Spherical F-Tilings by Triangles and r-Sided
Regular Polygons, r ≥ 5
Catarina P. Avelino
∗
Altino F. Santos
†
Department of Mathematics
UTAD, 5001 - 801 Vila Real, Portugal
Submitted: Dec 19, 2007; Accepted: Jan 24, 2008; Published: Feb 4, 2008
Mathematics Subject Classification: 52C20, 52B05, 20B35
Abstract
The study of dihedral f-tilings of the sphere S 2 by spherical triangles and equiangular spherical quadrangles (which includes the case of 4-sided regular polygons) was
presented in [3]. Also, in [6], the study of dihedral f-tilings of S 2 whose prototiles
are an equilateral triangle (a 3-sided regular polygon) and an isosceles triangle was
described (we believe that the analysis considering scalene triangles as the prototiles will lead to a wide family of f-tilings). In this paper we extend these results,
presenting the study of dihedral f-tilings by spherical triangles and r-sided regular polygons, for any r ≥ 5. The combinatorial structure, including the symmetry
group of each tiling, is given in Table 1.
Keywords: dihedral f-tilings, isometric foldings, spherical trigonometry
1
Introduction
Let S 2 be the Euclidean sphere of radius 1. By a dihedral folding tiling (f-tiling, for
short) of the sphere S 2 whose prototiles are a spherical r-sided regular polygon, P r , and
a spherical triangle, T , we mean a polygonal subdivision τ of S 2 such that each cell (tile)
of τ is congruent to P r or T and the vertices of τ satisfy the angle-folding relation, i.e.,
each vertex of τ is of even valency 2n, n ≥ 2, and the sums of alternate angles are equal;
that is,
n
n
X
X
α2i =
α2i−1 = π,
i=1
i=1
∗
([email protected]) Research Unit CM–UTAD of University of Trás-os-Montes e Alto Douro.
([email protected]) Supported partially by the Research Unit Mathematics and Applications,
through the Foundation for Science and Technology (FCT).
†
the electronic journal of combinatorics 15 (2008), #R22
1
where the angles αi around any vertex of τ are ordered cyclically. In this paper we
shall discuss dihedral f-tilings by spherical triangles and spherical r-sided regular polygons (r ≥ 5).
F-tilings are intrinsically related to the theory of isometric foldings of Riemannian
manifolds, introduced by S. A. Robertson [7] in 1977.
The classification of f-tilings was initiated by Ana Breda [1], with a complete classification of all spherical monohedral f-tilings. Later on, in 2002, Y. Ueno and Y. Agaoka [11]
have established the complete classification of all triangular monohedral tilings (without
any restrictions on angles).
Dihedral f-tilings by spherical parallelograms and spherical triangles was recently obtained in papers [3, 4, 5]. Robert Dawson has also been interested in special classes of
spherical tilings, see [8, 9, 10] for instance.
The study of dihedral f-tilings of the sphere by triangles and r-sided regular polygons
was initiated in 2004, [3], where the case r = 4 was considered. We believe that the case
r = 3 lead to a wide family of f-tilings and it is still in development. We will assume
throughout the text that r ≥ 5.
We shall denote by Ω (P r , T ) the set, up to an isomorphism, of all dihedral f-tilings of
S 2 whose prototiles are P r and T .
From now on P r (r ≥ 5) is a r-sided regular polygon of internal angle α and edge
length a and T is a spherical triangle of internal angles β, γ and δ, with edge lengths b
(opposite to β), c (opposite to γ) and d (opposite to δ), see Figure 1.
a
?
a
g
?
?
P
d
b
r
a
T
b
?
?
a
?
d
a
c
Figure 1: Prototiles: a spherical r-sided regular polygon and a spherical triangle
It follows straightway that
β+γ+δ >π
and
3π
5
≤
(r−2)π
r
< α < π.
In [3] it was established that any τ ∈ Ω (P r , T ) has necessarily vertices of valency four.
We shall describe the set Ω (P r , T ) by considering different cases separately depending on
the nature of T (isosceles or scalene). If T is an equilateral spherical triangle, then it is
easy to see that Ω(P r , T ) = ∅ (the proof is analogous to the case r = 4 in [3]).
In order to get any dihedral f-tiling τ ∈ Ω (P r , T ), we find useful to start by considering
one of its planar representations (P R), beginning with a common vertex to a spherical
regular polygon and a spherical triangle in adjacent positions.
the electronic journal of combinatorics 15 (2008), #R22
2
In the diagrams that follows it is convenient to label the tiles according to the following
procedures:
(i) The tiles by which we begin the P R of the tiling τ ∈ Ω (P r , T ) are a regular polygon
and a triangle in adjacent positions, labelled by 1 and 10 , respectively;
(ii) For j ≥ 2, the location of tile j can be deduced from the configuration of tiles
(1, 10 , 2, 3, . . . , j − 1) and from the hypothesis that the configuration is part of a
complete P R of a f-tiling (except in the cases indicated).
