 orbit (see Figure 2).
Thus we think of the Fagnano orbit as a degenerate period 6 and
include it in the class of its horizontal translations. Otherwise, orbits
of the same type have the same period.
The Fagnano orbit has many interesting properties [2]. For example:
Theorem 1. The Fagnano period 3 is the only periodic orbit with odd
period on an equilateral triangle.
4
ANDREW M. BAXTER1 AND RON UMBLE
Figure 2. The Fagnano and a period 6 orbit (dotted).
Proof. Let P Q be a periodic orbit with angle of incidence θ and assume
that Q is on line ℓ of the tessellation. Since P Q is periodic, the angle
between P Q and ℓ is θ. If ℓ is left-leaning, θ = 60◦ and P Q is the
Fagnano orbit. If ℓ is a right-leaning, θ = 30◦ and P Q has period 4
(see Figure 3). But if ℓ is a horizontal, P Q has even period by Theorem
3 below.
Figure 3. Angles of incidence between P Q and ℓ.
This peculiarity of the Fagnano orbit confirms our decision to include
it with the period 6.
The symmetry of the tessellation implies that a line crossing all three
of its families of parallels crosses exactly one with an angle of incidence
θ ∈ [30◦ , 60◦). Note that lines crossing one family at 60◦ , cross another
at 60◦ and are parallel to the third. Thus every orbit strikes some line
of the tessellation with an angle of incidence θ ∈ [30◦ , 60◦]. We call
θ the representative angle; any two orbits of the same type have the
same representative angle. The fundamental region Γ is bounded by
two rays with respective angles 30◦ and 60◦ and common pole P on
some horizontal. Periodic orbits are classified by those orbits in Γ with
initial point P .
PERIODIC ORBITS OF BILLIARDS ON AN EQUILATERAL TRIANGLE
5
Figure 4. The fundamental region Γ.
3. Rhombic Coordinates
In this section we introduce the analytical structure we need to count
the orbits of a given (even) period. Note that the orientation of a
periodic orbit is inconsequential since reversing orientation produces
the same orbit. Nevertheless, expressing a representation P Q as the
−→
vector P Q allows us to exploit the natural rhombic coordinate system
given by the tessellation. Position the origin at P ; then the x-axis is
the horizontal containing P . Take the line through P parallel to the
right-leaning lines to be the y-axis and the unit length to be the length
of a side of a triangle (see Figure 5).
Figure 5. Rhombic coordinates.
ANDREW M. BAXTER1 AND RON UMBLE
6
We have the following change of basis matrices:
Euclidean to Rhombic


1 − √13
0
√2
3


Rhombic to Euclidean


1
0
1
2
√
3
2


Henceforth, all ordered pairs represent vectors in the rhombic basis.
Then the fundamental region Γ = {(x, y) | 0 ≤ x ≤ y}. The geometric
length and representative angle θ of an orbit (x, y) ∈ Γ are given by
the formulas
√ !
p
3
y
.
(3.1) k(x, y)k = x2 + xy + y 2 and θ(x, y) = arctan
2x + y
There is a simple way to determine whether a given vector (x, y)
−→
represents a periodic orbit. Note that P Q is a periodic orbit if and only
if Q lies on a horizontal and is the image of P under some sequence of
reflections in lines of the tessellation. Periodic orbits in Γ terminate on
a horizontal since reflections of the initial side containing P generate a
tessellation by hexagons (see Figure 6).
Figure 6. Hexagonal tessellation by reflections of the
initial edge.
Theorem 2. An orbit (x, y) is periodic if and only if x, y ∈ Z and
x ≡ y (mod 3).
Proof. Let V be the set of all vector representations of periodic orbits
in Γ. Since the terminus of each vector in V lies on the hexagonal
tessellation, each vector in V can be written as a Z-linear combination
of (0, 3) and (1, 1). Thus (x, y) ∈ V iff (x, y) = a(0, 3) + b(1, 1) for some
a, b ∈ Z and the result follows.
PERIODIC ORBITS OF BILLIARDS ON AN EQUILATERAL TRIANGLE
7
Figure 7. Termini of vectors in V.
−→
The period of a periodic orbit P Q is the number of lines in the
tessellation that intersect P Q − {Q}. In fact, this number is quite easy
to compute.
Theorem 3. If (x, y) is a periodic orbit with y > 0, then

