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MATHEMATICAL COMMUNICATIONS
Math. Commun. 17(2012), 127–149
Spherical f-tilings by two non congruent classes of isosceles
triangles - I
Ana Maria Reis d’Azevedo Breda1,∗ and Patrcia dos Santos Ribeiro2
1
2
Department of Mathematics, University of Aveiro, 3 810 193 Aveiro, Portugal
Department of Mathematics, E.S.T. Setúbal, 2 910 761 Setúbal, Portugal
Received February 28, 2010; accepted February 24, 2011
Abstract. The theory of f-tilings is related to the theory of isometric foldings, initiated
by S. Robertson [8] in 1977. The study of dihedral f-tilings of the Euclidean sphere S 2
by triangles and r-sided regular polygons was initiated in 2004, where the case r = 4 was
considered [4]. In a subsequent paper [1], the study of all spherical f-tilings by triangles
and r-sided regular polygons, for any r ≥ 5, was described. Recently, in [2] and [3] a
classification of all triangular dihedral spherical f-tilings for which one of the prototiles
is an equilateral triangle is given. In this paper, we extend these results considering the
dihedral case of two non congruent isosceles triangles in a particular way of adjacency
ending up to a class of f -tilings composed by three parametrised families, denoted by Fk,α ,
Eα and Lk , k ≥ 3, α > π2 , respectively, and one isolated tiling, denoted by G. The
combinatorial structure including the symmetry group of each tiling is also given.
Dawson and Doyle in [6], [7] have also been working on spherical tilings, relaxing the edge
to edge condition.
AMS subject classifications: 52C20, 05B45
Key words: Dihedral f-tilings, isometric foldings, symmetry groups
1. Introduction
Spherical folding tilings or f-tilings for short, are edge-to-edge decompositions of the
sphere by geodesic polygons, such that all vertices satisfy the angle folding relation,
i.e. all vertices are of even valency and the sum of their alternate angles is π. An
f-tiling τ is said to be monohedral if it is composed by congruent cells and dihedral
if every tile of τ is congruent to one of two fixed sets X and Y (the prototiles of τ ).
We shall denote by Ω(X, Y ) the set, up to an isomorphism, of all dihedral f-tilings
of S 2 whose prototiles are X and Y .
The classification of all spherical folding tilings by rhombi and triangles was obtained in 2005, [5]. However, the corresponding study considering two triangular
(non-isomorphic) prototiles is not yet completed. This is not surprising, since it is
much harder. The cases studied, until now are the ones corresponding to prototiles
of the following type:
- an equilateral triangle and an isosceles triangle and
∗ Corresponding
author.
Email addresses:
[email protected] (P. S. Ribeiro)
http://www.mathos.hr/mc
[email protected] (A. M. d’Azevedo Breda),
c
⃝2012
Department of Mathematics, University of Osijek
128
A. M. d’Azevedo Breda and P. S. Ribeiro
- an equilateral triangle and a scalene triangle.
When the prototiles are two non congruent isosceles triangles, there are 3 distinct
types of adjacency, see Figure 1. Here, our interest is focused on dihedral f -tilings
by isosceles triangles with adjacency of type I.
a
a g
T1 b d T2
b
a
a g
I
a
c
d
c
a g
g
T1 b c T2
b
a
a d
b
d
II
c
g
T1 a c T2
a
c
a d
a
b g
d
III
Figure 1: Different types of adjacency
The type I edge-adjacency condition can be analytically described by the equation
cos β + cos2 α
cos δ + cos2 γ
=
.
(1)
sin2 α
sin2 γ
Observe that if α = γ, then β = δ, which is impossible. Therefore, we shall
consider, without loss of generality, α > γ.
From now on, T1 denotes an isosceles spherical triangle of angles α, α, β with sides
a (opposite to α) and b (opposite to β) and T2 another isosceles spherical triangle,
not congruent to T1 , of angles γ, γ, δ with sides c (opposite to γ) and d (opposite to
δ).
In order to get a dihedral f-tiling τ ∈ Ω(T1 , T2 ), we find it useful to start by considering a local representation, beginning with a common vertex to each one of the
isosceles triangles in adjacent positions. 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 local representation of a tiling τ ∈ Ω(T1 , T2 )
are labelled by 1 and 2, respectively;
(ii) For j ≥ 2, the location of tile j can be deduced from the configuration of
tiles (1, 2, . . . , j − 1) and from the hypothesis that the configuration is part
of a complete local representation of an f-tiling (except in the cases explicitly
indicated).
Let us begin by showing that it is impossible to have side b equal to side d and
side a equal to side c.
In fact, assuming that b = d and a = c, what is equivalent to have,
cos β + cos2 α
cos δ + cos2 γ
=
,
sin2 α
sin2 γ
and
cos γ(1 + cos δ)
cos α(1 + cos β)
=
,
sin α sin β
sin γ sin δ
(2)
Spherical f-Tilings by Two Non Congruent Classes of Isosceles Triangles - I
129
we may conclude, by the relation
sin α
sin γ
=
sin β
sin δ
(3)
that α and γ are in the same quadrant; α + β ̸= π, α + γ ̸= π, β + δ ̸= π and
γ + δ ̸= π.
Starting a configuration as shown in Figure 2, a decision about the angle λ1 ∈
{α, γ} must be taken.
bd
2
1
a g
a
g
v0 l 1
Figure 2: Local configuration
1.1. Suppose that λ1 = α. Then, 2α < π, since α = π2 implies, by equation (2),
that γ = α, which contradicts our assumption that α > γ. As, α < π2 then γ < π2 .
Looking at the possible order relations between α, β and γ, δ, one and only one
of the following conditions is satisfied,
- α < β and δ < γ,
- β < α and γ < δ,
- α < β and γ < δ.
Note that the condition β < α ∧ δ < γ is not considered, since it prevents the
existence of vertices of valency four and these must exit, see [4].