The paper is structured as follows. In Section 2 we obtain the class of all dihedral
spherical f-tilings by isosceles triangles, T , and r-sided regular polygons, P r . The case
when the prototile T is a scalene triangle is studied in Section 3. The paper is finished
in Section 4 with a briefly summary of the f-tilings obtained, where the combinatorial
structure of each tiling is presented.
2
Dihedral Spherical F-Tilings by Isosceles Triangles
and r-Sided Regular Polygons
In this section P r and T denote, respectively, a spherical r-sided regular polygon (r ≥ 5)
and a spherical isosceles triangle, where P r has angle α, and T has angles β, γ, γ, with
β 6= γ. As referred before, one has (r−2)π
< α < π and 2γ + β > π.
r
Any element of Ω (P r , T ) has at least two cells congruent, respectively, to P r and T ,
such that they are in adjacent positions and in one of the situations illustrated in Figure 2.
?
?
P
r
?
T
?
?
?
?
? g
?
b
P
g
r
?
A
b
g
T
?
?
g
B
Figure 2: Distinct cases of adjacency
We begin by noting that the edge length a of P r is uniquely determined by α (extension
of [2, Proposition 3]):
cos 2π
+ cos2
r
cos a =
sin2 α2
α
2
1 + cos α + 2 cos 2π
r
=
.
1 − cos α
(1)
The two distinct cases of adjacency illustrated in Figure 2 will be now analyzed separately in Proposition 2.2 and Proposition 2.3, respectively. The next lemma will be useful
in such propositions.
the electronic journal of combinatorics 15 (2008), #R22
3
Lemma 2.1 Let k1 α + k2 β + k3 γ = π be a sum of alternate angles around a vertex v of
a tiling τ ∈ Ω (P r , T ). Then at least one of the ki is 0.
Proof. If each ki ≥ 1, then
π = k1 α + k2 β + k3 γ ≥ α + β + γ > π,
which is impossible.
Proposition 2.2 Let P r and T be a spherical r-sided regular polygon and a spherical
isosceles triangle, respectively, such that they are in adjacent positions as illustrated in
Figure 2–A. Then, Ω(P r , T ) 6= ∅ iff
α + γ = π and β =
π
2
or
α + β = π and γ =
π
.
2
The first case leads to r = 6 and to a unique f-tiling, denoted by C, with α = arccos −2
.
3
A planar representation is illustrated in Figure 7. For its 3D representation see Figure 8.
In the second situation, for each r ≥ 5, there
is a single f-tiling given by an antiprism,
denoted by Arα , with α = arccos 1 − 2 cos πr = α0r . A planar representation is illustrated
in Figure 10. Some 3D representations are illustrated in Figure 11.
Proof. Suppose that P r and T are in adjacent positions as illustrated below (Figure 3).
Consider also that this configuration is contained in some element of Ω (P r , T ). With the
?
? g
?
1’
1
?
?
b
g
q
v
?
Figure 3: Planar representation
labelling used in Figure 3, we have θ 6= α, and so
θ=γ
or
θ = β.
or
α + γ < π.
We begin by considering the case θ = γ.
(i) If θ = γ, then necessarily
α+γ =π
1. Suppose firstly that α + γ = π. Then the initial P R is extended to get the one
illustrated in Figure 4. As α > (r−2)π
(r ≥ 5), then γ < 2π
. And so
r
r
β>
(r − 4)π
,
r
the electronic journal of combinatorics 15 (2008), #R22
4
?
? g
?
b v2
b
1’
1
g
? g
?
?
2
g q1
? v1
?
?
3
?
?
?
Figure 4: Planar representation
since 2γ + β > π. We show that the angle θ1 at vertex v1 (Figure 4), adjacent to γ, is
also γ. Clearly this angle cannot be α. If it is β, then α + β = π, since α + β + ρ > π,
for all ρ ∈ {α, β, γ} (observe that α + β + β > (3r−10)π
≥ π, for any r ≥ 5). Nevertheless
r
α + β = π = α + γ implies β = γ, which is a contradiction. Therefore θ1 = γ. Using
an analogous argument successive times around vertices surrounded by adjacent angles α
and γ we conclude that vertex v2 is exclusively surrounded by angles β. And so β = πk ,
for some k ≥ 2. As β > (r−4)π
, then r ≤ 7.
r
We will now consider separately the cases r = 5, r = 6 and r = 7.
If r = 5, then either β = π2 , β = π3 or β = π4 , whose corresponding extended planar
representations are illustrated in Figure 5.
? ?
?
1
1’
b b
b b
?
?
6
? v3
? q
2
g ?
g ?
g g
? ?
?
?
? ?
??
?
g
b b
g
? g
?
?
?
?
g
? q
? 2
b b
?
?
? g
? g
g
?
?
?
?
g
? g
?
?
?
?
?
?
?
??
g g
?