 2(x + y), x ≥ 0, y > 0
2y,
−y < x < 0
(3.2)
P eriod(x, y) =

−2x,
0 < y ≤ −x.
When y < 0, we have P eriod(x, y) = P eriod(x, −y).
Proof. We only give a detailed proof for (x, y) ∈ Γ. Assume x ≥ 0
−→
and y > 0. Let P Q = (x, y) be a periodic orbit and consider those
triangles in the tessellation whose interiors intersect P Q. The union
of the these triangles is a polygon that can be tiled by unit rhombuses with sides parallel to the coordinate axes. Thus P Q crosses
one tile whenever it crosses a horizontal or right-leaning line of the
tesselation. Since left-leaning lines subdivide each rhombus into triangles, P Q crosses a left-leaning line whenever it crosses a tile. Let
h, r, and l represent the number of horizontal, right-leaning and leftleaning lines crossed, respectively. Then by inspection, r = x and
h = y, and by the argument above l = r + h = x + y. It follows that
P eriod(x, y) = r + h + l = 2(x + y).
8
ANDREW M. BAXTER1 AND RON UMBLE
Figure 8. Rhombic tiling of (4, 7).
For the cases −y < x < 0 and 0 < y ≤ −x, cover the polygon
−→
determined by P Q with rhombuses as suggested in Figure 9.
Figure 9. Rhombic tilings of (−5, 7) (left) and (−9, 3) (right).
Let r, h, l be as above. For −y < x < 0, we see that h = r + l,
r = −x, and h = y; for 0 < y ≤ −x we get r = h + l and again r = −x
and h = y.
Finally, by symmetry of the tessellation we have P eriod(x, y) =
P eriod(x, −y) when y < 0, and the formulas now follow.
Theorem 3 essentially reduces the problem of counting orbits of period
2n to counting integer pairs (x, y) such that 0 ≤ x < y, x ≡ y (mod 3)
PERIODIC ORBITS OF BILLIARDS ON AN EQUILATERAL TRIANGLE
9
and x + y = n. We count these pairs in the next section and sharpen
the result in Section 5.
4. Periodic Orbits and Integer Partitions
Two interesting questions arise: (1) Is there a periodic orbit of period
2n for each n ∈ N? (2) If so, exactly how many distinct period 2n orbits
are there for each n? To begin, we simplify the problem by counting
the k-fold iterate of a period n orbit as an orbit of period kn. Then
question (1) has an easy answer. Obviously there are no period 2 orbits
since no triangle has a pair of parallel sides. For even values of n there
is the orbit ( n2 , n2 ), the n2 -fold iterate of the period 4 orbit (1, 1). For
− 1, n−1
+ 2).
odd values of n ≥ 3 there is the orbit ( n−1
2
2
Regarding question (2), observe that the distinct orbits (1, 10) and
(4, 7) both have period 22 (see Figure 10). To verify distinctness, compute their geometric lengths or representative angles using the formulas
in (3.1) above.
Figure 10. Period 22 orbits (1, 10) (left) and (4, 7) (right).
We answer question (2) by constructing a bijection between period
2n orbits and certain partitions of n. A partition of n is a multiset of
natural numbers that sum to n. For example, 3 can be partitioned as 3,
1+2, or 1+1+1 (without regard to order of addends). The addends of a
partition are called its parts. We are interested in partitions with 2 and
3 as parts; for example, we partition 11 as 2+2+2+2+3 or 2+3+3+3
but not as 3+8.
Lemma 1. There is a bijection
{period 2n orbits} ↔ {partitions of n with 2 and 3 as parts} .
10
ANDREW M. BAXTER1 AND RON UMBLE
Proof. Recall that (x, y), where 0 ≤ x ≤ y, represents a period 2n orbit
if and only if x ≡ y (mod 3) and x + y = n. If (x, y) is a period 2n
gives a partition of n with 2 and 3
orbit, then n = x + y = 2 · x + 3 · y−x
3
as parts. Conversely, given a partition of n with 2 and 3 as parts, write
n = 2a + 3b for some a, b ≥ 0. Then a + (a + 3b) = n and (a, a + 3b) is