1.1.1. Assume that α < β and δ < γ. Then, δ < γ < α < π2 and so β ≥ π2 .
If β ≥ π2 , then the angle λ2 is necessarily δ (any other combination violates the angle
folding relation). But, with λ2 = δ at vertex ve1 (see Figure 3) the sum α + γ + µ
also violates the angle folding relation, for any µ ∈ {α, β, γ, δ}.
l2
bd
2
1
a
a g
a
g ~
v
a 1
Figure 3: Local configuration
1.1.2. Suppose now that β < α and γ < δ. As β < α < π2 and γ < π2 , then δ ≥ π2 .
Taking into account that λ1 = α (see Figure 2), 2α + β > π and 2γ + δ > π, around
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A. M. d’Azevedo Breda and P. S. Ribeiro
vertex v0 the sum of the alternate angles containing λ1 is 2α + tγ = π, for some
t ≥ 1. Accordingly, γ < β < α < π2 ≤ δ.
Assuming that δ = π2 , then γ > π4 and consequently t = 1. Using this information
(2α + γ = π, δ = π2 ) in equations (1) and (2) we get α ≈ 337.5o and β ≈ 225o
contradicting α < π2 .
If δ > π2 , the angle λ3 in Figure 4, must be β and the vertex must be of valency
four, taking in consideration the order and angle folding relations. However, by
relation (3), it is impossible to have β + δ = π.
l3
bd
2
1
g
a
a g
a
a
Figure 4: Local configuration
e2 ∈ {α, β, γ, δ} must be
1.1.3. If α < β and γ < δ, a decision about the angle λ
taken, as shown in the next figure.
~
l2
bd
2
1
a
a g
a
g
a
Figure 5: Local configuration
e2 = α, then β + λ
e2 < π, by (3). In order to satisfy the angle folding
If λ
relation, the sum must be α + β + tγ = π, t ≥ 1, since α < β, 2α + β > π and
γ < α < β, β > π3 , δ > π3 . However, the angle arrangement leads us to the sums
δ + (t + 1)α or δ + γ + tα and both violate the angle folding relation (note that
γ < α).
e2 = β, then 2β < π, since β = π lead us to β+δ = π or β = δ, both impossible
If λ
2
e2 is
by (3). As γ < α < β < π2 ≤ δ, the sum of the alternate angles containing λ
2β + mγ = π, m ≥ 1. By the angle arrangement (Figure 6), we conclude that
the other sum of alternate angles is δ + β + ... + α = π or δ + β + ... + γ = π or
δ + α + ... + α = π or δ + α + ... + γ = π, all impossible by the order relation between
the angles and 2γ + δ > π.
e2 = γ, then β + λ
e2 < π, since β + λ
e2 = π implies that δ + α = π.
Supposing that λ
By relation (3), one has sin δ = ± sin γ, an absurdity. As β + γ < π, the angle λ4
Spherical f-Tilings by Two Non Congruent Classes of Isosceles Triangles - I
131
is either γ or α. In the first case, the sum δ + λ4 , see Figure 7, does not satisfy the
angle folding relation, by the order relation γ < α < β and γ < δ. In the second
case, the same impossibility arises.
e2 = δ, in order to fulfilled the angle folding relation, the sum β + λ
e2 +µ
Finally, if λ
must be β + δ + α = π or β + δ + γ = π, both impossible since 2α + β > π, 2γ + δ > π
and β, δ > π3 .
. .
.
g
g
. .
b
b
bd
g
.
g
b
bd
Figure 6: Angle arrangement
l4
g
bd
2
1
a
a g
a
g
a
Figure 7: Local configuration
1.2. If λ1 = γ in Figure 2, then α + λ1 < π by (3) and again γ < α < π2 . The order
relation β < α < π2 and δ < γ < π2 prevents the existence of vertices of valency four,
so we have to analyse the other three possibilities, namely,
- α < β and δ < γ,
- β < α and γ < δ,
- α < β and γ < δ.
1.2.1. In the first case, as π3 < γ < α < β and 2γ + δ > π, the sum α + γ + µ > π,
for all µ ∈ {α, β, γ, δ} and so the local configuration cannot be extended.
1.2.2. In the second case, as β < α < π2 and γ < π2 , then δ ≥ π2 and angle
e3 ∈ {α, β, γ}, in Figure 8.
λ
e3 = α, then by the configuration the sum of alternate angles containing δ and
If λ
e
λ3 is δ + α + mβ = π, m ≥ 1, contradicting 2α + β > π.
e3 = β, then the referred sum is δ + kβ = π, k ≥ 2 or δ + γ + nβ = π, n ≥ 1.
If λ
In both cases, we conclude that β < γ < α < π2 ≤ δ.
Assuming that δ = π2 , the sum containing the alternate angles α and γ at vertex ve2
is α + γ + qβ = π, q ≥ 1 or α + 2γ + rβ = π, r ≥ 0, since γ > π4 and α > π3 .
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A. M. d’Azevedo Breda and P. S. Ribeiro
~
l3
bd
2
1
g
g
a g
v~2 g
a
Figure 8: Local configuration
The first possibility leads to a sum of alternate angles of the form 2γ + qβ = π,
see Figure 9 (observe that at vertex ve3 , 2α + µ > π for all µ ∈ {α, β, γ, δ}) which
implies the absurdity of α = γ.
π
The second possibility leads us to γ = 5π
3 , β = 2 for r = 0 and 2α + β < π, for
r ≥ 1 contradicting β < γ < π2 and 2α + β > π.
b
bd
2
1
aa
v~ g
3
a g
g g
b
g
g
. ..
Figure 9: Local configuration
If δ > π2 , the vertices of valency four are surrounded by alternate angles δ and α
leading us to γ > π6 . The sum of alternate angles containing α and γ at vertex ve2 is
of the form α+γ +qβ = π, q ≥ 1 or α+2γ +rβ = π, r ≥ 0 or α+3γ +sβ = π, s ≥ 0.
In the first case, the angle arrangement is exactly the same as in the previous figure
and so an absurdity is achieved.
4π
In the second case, for r = 0 we get γ = 5π
3 , β = 3 and for r ≥ 1 one has 2α+β < π.
In both situations we find an impossibility.
5π
π
In the third case, for s = 0, we conclude that γ = 7π
4 , β = 4 contradicting γ < 2
and for s ≥ 1 one gets the absurdity 2α + β < π.
e3 = γ, the sum of alternate angles containing δ and λ
e3 must be δ + γ + kβ =
If λ
π, k ≥ 1. Accordingly, β < γ < α < π2 ≤ δ.