? g
g
g
g g
??
? ?
?
?
?
b
b
?
?
5
?
? gg
?
?
?
??
? ?
g g
?
4
2
3
7
? ?
g g
? g
? g
?
?
?
?
b
b
bb
bb
g g
??
??
?
?
g ?
g
?
g ?
g ?
b
b
q2
g
g ?
?
?
?
?
?
?
?
Figure 5: Planar representations
Nevertheless all these cases lead to a contradiction. We consider
only the first one
π
since the others are analogous. At vertex v3 in Figure 5 β = 2 , the angle θ2 must be
γ. And we get the configuration illustrated in Figure 6. But then we must have θ3 = α,
which is impossible (α + α > π).
the electronic journal of combinatorics 15 (2008), #R22
5
? ?
?
?
1
? g
? g
?
?
7
? ?
g g
1’
2
b b
b b
6
g ?
g ?
b
4
8
g g
? ?
3
? v3
? g
5
g
?
g
?
?
? ?
q3
Figure 6: Planar representation
If r = 6, then β = π2 and the P R illustrated in Figure 4 (with r = 6) is extended in a
unique way to get the configuration represented in Figure 7.
g
? ?
g
? ?
g g
27
g
g
?
23
?
g
22
? ? g
30 g ? ? g
?
24 g g 25
?
??
1
? ?
g g
g
20
1’
? g
? g
g ?
g ?
b b
b b
g
?
3
18
?
?
g
g
15
?
?
12
?
9
? g
? g
g ?
g ?
4
? ?
g g
g
g
? ? g
? ?
10
g g
? ?
5
g
g
11
b b
b b
8
?
14
?
g g
6
b b
b b
g g
? ?
g
?
g
g
7
2
21 19
g
bb
26
bb
g
13
?
bb
17 b b 16
g g
29
? ?
28
??
g g
b
bb
b
Figure 7: Extended planar representation of C
Now, if a is the edge length of P r (opposite to β), then, as γ ∈ (0, π3 ),
cos π2 + cos2 γ
sin2 γ
=
cot2 γ
=
cos π3 + cos2
sin2 α2
2 − cos γ
1 + cos γ
the electronic journal of combinatorics 15 (2008), #R22
α
2
⇐
⇒
⇐
⇒
6
cos γ
=
2
.
3
Hence γ = arccos 23 and α = π − γ ≈ 131.8◦ . We shall denote this f-tiling by C.
It is a straightforward
exercise to show that the edge lengths are a = b = arccos 54 and
√
c = arccos 2 5 5 (opposite to γ). A 3D representation of C is given in Figure 8.
Figure 8: F-tiling C
If r = 7, then we also have β = π2 , and we reach, as in case r = 5, to a contradiction
(see Figure 9(a), where a vertex with three angles α cannot be avoided).
?
?
??
?
?
1
?
1’
? g
? g
?
?
2
?
g
g
?
?
?
?
g
9
g g
? ?
?
?
b
b
b
?
g
v1
2
g
3
??
g g
? ?
?
?
g
?
g
10
4
1’
5
3
g
1
b
8
4
g g
??
?
?g
?
6
b
b
b b
g
?
7
??
gg
?
14
? g
? g
13
12
11
b
(a)
(b)
Figure 9: Planar representations
2. Suppose now that α + γ < π (Figure 3). In this case we have, by Lemma 2.1,
α + kγ = π, for some k ≥ 2. Thus, tile 4 (Figure 9(b)) is uniquely determined and so
the electronic journal of combinatorics 15 (2008), #R22
7
β≤
γ<
π
(see vertex v1 ).
2
2π
≤ π5 . But since
5k
On the other hand, since α > (r−2)π
≥ 3π
, we obtain kγ <
r
5
π
2γ + β > π, then β > 3π
>
,
which
is
impossible.
5
2
2π
,
5
i.e.,
(ii) Consider now that θ = β (Figure 3). We study separately the cases
α+β =π
and
α + β < π.
1. Suppose firstly that α + β = π. Taking in account the edge lengths, one gets
α + β = π = γ + γ, and so γ = π2 . The extension of the planar f-tiling is uniquely
determined, as illustrated in Figure 10.
Now, since a is the edge length of P r (opposite to β), and using (1), we obtain
Therefore,
1 + cos α + 2 cos 2π
r
= − cos α.
cos a =
1 − cos α
q
cos α − 2 cos α − 1 − 2 cos = 0 and so cos α = 1 − 2 + 2 cos 2π
= 1 − 2 cos πr .
r
Taking in account that α ∈ (r−2)π
,
π
, then we conclude that
r
α = arccos 1 − 2 cos πr = α0r .
2
2π
r
The edge length of T opposite to γ is c = π2 , for any r ≥ 5.
The f-tiling with such a P R will be denoted by Arα , with α = α0r (Figure 10). It is
a
?
g
g
g b
g
b
?