a period 2n orbit.
Partitioning an integer using only certain parts is well understood.
From OEIS sequence A103221 [8], we obtain the generating function
1
2
(1 − x )(1 − x3 )
and the explicit formula in our next theorem:
Theorem 4. For each n ∈ N, the number of distinct period 2n orbits
(including k-fold iterates) is exactly
n+2
n+2
−
.
O(n) =
2
3
5. Counting Prime Orbits
Let us sharpen our counting formula by excluding k-fold iterates
from the count.
Definition 2. Let (x, y) be a periodic orbit and let d ∈ N be the largest
value such that x/d ≡ y/d (mod 3). Then (x, y) is prime if d = 1;
otherwise (x, y) is a d-fold iterate.
Although it is difficult to compute d directly for large x and y, primality is surprisingly easy to check.
Theorem 5. An orbit (x, y) is prime if and only if either
(1) gcd (x, y) = 1 or
(2) (x, y) = (3a, 3b) for some a, b ∈ Z such that gcd (a, b) = 1 and
a 6≡ b (mod 3).
Proof. (1) If gcd (x, y) = 1, the orbit (x, y) is prime by definition. Conversely, suppose (x, y) is prime but 3 ∤ x and 3 ∤ y. Let g = gcd (x, y) ;
then x = gm and y = gn for some m, n ∈ N and gm ≡
gn (mod 3).
But 3 ∤ g, hence m ≡ n (mod 3). Therefore (m, n) = xg , yg is periodic. Since (x, y) is prime, g = 1.
(2) Suppose (x, y) = (3a, 3b) is periodic with gcd (a, b) = 1 and a 6≡ b
(mod 3). Let d be as in Definition 2. Then (x,
y) is a d-fold iterate of
x y
x y
,
for
d
=
1
or
d
=
3.
If
d
=
3,
then
,
= (a, b) is not a periodic
d d
d d
orbit since a 6≡ b (mod 3). Thus d = 1 and (x, y) is prime. Conversely,
PERIODIC ORBITS OF BILLIARDS ON AN EQUILATERAL TRIANGLE
11
suppose (x, y) = (3a, 3b) is prime. Then (a, b) is not periodic and a 6≡ b
≡ 3b
(mod 3),
(mod 3). Furthermore, suppose d|a and d|b. Since 3a
d
d
the orbit ( xd , dy ) is periodic. But (x, y) is prime so d = 1.
Identifying prime orbits of period 2n depends on the residue of n
modulo 4.
Example 1. Using Theorem 5, the reader can verify that the following
orbits of period 2n are prime:
• n = 2 : (1, 1)
• n = 2k + 1, k ≥ 1 : (k − 1, k + 2)
• n = 4k + 4, k ≥ 1 : (2k − 1, 2k + 5)
• n = 4k + 10, k ≥ 1 : (2k − 1, 2k + 11).
Our main result, which appears in Corollary 1 below, asserts that
for n = 1, 4, 6, 10, there is no prime orbit of period 2n. Thus Example
1 exhibits a prime orbit of every possible even period. Exactly how
many of each there are is the content of our main result, for which we
need the following simple lemma:
Lemma 2. If a period 2n orbit is a d-fold iterate, then d | n and each
iterate has period 2n
.
d
Proof. If (x, y) is a period 2n d-fold iterate, then ( xd , yd ) is periodic
although not necessarily prime. Thus d|x and d|y, and so d|(x+y) = n.
Furthermore, P eriod( xd , yd ) = 2( xd + yd ) = 2n
.
d
By Theorem 4, exactly O(n/d) period 2n orbits are d -fold iterates.
Hence we can account for all iterates by examining the divisors of n.
Let n = pr11 pr22 · · · prmm be the prime factorization. Then O(n/pi ) period 2n orbits are pi -fold iterates. Summing these approximates the
number of orbits that are iterates. Since the number of (pi pj )-fold
itP
erates have been counted twice, we must subtract i<j O pinpj . But
this subtracts too many; we must add the number of (pi pj pk )-fold iterates for i < j < k. Continue this process, alternately adding and
subtracting per the Principle of Inclusion-Exclusion, until the number
of p1 p2 · · · pm -fold iterates is obtained. The final tally of these 2m − 1
summands is
X
I(n) =
−µ (d) O (n/d)
d>1,d|n
= O (n) −
X
d|n
µ (d) O (n/d) ,
ANDREW M. BAXTER1 AND RON UMBLE
12
where