Suppose firstly, that δ = π2 . Then, γ > π4 and in Figure 8 the sum of alternate
angles containing α and γ, at vertex ve2 , is of the form 2α + γ = π or α + 2γ + rβ =
π, r ≥ 0 or α + γ + qβ = π, q ≥ 1.
The first possibility leads us to conclude that γ < β, contradicting the order relation
between the angles.
π
The second possibility, for r = 0 we get γ = 5π
3 , β = 2 and for r ≥ 1 one has
2α + β < π. In both situations an absurdity is achieved.
133
Spherical f-Tilings by Two Non Congruent Classes of Isosceles Triangles - I
What remains is the third possibility and extending the configuration in Figure 8,
we end up at a vertex (e
v4 ), surrounded by alternate angles δ and α, as illustrated
in Figure 10. Taking into account the order relation between the angles, this vertex
must be of valency four, contradicting the fact that α < π2 = δ.
g
g
4
bd
2
1
d a
~
v a
4
g
a g
a
b
8
6
g
b b
. .. 5
3
7
Figure 10: Local configuration
Note that in the previous configuration the tile labeled 8 is unique, otherwise, around
vertex ve2 , we get sums α + γ + qβ = π and 2γ + qβ = π implying that α = γ, which
is an impossibility.
If δ > π2 and taking into account that the sum of alternate angles containing δ
e3 satisfies δ + γ + kβ = π, k ≥ 1, vertices of valency four must be surrounded
and λ
by alternate angles δ and α. Consequently, they satisfy the condition δ + α = π. As
2γ + δ > π, one gets γ > π6 and so at vertex ve2 the sum containing the alternate
angles α and γ must be of the form 2α + γ = π or α + γ + qβ = π, q ≥ 1 or
α + 2γ + rβ = π, r ≥ 0 or α + 3γ + sβ = π, s ≥ 0.
In the first case we get γ < β, contradicting once again the order relation between
the angles.
In the second case, extending the configuration in Figure 8, tile 7 has two possibilities, as shown in Figure 11:
. .
. .
.
8
d a
d a
2
g b
a g
g
a
b
b
7
. ..
aa
g
g
g
g
d a
g
12
db b
a
g
b b
9
11
.
..
2
1
4
3
6
5
bd
g
1
4
g
g
.
bd
g
d
a g
g
b
g
b
7
3
6
5
. ..
g
g
d
a
a
10
13
I
II
Figure 11: Local configuration
Observe that in Figure 11-I, tile 9 is unique, in order to avoid two alternate
angles α, whose sum 2α + µ does not satisfy the angle folding relation for any
µ ∈ {α, β, γ, δ}.
The extended configuration ends up at a vertex surrounded by angles δ, α, γ, β...., β.
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A. M. d’Azevedo Breda and P. S. Ribeiro
At this vertex the sums of alternate angles satisfy δ + γ + kβ = π = α + (k + 1)β.
Therefore, we get δ+γ = α+β and since vertices of valency four satisfy the condition
δ + α = π, one has γ + 2δ = π + β. As β < γ, then 2δ < π, an absurdity.
In Figure 11-II, one has both sums α + γ + qβ = π and γ + γ + qβ = π implying
that α = γ, also an absurdity.
4π
In the third case, α + 2γ + rβ = π, r ≥ 0, we get γ = 5π
3 and β = 3 for r = 0
and for r ≥ 1 we get 2α + β < π, both situations impossible.
5π
In the last case, α + 3γ + sβ = π, s ≥ 0, one has γ = 7π
4 , β = 4 for s = 0 and
π
2α + β < π for s ≥ 1, contradicting γ < 2 and the fact that 2α + β > π.
e3 (as
1.2.3. Finally, assume that α < β and γ < δ and consider again the angle λ
shown in Figure 8).
e3 = α, α + δ < π (otherwise, we get β = δ or α = β) and consequently, the
If λ
sum of alternate angles containing α and δ is one of the following ones: mα + δ =
π, m ≥ 2 or nα + 2δ = π, n ≥ 1 or α + δ + β = π. None of them can occur, since
β, δ > π3 , 2γ + δ > π and γ < α.
e3 = β, by relation (3), δ + β < π and consequently δ + β + α = π or
If λ
δ + β + γ = π, both conditions are impossible (2α + β > π, 2γ + δ > π).
e3 = γ, the sum δ + γ + µ = π, µ ∈ {α, β, γ, δ} is not satisfied, since γ < α <
If λ
β, γ < δ and 2γ + δ > π.
e3 = δ, then δ < π (otherwise, β = δ, an impossibility). However, 2δ + µ > π,
If λ
2
for all µ ∈ {α, β, γ, δ}.
As a direct consequence of the type of adjacency I, we get the following result:
Lemma 1 (Elimination Lemma). In any local configuration of a dihedral f-tiling
whose prototiles are isosceles triangles with adjacency of type I, there are no vertices
surrounded by sequences of alternate angles containing:
(a) p angles β and q angles δ with p, q ≥ 1, or
(b) one angle α (resp. γ) and q angles δ (resp. β), q ≥ 1, or
(c) one angle α (resp. γ), p angles β and q angles δ with p, q ≥ 1.
2. Triangular dihedral f-tilings with adjacency of type I
Starting a local configuration of τ ∈ Ω(T1 , T2 ) with two adjacent cells congruent to
T1 and T2 , respectively, see Figure 12, a choice for angle θ1 ∈ {γ, δ} must be made.
a g
2
1
b
a g
d
q1
Figure 12: Local configuration
Spherical f-Tilings by Two Non Congruent Classes of Isosceles Triangles - I
135
In order to facilitate the construction of dihedral f-tilings, we distinguish the
different order relations between the angles. Therefore, in the first subsection, we
shall consider the order relation α > β and γ > δ, in the second subsection, α > β
and δ > γ, in the third subsection β > α and γ > δ and in the fourth subsection
β > α and δ > γ.
2.1. α > β and γ > δ
With the above terminology one has:
Proposition 1. If θ1 = γ, then Ω(T1 , T2 ) ̸= ∅ if and only if β = δ = πk , k ≥ 3 and
α + γ = π, α ∈] π2 , 2π
3 [. In this case, we get a 2-parameter family of dihedral f-tilings
denoted by Fk,α with D2k , the dihedral group of order 4k, as a group of symmetry.