?
b g
g
a
g
a
g
b
?
g
a
?
g
b
g
a
?
g
a
g
?
g b
a
g
?
?
a
g
b
g
b g
?
b
g
g
?
a
Figure 10: Planar representation of Arαr0 , α0r = arccos 1 − 2 cos πr , r ≥ 5
easy to see that α0r is an increasing function in r. We have α05 ≈ 128.2◦ , α06 ≈ 137.1◦ and
lim α0r = π. 3D representations of A5α5 and A6α6 are given in Figure 11.
r→+∞
0
0
2. Suppose now that α + β < π. Then, by Lemma 2.1, α + kβ = π, for some k ≥ 2.
As α > (r−2)π
≥ 3π
, then β < 2π
≤ π5 . And so γ > 2π
, since 2γ + β > π.
r
5
kr
5
According to the edge lengths (Figure 12), the sum of alternate angles containing γ is
γ + (k − 1)β + γ, which is greater than π, and so we reach a contradiction.
the electronic journal of combinatorics 15 (2008), #R22
8
Figure 11: F-tilings Arαr0 , α0r = arccos 1 − 2 cos πr , cases r = 5 and r = 6
?
? g
?
1
?
b
1’
?
g
2
b
b b
?
4
3
Figure 12: Planar representation
Proposition 2.3 Let P r and T be a r-sided regular polygon and an isosceles triangle,
respectively, such that they are in adjacent positions as illustrated in Figure 2–B. Then,
Ω(P r , T ) 6= ∅ iff r = 5, α + γ = π and β = α. In this situation, there is a single tiling
. A planar representation and a 3D representation
given by an antiprism A5α , with α = 4π
5
5
of A 4π are given in Figure 14(a) and Figure 14(b), respectively.
5
Proof. Let v be a common vertex to adjacent tiles, congruent to P r and T , respectively,
and surrounded by α and γ. The configuration of a such f-tiling near v must be the one
illustrated in Figure 13.
?
v1
?
?
b
1
?
g
1’
?
v
?
q
g
g
g
2
b
Figure 13: Planar representation
the electronic journal of combinatorics 15 (2008), #R22
9
At vertex v we must have
α+γ =π
or
α + γ < π.
(i) Suppose that α + γ = π. With the labelling of Figure 13, θ = β or θ = γ. If
θ = β, then α + β < π (otherwise β = γ). And so, by Lemma 2.1, α + kβ = π, for some
k ≥ 2. Hence, β < 2π
≤ π5 . Therefore, 2γ + β < 4π
+ π5 ≤ π, which is a contradiction.
kr
r
We conclude that θ = γ and vertex v1 is surrounded by the cyclic sequence of angles
(α, β, γ, γ), with α + γ = π = β + γ, and so α = β.
The extension of the planar f-tiling is now uniquely determined. As
−
we may conclude that
1 + cos α + 2 cos 2π
cos α
r
=
,
1 − cos α
1 − cos α
cos α = − 21 − cos 2π
r
and so
r = 5;
√
observe that if r > 5, then − 21 − cos 2π
< −1. Therefore, cos α = − 12 − cos 2π
= − 1+4 5
r
5
and so α = 4π
= β.
5
√
√
The edge lengths of the prototiles of A54π are a = c = arccos 55 and b = arccos −5 5 . A
5
planar representation of A54π is given in Figure 14(a). Its 3D representation is illustrated
5
in Figure 14(b).
a
?
?
g
g
b
g
b
g
b
?
g
a
a
b
g ?
g
a
g
?
g
g
? g
b
a
a
b
a
g
g
b ?
g g
b
g
? g
b g
a
?
a
a
b
g
g
?
(a) Planar representation
(b) 3D representation
Figure 14: F-tiling A54π
5
(ii) Suppose now that α + γ < π. Thus, by Lemma 2.1, α + kγ = π, with k ≥ 2. And
so β > α > γ, since β + 2γ > π. It follows that the sum of alternate angles containing β
at vertex v1 (Figure 13) cannot be defined, which is impossible.
the electronic journal of combinatorics 15 (2008), #R22
10
3
Dihedral Spherical F-Tilings by Scalene Triangles
and r-Sided Regular Polygons
Here T stands for a scalene spherical triangle of angles β, γ and δ, with β > γ > δ
(β + γ + δ > π), and P r (r ≥ 5) is a spherical r-sided regular polygon of angle α and side
a.
If τ ∈ Ω(P r , T ), then there are necessarily two cells of τ congruent to P r and T , respectively, such that they are in adjacent positions and in one of the situations illustrated
in Figure 15.
The three distinct cases of adjacency will be now analyzed in Proposition 3.1, Proposi?
?
?
?
P
?
?
b
T
?
?
?
d
r
?
g
P
r
g
T
? d
?
A
P
r
?
B
?
? g
?
b
?
?
d
T
b
C
Figure 15: Distinct cases of adjacency
tion 3.2 and Proposition 3.3, respectively.