d=1
 1,
r
(−1) , d = p1 p2 · · · pr for distinct primes pi
µ (d) =

0,
otherwise
is the Möbius function. This proves:
Theorem 6. For each n ∈ N, exactly I (n) period 2n orbits are iterates.
More importantly, since each of the O(n) orbits is either iterate or
prime, this gives the number of prime period 2n orbits.
Corollary 1. For each n ∈ N, the number of prime period 2n orbits is
exactly
X
P(n) = O(n) − I(n) =
µ(d)O(n/d).
d|n
Table 1 in the Appendix displays sample values for O, P, and I. Interesting special cases arise for certain values of n. For example:
Corollary 2. Every period 2n orbit is prime if and only if n is prime.
Proof. This follows directly from the fact that O(n) = 0 if and only if
n = 1.
6. Concluding Remarks
Interesting consequences of our main result in Corollary 1 include
(1) P(n) = 0 if and only if n = 1, 4, 6, 10 and (2) P(n) = O(n) if and
only if n is prime. Many interesting open questions remain; we mention
two:
(1) Every acute triangle admits a period 6 orbit resembling (0,3), and
every isosceles triangle admits a period 4 orbit resembling (1,1). Further, empirical evidence suggests that every acute isosceles triangle admits a period 10 orbit resembling (1,4). Thus we ask: To what extent
do the results above generalize to acute isosceles triangles?
Pn
Pn
(2) Define OO(n) =
i=1 O(n) and PP(n) =
i=1 P(n). Inspection of graphs suggests that O(n) and P(n) are approximately linear while OO(n) and PP(n) are approximately quadratic. DeTurck
asked: Are O(n) and P (n) in some sense derivatives of OO(n) and
PP(n)
P P (n)? Furthermore, note that the function QQ(n) = OO(n)
< 1.
Does lim QQ(n) exist? If so, this limit is the fraction of all periodic
n→∞
orbits (counted by O) that are prime.
PERIODIC ORBITS OF BILLIARDS ON AN EQUILATERAL TRIANGLE
7. Appendix
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
2n O(n) P(n) I(n)
2
0
0
0
4
1
1
0
6
1
1
0
8
1
0
1
10
1
1
0
12
2
0
2
14
1
1
0
16
2
1
1
18
2
1
1
20
2
0
2
22
2
2
0
24
3
1
2
26
2
2
0
28
3
1
2
30
3
1
2
32
3
1
2
34
3
3
0
36
4
1
3
38
3
3
0
40
4
2
2
42
4
2
2
44
4
1
3
46
4
4
0
48
5
1
4
50
4
3
1
52
5
2
3
54
5
3
2
56
5
2
3
58
5
5
0
60
6
2
4
n
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
2n O(n) P(n) I(n)
62
5
5
0
64
6
3
3
66
6
3
3
68
6
2
4
70
6
4
2
72
7
2
5
74
6
6
0
76
7
3
4
78
7
4
3
80
7
2
5
82
7
7
0
84
8
2
6
86
7
7
0
88
8
4
4
90
8
4
4
92
8
3
5
94
8
8
0
96
9
3
6
98
8
7
1
100
9
4
5
102
9
5
4
104
9
4
5
106
9
9
0
108 10
3
7
110
9
6
3
112 10
4
6
114 10
6
4
116 10
4
6
118 10
10
0
120 11
2
9
Table 1: Sample Values for O(n), P(n), I(n).
13
14
ANDREW M. BAXTER1 AND RON UMBLE
References
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[12] Wilf, H., “generatingfunctionology,” Academic Press, Inc., 2nd Ed. (1994).
Department of Mathematics, Rutgers University, New Brunswick,
NJ 08901
E-mail address: [email protected]
Department of Mathematics, Millersville University of Pennsylvania, Millersville, PA. 17551
E-mail address: [email protected]