The action of D2k , respectively, on the faces and vertices of Fk,α gives rise to 2
classes of isohedrality and 3 classes of isogonality. A 3D representation of a member
of this family F3,α is given in Figure 15.
Proof. In order to have Ω(T1 , T2 ) ̸= ∅, necessarily α + θ1 ≤ π.
1. Let us assume that α + θ1 = π, with θ1 = γ. By adjacency condition (1) and
taking into account that α > γ, one has α > π2 > γ > δ and β = δ.
Expanding the configuration illustrated in Figure 12, we obtain the following one
with θ2 ∈ {α, β}.
ag
1
q2
2
ag
ag
b
b
4
d
d
3
ag
Figure 13: Local configuration
1.1. Suppose firstly, that θ2 = α. Then, θ2 + β < π, otherwise, from α + γ = π, we
get β = δ = γ, which is impossible. Taking into account the Elimination Lemma,
the sum containing the alternate angles θ2 and β must be of the form α + kβ = π,
for k ≥ 2.
If α + kβ = π, k ≥ 2, the configuration extends a bit more, to the one shown in
Figure 14. But due to incompatibility of the sides length, it is impossible to pursue
the construction of the dihedral f-tiling.
1.2. Assume that θ2 = β.
If β is a submultiple of π, say β = πk , then k ≥ 3 and we may expand (in a unique
way) the configuration given in Figure 13 by obtaining a global representation of
a tiling τ ∈ Ω(T1 , T2 ) denoted by Fk,α . For each k ≥ 3 and α ∈] π2 , 2π
3 [, Fk,α has
4k isosceles triangles (2k of each type). In Figure 15, we show a 2D and a 3D
representation of F3,α .
The symmetry group of Fk,α is D2k , the dihedral group of order 4k generated
by the rotation, Rzπ , around the zz axis, through πk and the reflection ρyz , on the
k
136
A. M. d’Azevedo Breda and P. S. Ribeiro
plane yOz. D2k acts on the set of vertices of Fk,α partitioning it into three classes.
Each vertex of valency 2k forms a class of isogonality and the vertices of valency
four are all in the same vertex transitive class. In relation to the faces of Fk,α , the
action of the dihedral group D2k , gives rise to two classes of isohedrality, each class
containing 2k congruent isosceles triangles.
g
a a
a
a
7
6
a g
1
.
.
b b b
b
.
a
2
a g
a g
d
d
3
4
5
ag
b
b
a
a
?
.
. .
8
a
Figure 14: Local configuration
b
a
10
g a a
a
g
g g
9
11
b b b b b
5
6
aa
gg
2
d d
d
a
7 a
g
g
3
1
aa
gg
4
d
a a
g g
8
12
d
d
Figure 15: 2D and 3D representation of F3,α
Assuming now kβ ̸= π, k ≥ 2, then, in order to satisfy the angle folding relation,
one should have kβ + α = π for k ≥ 2, by the Elimination Lemma. In this case, the
angle α has several possible positions to be placed and we shall study every one of
these possibilities.
1.2.1. Assume firstly, that the angle α is placed in the tile numbered 6, as in Figure 16-I (by a symmetry argument it is not necessary to consider the case in which
angle α is placed next to tile 5). In Figure 16-I the angle labelled θ3 is either β or
α.
By the Elimination Lemma, the sum of alternate angles β and γ does not satisfy
the angle folding relation. Therefore, θ3 = β.
Expanding a bit more the configuration in Figure 16-I, we get a vertex surrounded
by alternate angles β and γ, whose sum β + γ + µ, µ ∈ {α, β, γ, δ} leads to the
angle folding relation infringement or it is wiped out by the Elimination Lemma
(Figure 16-II).
1.2.2. If the angle α is placed in the tile labelled 10 (Figure 17), the vertices located
on the border of the configuration in Figure 17 are all of valency four giving rise to
Spherical f-Tilings by Two Non Congruent Classes of Isosceles Triangles - I
a
6
a
.
. .
q3
1
b
b
b
2
a g
a g
5
7
d
d
.
. .
a
I
a g
1
a b b
b
b
5
ag
a
g
a
a
6
3
4
g
g
ba
a g
137
2
a g
a g
d
d
3
4
ag
a
g
II
Figure 16: Local configuration
a vertex surrounded by alternate angles β and γ, which is an impossibility, as we
have seen before.
g
aa
a
a
a
8
a
5
a g
1
9
b b b
b
a b
b
.
..
10
.
2
a g
a g
b
d
d
3
4
..
7
6
a
aa
ag
a
g
Figure 17: Local configuration
2. Assume that α + θ1 < π, with θ1 = γ (Figure 12). Taking into account that
α > γ, α > β and γ > δ, then δ < γ < π2 , α + δ < π, β + γ < π and β + δ < π.
Since vertices of valency four must exist, then α ≥ π2 .
2.1. Assume that α = π2 . Then, β < π2 and vertices of valency four are surrounded
exclusively by angles α. Taking into account that π2 = α > γ >, 2α + β > π and
2γ + δ > π, the sum of the alternate angles α and γ satisfies α + γ + kβ = π, k ≥ 1.
Extending the configuration in Figure 12, a decision must be taken about the angle
θ4 ∈ {α, γ}, Figure 18.
v1
b a
g a
q4
a
3
g a b
b
g
a
b
2
g
a
4
5
d
d
1
a
6
a
.
.
.
b
a
7
a
Figure 18: Local configuration
If θ4 = α, we are led to a vertex surrounded by alternate angles α and β, whose
138
A. M. d’Azevedo Breda and P. S. Ribeiro
sum is of the form α + β + tβ = π, t ≥ 1 or α + γ + kβ = π, k ≥ 1, since other sums
are impossible by using the Elimination Lemma and the fact that α > γ. Assuming
that α + γ + kβ = π, k ≥ 1, the angle arrangement is illustrated in the next figure,
where we reach an incompatibility:
a
a
.
. .
b
. .
.
b
g
?
Figure 19: Angle arrangement
Accordingly, the vertex v1 surrounded by the angles β, α, α satisfies α + tβ =
π, t ≥ 2 and by expanding the local representation in Figure 18, we are led to
another incompatibility of the sides length, as shown in Figure 20-I.