Proposition 3.1 Let τ ∈ Ω(P r , T ). Then τ does not contain a pair of tiles in the position
of Figure 15–A.
Proof. Suppose that τ ∈ Ω(P r , T ) has two cells in adjacent positions as illustrated in
Figure 15–A. Let θ1 be the angle adjacent to γ and opposite to α and let θ2 be the angle
adjacent to δ and opposite to α as indicated in Figure 16(a). It follows that
θ1 ∈ {β, γ}
and
θ2 ∈ {β, δ}.
If θ1 = β, then α + β = π (otherwise, α + β + ρ ≥ α + β + δ > π, for all ρ ∈ {α, β, γ, δ}).
Taking in account the edge lengths, we obtain α + β = π = γ + δ < π, which is a
contradiction.
Thus, θ1 = γ and analogously we obtain θ2 = δ. Consequently β = π2 and a corresponding P R is illustrated in Figure 16(b).
Since α > β, we have, at vertex v1 , α + γ = π. Consequently, at vertex v2 , we obtain
α + kδ = π, for some k ≥ 2. As β + γ + δ > π, β = π2 and γ = kδ, k ≥ 2, we obtain
γ > π3 . On the other hand, since α > (r−2)π
, we have γ < 2π
. And so r = 5.
r
r
2π
Now, π2 < γ + δ < 2π
+
,
and
so
5
5k
k = 2 or k = 3.
the electronic journal of combinatorics 15 (2008), #R22
11
?
?
g
v2
?
?
d
q2
1’
1
?
?
g
?
?
b
b b4
b
1’
?
g
?
g
3
1
b
q1
d
d
2
g
v1
d d
?
?
(b)
(a)
Figure 16: Planar representations
?
?
6
d
?
?
d
1
1
3
b b
b b
? g
? g
5
?
q
?
?
8
d
5
?
?
d
d
v
4
d
?
(a)
g
g
g
d
g
d
10
11
g
d
b
?
?
9
3
b b
b b
? g
? g
2
d
d
b
1’
g
g
4
2
?
?
d
1’
?
?
d
? d
d
7
(b)
Figure 17: Planar representations
With the labelling used in Figure 17(a), the angle θ must be α. In fact, the analysis of
the edge lengths implies θ 6= β and θ 6= δ. If θ = γ, then γ + γ + δ = π and consequently
we reach a contradiction at vertex v in Figure 17(b) (see edge lengths of tile 7). Thus,
θ = α and we obtain the P R below (Figure 18).
We study now the cases k = 2 and k = 3 separately.
We shall consider firstly that k = 3, i.e., α + 3δ = π. The P R illustrated in Figure 18
is extended to get the one represented in Figure 19 (observe that, after we set up tile 17,
we get 2γ < π, and so 2γ + δ = π, at vertex w1 ). Taking in account the edge lengths, we
have no way to have the angle folding relation full filled at vertex w2 .
Consider now the case k = 2. With the labelling of Figure 18, θ 0 = δ or θ 0 = γ. If
0
θ = δ, then we obtain the P R illustrated in Figure 20, where a vertex surrounded by the
cyclic sequence of angles (α, α, δ, δ, γ, γ) takes place, leading to a contradiction. The case
θ 0 = γ is similar and also leads to a contradiction.
the electronic journal of combinatorics 15 (2008), #R22
12
?
?
?
? ?
?
1
6
dd
1’ 3
’
? q
?
? g b b g q= ?
? g bb g ?
?
?
2 4
dd
5
7
? ?
?
?
?
?
Figure 18: Planar representation
?
?
?
?
?
12
? ?
?
?
1
5
?
?
?
?
? g
g
d
d
2
4
?
g
b
b
g
?
?
g
b
b
g
?
? dd
d
?
d
1’ d
3 d
13
16
d
dg
dg
?
??
’
q
?
w2
b
11 b b b 14 19
10 g g d b
wg d b
9 1 g
b
18
8 b bb
17
g
? g
?
6
7
d
?
?
15
?
?
Figure 19: Planar representation
Proposition 3.2 Let τ ∈ Ω(P r , T ) and suppose that P r and T are in adjacent positions
r
as illustrated in Figure
15–B. Then, τ is an antiprism Aα , where α + γ = π = β + δ,
4π
r = 5 and α ∈ 5 , π .
Proof. Suppose that there are two cells in adjacent positions as illustrated in Figure 15–B.
Let θ be the angle adjacent to β and opposite to α (Figure 21(a)). It follows that
θ=β
or
θ = γ.
(i) If θ = β, then necessarily α + β = π and the cyclic sequence of angles around v
is (α, β, β, α). Now, the vertices surrounded by α and δ cannot have valency 4, and so
α + kδ = π, for some k ≥ 2. A vertex surrounded by, at least, four consecutive angles γ
the electronic journal of combinatorics 15 (2008), #R22
13
?
?
?
?
1
?
?
5
?
?