If θ4 = γ, then the configuration extends a bit more, but we are led to a similar
impossibility, Figure 20-II.
a
. .
aa
.
a
10
b
a
11
a
13
b b
..
? g
b
12 b . b b
a
a 8
b a
g a
a 9
5 a aa
d
1
d
2
4
g a b
a
3
a a
b
g
g
a
a
.
b
7
d
g
.
d
b a
g a
7
.
b
a
a
a
9
b
a
.
4
a
.
.
.
b
a
6
a
g
. .
g a b
b
g
a
b
3
?
8
2
d
g
a
5
1
d
g
a
6
a
I
II
Figure 20: Local configuration
2.2. If α > π2 , as δ < γ < π2 , vertices of valency four are surrounded by alternate
angles α and β since α + δ < α + γ < π. Additionally, the sum containing the
alternate angles α and γ is α + γ + mδ = π, m ≥ 1, which is impossible by the sides
length.
Proposition 2. If θ1 = δ (Figure 12), then Ω(T1 , T2 ) = ∅.
Proof. If θ1 = δ, then α + θ1 < π (observe the incompatibility of the sides length if
α + δ = π). The sum containing these two angles α + δ + µ violates the angle folding
relation for any µ ∈ {α, β, γ, δ}, taking into account the Elimination Lemma, α > γ
and 2γ + δ > π.
Spherical f-Tilings by Two Non Congruent Classes of Isosceles Triangles - I
139
2.2. α > β and δ > γ
Using the same terminology as in Figure 12, we get:
Proposition 3. If θ1 = γ, then Ω(T1 , T2 ) is composed by one parameter continuous
family of an f-tiling Eα , π2 < α < π, one discrete family of isolated dihedral f-tilings
Lk , k ≥ 3, and one isolated dihedral f-tiling G such that the sum of alternate angles
around vertices are respectively of the form:
π
for Eα ;
2
π
π
α + 2γ = π, δ =
and β = , k ≥ 3 for Lk ;
2
k
α + γ = π, β = δ =
√
6
π
π
2α + γ = π, α = arccos
, δ=
and β =
for G.
6
2
3
3D representations of a member of Eα , L3 , and G are given in Figures 22, 27 and
28, respectively.
Proof. Assume that θ1 = γ, in Figure 12. Then, α + θ1 ≤ π.
1. If α + θ1 = π, with θ1 = γ, then by adjacency condition (1), one has β = δ and
since δ > γ, then β = δ > π3 . Adding a new cell to the configuration in Figure 12, a
decision must be made about the angle θ5 ∈ {α, β}, as in Figure 21.
ag
1
q5
2
ag
a g
b
b
4
d
d
3
ag
Figure 21: Local configuration
1.1. Suppose firstly that θ5 = α. Then, θ5 + β < π, otherwise β = γ = δ,
which is an absurdity. As α + β < π and α > β, then β < π2 and consequently
γ < δ = β < π2 < α. Therefore, the sum α + β + µ violates the angle folding relation
for any µ ∈ {α, β, γ, δ}.
1.2. If θ5 = β, then β ≤ π2 .
1.2.1. Suppose that β = π2 = δ. Then, π4 < γ < π2 and α > π2 . The configuration expands and we get a global representation of an f-tiling denoted by Eα and
illustrated in Figure 22. It is composed by four isosceles triangles of each.
1.2.2. If β = δ < π2 , then γ < π2 < α.
Taking into account the order relation between the angles and the Elimination
Lemma, the sum 2β + µ violates the angle folding relation, for any µ ∈ {α, β, γ, δ}.
2. If α + θ1 < π, with θ1 = γ, a decision about the angle θ6 ∈ {γ, δ} must be made.
140
A. M. d’Azevedo Breda and P. S. Ribeiro
b
bb b
4
1
a a
g g
a
g
a
g
2
5
a a
g g
3
a
g
a
g
6
d d d
d
7
8
Figure 22: 2D and 3D representation of a member of Eα
v2
a g
2
1
a g
g
b
3
d
d q
6
g
Figure 23: Local configuration
2.1. If θ6 = γ and δ + θ6 = π, the configuration ends up at a vertex (v2 in
Figure 24) surrounded by angles α, γ, δ, whose sum α + δ must be of the form
α + δ + kβ = π, k ≥ 1, since δ + γ = π, δ > π3 and α > π3 . However, it is impossible
to arrange the angles in order to satisfy this sum, due to the Elimination Lemma.
v2
ag
1
b
d
5
2
ag
g
d g
d g
3
g
g
g
4
d
Figure 24: Local configuration
2.2. Suppose that θ6 = γ and δ+θ6 < π. Taking into account that δ > γ, 2γ +δ > π
and α > π3 , the sum δ + γ satisfies δ + γ + kβ = π, k ≥ 1. However, it is impossible
to arrange the angles in order to satisfy the sum δ + γ + kβ = π, k ≥ 1.
2.3. Assume now that θ6 = δ. We shall study the cases δ = π2 and δ < π2 separately.
2.3.1. If δ = π2 , then π4 < γ < π2 . Since α > γ, one of the sums of the alternate
angles at vertex v3 in Figure 25 is α + 2γ = π or 2α + γ = π or α + γ + kβ = π, k ≥ 1
or α + 2γ + pβ = π, p ≥ 1.
2.3.1.1. If α + 2γ = π at vertex v3 , then π4 < γ < π3 . We may expand the configuration in Figure 25 and get the one shown in Figure 26-I. Note that in the construction
141
Spherical f-Tilings by Two Non Congruent Classes of Isosceles Triangles - I
ag
1
b
g
5
2
ag
g
v3
d d
d d
3
g
g
g
4
g
Figure 25: Local configuration
of the configuration, vertices surrounded by a sequence of angles containing the subsequence (γ, γ, γ) must be of valency six, with the sum 2γ + α = π. Additionally,
one of the sums of alternate angles at vertices surrounded by a sequence of angles
containing the subsequence (α, γ, γ) is α + γ + kβ = π, k ≥ 1 or α + 2γ = π.
12
g
d
dd d
ag
d
dd d
6
10
7
b
b
1
ag g
ag g
gg a
gga
3
4
d d
d
d
ag
g
5
b
b
1
9
11
g
a
10
7
g
ga
8
2
v4
6
12
g
ag
b
b
8
ag g
ag g
gg a
gga
3
4
2
.
..
g
ag
ba g
13
a
d
d d
d
5
g
ga
9
11
b
b
a
g
g
a
b
14
ab
v a
5
15
a
I
II
Figure 26: Local configuration
If α + γ + kβ = π, k ≥ 1, at vertex v4 , for instance, we extend in a unique way
the configuration and we are led to another vertex (v5 ) surrounded by angles α, β, α,
whose sum 2α satisfies 2α + tγ = π, t ≥ 1, Figure 26-II. Therefore, from α + 2γ = π,
we conclude that t = 1, which is impossible. Accordingly, every vertices surrounded
by angles α, γ, γ are of valency six, with the sum α + 2γ = π and the configuration
expands in a unique and symmetric way.
Figure 27 shows a global representation of an f-tiling τ ∈ Ω(T1 , T2 ), with three
types of vertices: vertices of valency four surrounded exclusively by angles δ, vertices
of valency six surrounded by alternate angles α, γ, γ and vertices of valency six
surrounded exclusively by angles β. It is denoted by L3 and composed of twelve
isosceles triangles of angles α, α, β and twenty four isosceles triangles of angles γ, γ, δ.
For k ≥ 3, we get a family of discrete dihedral f-tilings denoted by Lk with 4k
isosceles triangles of angles α, α, β and 8k isosceles triangles of angles γ, γ, δ.
2.3.1.2. If 2α + γ = π at vertex v3 , then π4 < γ < β < α < π2 = δ and all the
142
A. M. d’Azevedo Breda and P. S. Ribeiro
d
d
dd
g
g
g
31
g
a a
28
34
gg
d
d
18 d d
g g 15
g
g gg
g
a g
aa
a
27
14
g
13
16
30
a
ga
g
33
29
b bb
b b
b
17
ga
gg
5
a
1
2
ga a
ggg
3
g
26
g
g
a
ga
b 11
b b
b b b
25
23
32
d
d
d d
36
a a
g
g g
g
35
gg
a
ag
9
8
g
g
g g
aa
g
19
g
6 g
g
d
d
d
10
ag
a g g
g
22
g
a
20
7
dd
dd
4
a
d
12
21
dd
d
d
24
Figure 27: 2D and 3D representation of L3
sums containing angles α and γ must satisfy 2α + γ = π, since other possibilities
(α + 2γ = π and α + γ + β = π) imply that α = γ and α = β, which are both
impossible. Expanding the configuration we get a global representation of a tiling
denoted by G ∈ Ω(T1 , T2 ), see Figure 28.
G is composed of forty eight isosceles triangles of angles α, α, β and
twenty four
√
isosceles triangles of angles γ, γ, δ, with β = π3 , δ = π2 , α = arccos 66 ≈ 65.905◦
and γ ≈ 48.19◦ .
2.3.1.3. Assuming that α + γ + kβ = π at vertex v3 , the angle θ7 in Figure 29-I, is
β, since θ7 = α implies that 2α + γ = π and so π4 < γ < β. Consequently, k = 1
and α = β, which is impossible. Extending the configuration in Figure 29-I with
θ7 = β, we end up at a vertex surrounded by alternate angles α, γ, γ, whose sum is
α + 2γ ≤ π (Figure 29-II).
In case α + 2γ = π, we may add some new cells to the configuration above
and the sum of the alternate angles at vertex v6 (see Figure 30) is of the form
3γ + qβ = π, q ≥ 1, which is impossible by the Elimination Lemma. Therefore
α + 2γ < π and it remains the possibility α + 2γ + pβ = π, p ≥ 1, which is also
impossible due to the incompatibility of the edge sides.
2.3.1.4. If α + 2γ + pβ = π, p ≥ 1, at vertex v3 , then α < α + pβ < δ = π2 so
2γ ≤ 2γ + (p − 1)β < α < π2 taking into account that 2γ + δ > π and 2α + β > π.
Therefore, γ < π4 (since 2γ < α < π2 ), contradicting δ = π2 and 2γ + δ > π.
2.3.2. In case δ < π2 , then π4 < γ < π2 and so α ≥ π2 or β ≥ π2 . Either way, α ≥ π2 .
The sum of alternate angles α and γ in Figure 23 is α + γ + kβ = π, k ≥ 1, since
other possibilities (2α + γ = π or 2γ + α = π) contradict the facts α ≥ π2 and α > γ.
Extending the configuration in Figure 23, with θ6 = δ, we end up at a vertex whose
angles arrangement is impossible, see Figure 31-I.
Spherical f-Tilings by Two Non Congruent Classes of Isosceles Triangles - I
d
d
dd
69
70
a
gg
aa
a
46
47
72
67
143
64
aa
71
a gg
63
a
g
g
26
25
28
29
g a
b
a g
b bb
dd
b b
g a
a
a a a 61 b b
dd
a gg
a a
b b 62
b
20 19 g a a 27 b a a
a a a39 30 a a
40a
a
a
66
65
g
g
a
gg
a g 37 ga a
a g 38 g a
9
15
a
a
d
d
16 a g
10
g aa
d d
a
aa
d d
b b
b b
33
35
d
11 b
b 17
d
b 5
4 b
48
a
49
bb
bb
ag
aa
36 g g a 12
18 a g 34
1 a a 6
a
a
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a
a
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gg
aa
60 g ga 31
2
3 g
g
32 a a 59
a 22
a
14 aa
g
aa
ga
bb
aa
b b
dd
b
b 50
b b b
d
d
41
b
13
21
7
54 a
8
aa 53
b
b
a a
b
a
b
a
bb
bb
a
g
a gg a
42
g
43
b b
b b
g
68
aa
g
23
24
b
aa
g g
a
45
b
a
44
55
56
dd
dd
51
52
gg
aa
a
57
a
58
aa
g a
a
g
Figure 28: 2D and 3D representation of G
ba
bb
.
a
.
.
g
ga b
2
d
d d
5 d
4
a
b
9
1
a
gg
ba
a
6
a
b
3
g
ga
7
8
b
q7
9
bb
.
a
.