?
?
?
d
?
? g 1’ d
g b
3
2 b b bg
g ?
4 ?
d
d
? ?
12
6
?
g
? q=d 8 b
’d
?
9 bg
d d
10 g g
11 b b
7
?
? g
15
?
?
?
?
g
?
b 13 d
?
b 14 d
gg
g
17
d
d
b b 18
19
?
16
d ?
Figure 20: Planar representation
?
b
?
?
3
q
1
q=b
? b
?
?
b
1’
g
7
?
?
d
g g
g g
gg
2
v
?
?
b
4
? d
d
d
6
1’
5
d d
?
b
b
? d
?
1
?
?
?
?
(a)
(b)
Figure 21: Planar representations
takes place. Taking in account the edge lengths, we conclude that γ = π3 , as represented
in Figure 21(b).
Observe now that if r ≥ 6, then α > 2π
, and so β < π3 = γ, which is impossible.
3
Therefore r = 5. In this case, β < 2π
, and so δ > 4π
, since β + γ + δ has to be
5
15
greater than π. On the other hand, we have β = kδ (k ≥ 2), and so δ < π5 , which is a
contradiction.
(ii) Suppose now that θ = γ. We consider separately the cases
α+γ =π
and
α + γ < π.
1. If α + γ = π, beginning the P R from vertex v (Figure 21(a)) and taking in account
the electronic journal of combinatorics 15 (2008), #R22
14
the edge lengths of P r and T , we conclude that an extended P R of such f- tiling(s)
corresponds to antiprism(s) τ = Arα such that α + γ = π = β + δ. We will see that,
under these angles relations, we have necessarily r = 5. The refereed P R is shown in
Figure 22(a).
Having in consideration the relations between angles, we have β > α > π2 , γ = π − α
and δ = π − β.
a
?
?
b
d
g
b
d
a
a
b
g ?
d
a
?
g
g
d ? b
d
a
? q= g
b
a
d
g
b
a
g
b ?
g d
b
?
d
? d
b g
a
a
a
d
g
?
b
(a) Planar representation
(b) 3D representation
Figure 22: A5α , α ∈
4π
,π
5
One has
1 + cos α + 2 cos 2π
cos α + cos2 β
r
=−
,
1 − cos α
sin2 β
where a is the edge length of T opposite to γ. As α, β ∈ π2 , π , we have
cos a =
(2)
1 + cos α + 2 cos 2π
cos α + cos2 β
r
+
1 − cos α
sin2 β
=
0
⇐
⇒
1 + cos α + 2 cos 2π
r
− 1 sin2 β
1 − cos α
=
− cos α − 1
⇐
⇒
sin2 β
=
sin2 α
−2 cos α + cos 2π
r
⇐
⇒
sin β =
Taking in account that β > α, we obtain
the electronic journal of combinatorics 15 (2008), #R22
sin α
q
.
2π
−2 cos α + cos r
(3)
15
sin β
<
sin α
⇐
⇒
q
−2 cos α + cos 2π
r
>
1
⇐
⇒
cos α
<
− 12 − cos 2π
,
r
leading us toconclude that r = 5 (otherwise, last inequality does not make sense), and
so α ∈ 4π
, π . Using (3), we may also conclude that 4π
< β < π and
5
5
lim β =
4π
α→
5
4π
5
and lim β = π.
α→π
In Figure 22(b) is illustrated a 3D representation of A5α ,
4π
5
< α < π.
2. If α + γ < π (Figure 21(a)), we must have
β > α > γ > δ.
Taking in account that there are necessarily vertices of valency four, it follows that
β + δ = π or β + γ = π.
If β + δ = π then, taking in account the edge lengths, we obtain, at vertex v,
β + δ = π = γ + α < π, which is impossible. It follows that β + γ = π and the
sums of alternate angles at vertex v must be β + kδ = π = α + γ + (k − 1)δ, for some
k ≥ 2, as represented in Figure 23.
g
5
b
?
g
4
d
?
b
3
d
v d
? g
b
2
1’
1
d
b
g
?d
?
?
Figure 23: Planar representation
The current conditions between angles allows to write α, β and γ as functions of δ:
α = π − (2k − 1)δ,
β = π − kδ,
γ = kδ,
k ≥ 2.
Next we prove that these relations between angles lead to a contradiction. In fact, as
2π
α > (r−2)π
, we have δ < r(2k−1)
.
r
Since
1 + cos α + 2 cos 2π
cos γ + cos β cos δ
r
=
,
1 − cos α
sin β sin δ
the electronic journal of combinatorics 15 (2008), #R22
16
we conclude that
1 − cos((2k − 1)δ) + 2 cos 2π
cos(kδ)(1 − cos δ)
r
=
.