.
g
ga b
1
a
gg
a
a
2
d
d d
5 d
4
gg
gg
g 11
d
a
6
3
7
8
a
a
ga
b
g a
q7 =b
g
10
a
II
I
Figure 29: Local configuration
If θ6 = γ (Figure 31-II), then the sum containing the alternate angles δ and α
is of the form α + δ + γ = π or α + δ + pβ = π, p ≥ 1. However, the first case
leads to α < γ, which is an absurdity and the second case is impossible using the
Elimination Lemma.
Proposition 4. The symmetry group of:
i) Eα , π2 < α < π, is the dihedral group D4 , whose action on the set of vertices
and faces gives rise to three classes of isogonality and two classes of isohedrality, respectively;
144
A. M. d’Azevedo Breda and P. S. Ribeiro
ba
g
a
14
d
d
15
b a
13 b
9
bb
.
a
.
.
g
ga b
1
g aa
g gg
2
d
d d
5 d
4
a
6
3
7
8
a
a
ga
b
g a
b
q7=
g g
g g
10
g
v6 g g 11 12 a
g
d d
16 d d
g
Figure 30: Local configuration
ba
.
. .
a
a
g
8
ga
b
b 10
b
a
a
6
bb
.
a
.
.
g
ga b
2
d
d
g
a
?
4
a
5
1
7
3
g
4
g
a
b
bb
.
a
.
.
g
ga b
2
d
d
d
ba
6
1
a
g
a
5
a
b
3
d
ga
7
8
b
a
a
g
9 a
a
I
Figure 31: Local configuration
II
ii) Lk , k ≥ 3 is the group D2k × Z2 , partitioning the set of vertices and faces into
three classes of isogonality and two classes of isohedrality, respectively;
iii) G is the group S4 × Z2 , whose action on the set of vertices and faces give rise
to three classes of isogonality and two classes of isohedrality, respectively.
Proof. For each α ∈] π2 , π[, Eα has as a symmetry group the group generated by ρxz ,
the reflection on the coordinate plane xOz and Rzπ , the rotation through π2 around
2
the zz axis, that is the dihedral group of order 8, D4 . It acts on the set of vertices of
Eα giving rise to three classes of isogonality. The two vertices surrounded by β = δ
angles form two of these classes. The remaining four vertices are all in the same
isogonality class. In relation to the set of faces of Eα , D4 acts partitioning it into
two classes. Each class contains four congruent isosceles triangles.
The symmetries of Lk , k ≥ 3, fixing the vertex (0, 0, 1) is the dihedral group
D2k of order 4k generated by ρyz , the reflection on the plane yOz and the rotation
Rzπ . Besides, the reflection on the plane xOy, ρxy , sends (0, 0, 1) to (0, 0, −1).
k
Consequently the group of symmetries of Lk , k ≥ 3, is the group D2k × Z2 . There
are three classes of isogonality determined by the action of this group, each one
containing all vertices of the same type. As in the previous case, there are two
classes of isohedrality, each one containing congruent tiles of the prototiles.
Spherical f-Tilings by Two Non Congruent Classes of Isosceles Triangles - I
145
The f -tiling G has the same symmetries as an octahedron and so S4 × Z2 is its
symmetry group. It follows immediately that the action of S4 ×Z2 on the vertices and
faces of G gives rise to three classes of isogonality and two classes of isohedrality.
Proposition 5. If θ1 = δ (Figure 12), then Ω(T1 , T2 ) = ∅.
Proof. If θ1 = δ, in Figure 12, then α + θ1 < π, since α + θ1 = π implies that the
vertex is of valency four, which is incompatible with the edge sides.
Taking into account that α > π3 and δ > π3 , then the sum α + δ + µ = π, for some µ
is of the form α + δ + γ = π or α + δ + γ + pβ = π, p ≥ 1. Both cases are impossible,
since α > γ and 2γ + δ > π.
2.3. β > α and γ > δ
Suppose now that β > α and γ > δ. Then, one has β > α > γ > δ and also
β > α > γ > π3 .
Proposition 6. If β > α and γ > δ, then Ω(T1 , T2 ) = ∅.
Proof. Assume that β > α and γ > δ.
1. If θ1 = γ in Figure 12, then α + θ1 ≤ π. Suppose first that α + θ1 = π.
Then, δ < γ < π2 < α < β and by the adjacency condition (1), we conclude that
cos β = cos δ, which is an impossibility.
If α + θ1 < π, then in order to satisfy the angle folding relation, the sum α + θ1
is of the form α + γ + kδ = π, for some k ≥ 1, since β > α > γ > π3 . However,
α + γ + kδ > γ + γ + kδ > π, which is an absurdity.
2. If θ1 = δ (Figure 12), then α + θ1 < π due to incompatibility of the edge sides.
As β > α > γ > δ and 2γ + δ > π, then the sum α + θ1 must be of the form
α + kδ = π, for some k ≥ 1. However, by the Elimination Lemma, the sequence of
alternate angles (α, δ, δ, ..., δ) is impossible.
2.4. β > α and δ > γ
Assume finally that β > α and δ > γ.
Proposition 7. If θ1 = γ (Figure 12), then Ω(T1 , T2 ) = ∅.
Proof. Assume that θ1 = γ in Figure 12, then α + θ1 = π or α + θ1 < π. We will
study both cases separately.
1. If α + γ = π, then β + δ > π, α > π2 > γ and by the adjacency condition (1), we
conclude that β = δ > π2 . Adding some new cells to the local configuration, we get
a vertex surrounded by the sequence β, β, α and the sum of the alternate angles β
and α does not satisfy the angle folding relation.
2. Suppose that α + θ1 < π, with θ1 = γ. A decision about the angle θ8 ∈ {γ, δ}
must be made, see Figure 33.
2.1. If θ8 = γ and δ + θ8 = π, then δ > π2 > γ. By the adjacency condition (1), we
get β > π2 and consequently the sum α+δ at vertex v2 satisfies α+δ+kα = π, k ≥ 1.
Since 2γ+δ > π, one has α+δ+kα = π < 2γ+δ and consequently, 2α ≤ α+kα < 2γ,
which is an absurdity.
146
A. M. d’Azevedo Breda and P. S. Ribeiro
b
a
5
a
ag
1
2
ag
ag
b
b
4
d
d
3
ag
Figure 32: Local configuration
v2
ag
1
2
q
8
b
ag
g
d
d
3
g
Figure 33: Local configuration
2.2. If θ8 = γ and δ + θ8 < π, then γ < π2 and in order to satisfy the angle folding
relation, δ + γ + kα = π, k ≥ 1 or δ + γ + β = π or δ + γ + α + β = π. In case
δ + γ + kα = π, k ≥ 1, then α < γ, which is impossible, since α > γ. If δ + γ + β = π
or δ + γ + α + β = π, then γ > β > π3 , contradicting the sums.
2.3. If θ8 = δ, then 2δ ≤ π.
2.3.1. Assume first that δ = π2 . Then, π4 < γ < π2 and, by the adjacency condition
(1), α < π2 and β ̸= π2 . The configuration in Figure 33 expands to the one shown in
Figure 34-I, with θ9 ∈ {α, γ}.
In case θ9 = α, then angle θ10 = β in Figure 34-II, otherwise we would have
π
π
2α + γ = π and so π3 < α < 3π
8 < 2 . Consequently β > 2 , from the adjacency
condition (1), and the configuration lead us to a vertex surrounded by angles α, β, β,
whose sum α + β + µ violates the angle folding relation for any µ ∈ {α, β, γ, δ}.
Therefore, α + γ + β = π, since the sum α + γ + β + µ does not satisfy the angle
folding relation, for any µ ∈ {α, β, γ, δ}.
ag
1
b
g
g
ag
4
2
ag
g
q9
1
d
d
d d
3
g
g
4
2
d
d
d d
ag
g
a
b
q10
3
5
b
g
a g
I
II
Figure 34: Local configuration
The configuration in Figure 34-II now takes the form in Figure 35.
Spherical f-Tilings by Two Non Congruent Classes of Isosceles Triangles - I
a
a
a
147
b
8
7
1
b ba
ag g
a
b
6
a
gg
2
4
3
ag
d
g
g
d d
d
g
5
Figure 35: Local configuration
Note that all the sums of alternate angles α and γ satisfy α + γ + β = π, since the
other possibilities (α + 2γ = π or 2α + γ = π) contradict the angle folding relation
and the fact that α < β. So the above configuration is now the one shown below:
a
a
a
b a
10
8
7
1
b ba
ag g
a
b
a
6
ag
g
ba
a
a
4
d
a
b
11
2
g
d d
d
g
5
a
a
a
b 13
b
12
14
16
a
9
b
a b
gga
3
15 b
a
a
a
b a
a
Figure 36: Local configuration
In the above configuration, the sum containing two alternate angles α is 3α = π
or 2α+γ = π, which is an impossibility, since in the first case we get 3α = π = 2α+β
which is impossible and in the second case, α = β.
2.3.2. If γ < δ < π2 , then the sum containing 2δ must satisfy 2δ + γ = π or
2δ + γ + sα = π, s ≥ 1. In both cases, from 2γ + δ > π, we get δ < γ, which is an
absurdity.
Proposition 8. If θ1 = δ (Figure 12), then Ω(T1 , T2 ) = ∅.
Proof. If θ1 = δ, then α + θ1 < π, since the sum α + θ1 = π is incompatible with
the length of the sides. A decision must be made about the angle θ11 ∈ {γ, δ} in
Figure 37.
1. If θ11 = γ, then γ < π2 . The angle labelled θ12 = α, otherwise we would have
α + δ + γ ≤ π and from 2γ + δ > π, we conclude that α < γ, which is an absurdity.
Since θ12 = α, then 2α + δ ≤ π. But taking into account that α > γ and 2γ + δ > π,
then 2α + δ > π contradicting 2α + δ ≤ π.
2. If θ11 = δ, then the vertex is of valency greater than four, due to the incompatibility of the length of the sides. Accordingly, α + γ < α + δ < π, γ + δ < π, γ < π2
148
A. M. d’Azevedo Breda and P. S. Ribeiro
ag
1
b
2
ag
d
d
g
3
q12 q11
g
Figure 37: Local configuration
and so vertices of valency four are surrounded exclusively by angles α or by angles
β or by angles δ or by alternate angles α and β. Either way, β ≥ π2 or δ = π2 .
2.1. Assuming that β = π2 , then π4 < α < π2 and consequently, the sum α + δ + µ
does not satisfy the angle folding relation, for any µ ∈ {α, β, γ δ}, since α > γ, δ > γ
and 2γ + δ > π.
2.2. If β > π2 , then vertices of valency four are surrounded by alternate angles
α and β or exclusively by angles δ. If α + β = π and taking into account that
α + γ < α + δ < π, one has γ < δ < β. By the order relation between the angles
α > γ and 2γ + δ > π, the sum containing the alternate angles γ and δ does not
satisfy the angle folding relation.
2.3. If δ = π2 , then π4 < γ < π2 and by compatibility of the edge sides and taking
into account that α > γ and 2γ + δ > π, the sum containing alternate angles γ and
δ violates the angle folding relation.
All partial results obtained are summarized in the next theorem:
Theorem 1. If β > α, then Ω(T1 , T2 ) = ∅ and if α > β, then Ω(T1 , T2 ) is composed
of 2-parameter family of dihedral f -tilings denoted by Fk,α , k ≥ 3, α ∈] π2 , 2π
3 [, one
parameter continuous family Eα , π2 < α < π, one discrete family of isolated dihedral
f-tilings Lk , k ≥ 3, and one isolated dihedral f-tiling G such that the sums of the
alternate angles around vertices are of the form:
π
π 2π
, α ∈] ,
[, k ≥ 3 for Fk,α ;
k
2 3
π
α + γ = π and β = δ = for Eα ;
2
π
π
α + 2γ = π, δ = and β = , k ≥ 3 for Lk ;
2
√ k
6
π
π
2α + γ = π; α = arccos
, δ = and β = for G, respectively.
6
2
3
α + γ = π and β = δ =
3D representations of a member of F3,α , Eα , L3 , and G are given in Figures 15,
22, 27 and 28, respectively.
Acknowledgments
The author would like to thank the referees for their helpful suggestions.
Spherical f-Tilings by Two Non Congruent Classes of Isosceles Triangles - I
149
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