1 + cos((2k − 1)δ)
sin(kδ) sin δ
2π
Considering the (continuously differentiable) function ψ defined in 0, r(2k−1)
by
1 − cos((2k − 1)δ) + 2 cos 2π
cos(kδ)(1 − cos δ)
r
ψ(δ) =
−
,
1 + cos((2k − 1)δ)
sin(kδ) sin δ
2π
we have that ψ has no zeros in 0, r(2k−1)
. Indeed, the first derivative of ψ is
ψ 0 (δ) = (2k − 1) cos2
π
r
sin((2k − 1)δ) sec4
(2k−1)δ
2
+ 12 φ(δ) csc(δ) csc2 (kδ) tan 2δ ,
4kπ
where φ(δ) = 2k sin δ − sin(2kδ). As 0 < δ < 2kδ < r(2k−1)
< π and using the fact that
the cosine is a decreasing function on (0, π), we conclude that
2π
φ0 (δ) = 2k(cos δ − cos(2kδ)) > 0, ∀δ ∈ 0, r(2k−1)
⊂ (0, π).
2π
Then φ is an increasing function and since φ(0) = 0, we obtain φ(δ) > 0 in 0, r(2k−1)
.
Thus we can
that ψ 0 (δ) is positive and so ψ is an increasing continuous
easily observe
2π
1
function in 0, r(2k−1)
. Since lim+ ψ(δ) = cos 2π
− 2k
> 0 (k ≥ 2), we obtain ψ(δ) > 0,
r
δ→0
2π
2π
for all δ ∈ 0, r(2k−1)
, and so ψ has no zeros in 0, r(2k−1)
.
Proposition 3.3 Suppose that P r and T are in adjacent positions as illustrated in Figure 15–C. Then, in this situation, Ω(P r , T ) 6=∅ iff α + δ = π = β + γ. It is composed by
r
5 4π
a family of antiprisms A
if r = 5, and α ∈ (α0r , π) if r ≥ 6, where
α , with α ∈ α0 , 5
α0r = arccos 1 − 2 cos πr .
Proof. Let v be a common vertex to adjacent tiles, congruent to P r and T , and surrounded
by α and β. The P R near v is illustrated in Figure 24(a). With the labelling of this Figure,
we must have θ = β or θ = δ.
(i) Consider firstly that θ = β. Then necessarily α + β = π, implying α > β > γ > δ.
Consequently vertex v1 (Figure 24(b)) is not of valency four, and so α + kδ = π, for some
k ≥ 2.
Taking in account the edge lengths of P r and T , the sum of alternate angles containing
γ at vertex v1 is γ + (k − 1)δ + β, which is greater than π.
(ii) Consider now that θ = δ (Figure 24(a)). If α + δ < π, then it is a straightforward
exercise to show that the sum of alternate angles containing β at vertex v cannot be
defined (see edge lengths of P r and T ), which is an impossibility. And so α + δ = π.
The P R represented in Figure 24(a) is now extended in a unique way as illustrated in
Figure 25, which represents an antiprism, where
α + δ = β + γ = π, with α > β >
the electronic journal of combinatorics 15 (2008), #R22
π
2
> γ > δ.
17
?
? g
?
?
1’
1
? g
?
1’
?
?
d g
b
? b
?
?
1
d
2
4
g d
?
?
?
b
d
v1 d
q
b
5
3
v
?
?
?
?
(a)
(b)
Figure 24: Planar representations
a
?
g d
g
b
b
?
d
b
d
g
?
a
g
a
b
d
? g
a
?
g
d
b
a
?
b
g
b
?
?
d
d
b
?
g d
a
a
?
a
b
g
b
a
d g
?
Figure 25: Planar representation of Arα , α ∈ α05 , 4π
if r = 5, and α ∈ (α0r , π) if r ≥ 6
5
As seen before, the equalities (2) and (3) are verified. Using the equality (3), it follows
that β < α iff cos α > − 12 − cos 2π
. This condition is always verified if r ≥ 6. And, if
r
4π
r = 5, then we have β < α iff α < 5 .
q
We also have 0 < sin α < −2 cos α + cos 2π
. Consequently
r
sin2 α < −2 cos α − 2 cos 2π
,
r
and so
α > α0r = arccos 1 − 2 cos πr , r ≥ 5.
√
√ 3D representations of A5α , α ∈ arccos 1−2 5 , 4π
, and A6α , α ∈ arccos 1 − 3 , π , are
5
the electronic journal of combinatorics 15 (2008), #R22
18
√
√ and A6α , α ∈ arccos 1 − 3 , π
Figure 26: A5α , α ∈ arccos 1−2 5 , 4π
5
illustrated in Figure 26.
With the notation used in the paper, we have:
• if α → (α0r )+ , then β → ( π2 )+ and γ → ( π2 )− , for any r ≥ 5;
−
−
• if r = 5 and α → 4π
, then β → 4π
, with α > β, and so γ →
5
5
+
4π +
• if r = 5 and α → 4π
,
then
β
→
, with α < β, and so γ →
5
5
π +
,
5
π −
,
5
δ→
δ→
• if r > 5 and α → π − , then β → π − , with α > β, and so γ → 0+ , δ → 0+ ;
4
π +
;
5
π −
;
5
Summary
In Table 1 is shown a complete list of all spherical dihedral f-tilings whose prototiles are
an isosceles triangle T of angles β, γ, γ (β 6= γ) or a scalene triangle T of angles β, γ, δ
(β > γ > δ), and a r-sided regular polygon P r (r ≥ 5; for r = 4 see [3]) with angle α.
Our notation is as follows:
• α0r = arccos 1 − 2 cos πr , r ≥ 5;
• β = βαr , r ≥ 5, is the solution of equation (3);
• |V | is the number of distinct classes of congruent vertices;
• M and N are, respectively, the number of triangles congruent to T and the number
of r-sided regular polygons congruent to P r , used in the dihedral f-tilings;
• G(τ ) is the symmetry group of each tiling τ ∈ Ω (P r , T ); by Dn we mean the dihedral
group of order 2n; the octahedral group is Oh ∼
= C2 × S4 (the symmetry group of
the cube).
the electronic journal of combinatorics 15 (2008), #R22
19
F-Tiling
α
β
γ
Arαr , r ≥ 5
αr0
π − αr0
π
2
0
4π
A5α
A5
α50 , 5
4π
5
A5α
4π
5 ,π
(αr0 , π)
arccos −2
3
4π
5
Arα ,
r≥6
C (r = 6)
|V |
M
N
G(τ )
–
1
2r
2
D2r
π−α
1
10
2
D5
–
1
10
2
D5
1
10
2
D5
1
2r
2
Dr
2
24
8
Oh
βα5
4π
5
π−
βα5
π−α
π − βα5
π−α
–
βαr
π
2
βα5
δ
π
5
π−
βαr
π−α
Table 1: The Combinatorial Structure of the Dihedral F-Tilings of S 2 by r-Sided Regular
Polygons, r ≥ 5, and Triangles
In Figure 27 we illustrate in 3D the dihedral f-tilings obtained in the paper. They
consist of:
• a continuous family of pentagonal antiprisms (A5α )α∈[α5 ,π) , in which T is isosceles iff
0
α = α05 or α = 4π
; in Figure 27 we √have considered α05 < α1 <
5
α05 = arccos 1 − 2 cos π5 = arccos 1−2 5 ;
4π
5
< α2 < π and
• for each r ≥ 6, a continuous family of antiprisms (Arα )α∈[αr ,π) , in which T is isosceles
0
iff α = α0r ; we illustrate A6α6 and A6α , with α06 < α < π;
0
• an sporadic f-tiling C in which the prototiles are a 6-sided regular polygon of angle
α = arccos −23 and an isosceles triangle of angles π2 , π − α and π − α.
References
[1] A. M. Breda, A Class of Tilings of S 2 , Geometriae Dedicata, 44 (1992), 241−253.
[2] A. M. Breda and A. F. Santos, Well centered sherical quadrangles, Beiträge zur
Algebra und Geometrie, 44 (2003), 539−549.
[3] A. M. Breda and A. F. Santos, Dihedral f-tilings of the sphere by spherical triangles
and equiangular well-centered quadrangles, Beiträge zur Algebra und Geometrie, 45
(2004), 447−461.
[4] A. M. Breda and A. F. Santos, Dihedral f-tilings of the sphere by rhombi and triangles, Discrete Math. Theorical Computer Sci., 7 (2005), 123−140.
[5] A. M. Breda and A. F. Santos, Dihedral f-tilings of the sphere by triangles and
well-centered quadrangles, Hiroshima Math. J., 36 (2006), 235−288.
[6] A. M. Breda, P. S. Ribeiro and A. F. Santos, A class of spherical dihedral f-tilings,
2007, submitted.
the electronic journal of combinatorics 15 (2008), #R22
20
Figure 27: Dihedral F-Tilings of S 2 by r-Sided Regular Polygons (r ≥ 5) and Triangles
[7] S. A. Robertson, Isometric folding of Riemannian manifolds, Proceedings of the Royal
Society of Edinburgh, 79 (1977), 275−284.
[8] R. J. Dawson, Tilings of the sphere with isosceles triangles, Disc. and Comp. Geom.,
30 (2003), 467−487.
[9] R. J. Dawson and B. Doyle, Tilings of the sphere with right triangles I: the asymptotically right families, Electronic Journal of Combinatorics, 13 (2006), #R48.
[10] R. J. Dawson and B. Doyle, Tilings of the sphere with right triangles II: the (1, 3, 2),
(0, 2, n) subfamily, Electronic Journal of Combinatorics, 13 (2006), #R49.
[11] Y. Agaoka and Y. Ueno, Classification of tilings of the 2-dimensional sphere by
congruent triangles, Hiroshima Math. J., 32 (2002), 463−540.
the electronic journal of combinatorics 15 (2008), #R22
21