Spherical f-Tilings by Scalene Triangles and Isosceles Trapezoids III Catarina P. Avelino ∗ Altino F. Santos † Department of Mathematics UTAD, 5001 - 801 Vila Real, Portugal Submitted: Jun 23, 2009; Accepted: Jul 13, 2009; Published : Jul 24, 2009 Mathematics Subject Classification: 52C20, 52B05, 20B35 Abstract The study of the dihedral f-tilings of the sphere S 2 whose prototiles are a sca- lene triangle and an isosceles trapezoid was initiated in [7, 8]. In this paper we complete this classification presenting the study of all dihedral spherical f-tilings by scalene triangles and isosceles trapezoids in the remaining case of adjacency. A list containing all the f-tilings obtained in this paper is presented in Table 1. It is composed by isolated tilings as well as discrete and continuous families of tilings. The combin atorial structure is also achieved. Keywords: d ihedral f-tilings, combinatorial properties, 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 ar e a spherical isosceles trapezoid, Q, and a spherical triangle, T , we mean a polygonal subdivision τ of S 2 such that each cell (tile) of τ is congruent to Q 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, a nd the sums of alternate angles ar e equal; that is, n  i=1 θ 2i = n  i=1 θ 2i−1 = π, where the angles θ i around any vertex of τ are ordered cyclically. ∗ (cavelino@utad.pt) † (afolgado@utad.pt) Supported partially by the Research Unit CM–UTAD of University of Tr´as-os- Montes e Alto Douro, through the Fo undation for Science and Technology (FCT). the electronic journal of combinatorics 16 (2009), #R87 1 Folding tilings are intrinsically related to the theory of isometric foldings on Rie- mannian manifolds. In fact, the set of singularities of any spherical isometric folding corresponds to a folding tiling of the sphere, see [9] for the foundations of this subject. The study of dihedral f-tilings of the sphere started in 2004 [1, 2, 3], where the clas- sification of all dihedral f-tilings by spherical parallelograms and spherical triangles was obtained. Later on, in [5], the classification of all dihedral f-tilings of the sphere by tri- angles and r-sided regular polygons (r ≥ 5) was achieved. In a subsequent paper [4], is presented the study of all dihedral spherical f-tilings whose prototiles are an equilateral triangle and an isosceles triangle. Robert Dawson and B. Doyle have also b een interested in special classes of spherical tilings, see [10, 11] f or instance. In this paper we shall discuss dihedral f-tilings by spherical scalene triangles, T , and spherical isosceles trapezoids, Q, with a certain adjacency pattern. We present in Table 1 a list containing all the f-tilings obtained in this paper. We shall denote by Ω (Q, T ) the set, up to an isomorphism, of all dihedral f-tilings of S 2 whose prototiles are Q and T . From now on Q is a spherical isosceles trapezoid of internal angles α 1 and α 2 (α 1 > α 2 ) and edge lengths a, b and c (b > c), and T is a spherical scalene triangle of internal angles β, γ and δ (β > γ > δ), with edge lengths d (opposite to β), e (opposite to γ) and f (opposite to δ), see Figure 1. T g d b d f e a 2 a 2 a 1 a 1 a a b c Q Figure 1: Prototiles: a spherical isosceles trapezoid and a spherical scalene triangle It follows immediately that β + γ + δ > π and α 1 + α 2 > π, with α 1 > π 2 . In o rder to get any dihedral f -tiling τ ∈ Ω (Q, T ), we find useful to start by considering one of its local configurations, beginning with a common vertex to two tiles of τ in adjacent positions. In the diagrams that follows it is convenient to label the tiles according to the following procedures: (i) We begin the configuration of a tiling τ ∈ Ω (Q, T ) with an isosceles trapezoid, labelled by 1; then we label with 1 ′ an isosceles trapezoid or a scalene triangle adjacent to tile 1 and sharing the side of length c; (ii) For j ≥ 2, the location of tile j can be deduced from the configuration of tiles (1, 1 ′ , 2, 3, . . . , j − 1) and from the hypothesis that the configuration is part of a complete f-t iling (except in the cases indicated). the electronic journal of combinatorics 16 (2009), #R87 2 2 Dihedral Spherical f-Tilings by Scalene Triangles and Isosceles Trapezoids Any element of Ω (Q, T ) has at least two cells such that they are in adjacent positions and in one of the situations illustrated in Figure 2. The cases of adjacency I − IV were Q T I b a 1 a 1 a 2 a 2 Q a 1 a 1 a 2 a 2 Q a 1 a 1 a 2 a 2 II IV Q a 1 a 1 a 2 a 2 Q a 1 a 1 a 2 a 2 V g Q T b a 1 a 1 a 2 a 2 g III Q T b a 1 a 1 a 2 a 2 g d d d Figure 2: Distinct cases of adjacency analyzed in [7, 8 ]. In this paper we complete the study of all dihedral spherical f-tilings by scalene triangles and isosceles trapezoids through the analysis of the case of adjacency V . In the following results we will use the fact that we cannot have a spherical trapezoid adjacent to a spherical triangle and sharing the side of length c, and also that we cannot have two trapezoids sharing the sides of length a and c, respectively. These situations have already been studied before ([7, 8]) and do not give rise to any f-tiling when there are two trapezoids sharing the side of length c. Consequently, a trapezoid must always have an adjacent trapezoid sharing the side of length c; and, on the other hand, we cannot have two trapezoids sharing the sides of length a and b (nor b and c). Suppose that any element of Ω (Q, T ) has at least two cells such that t hey are in adjacent positions as illustrated in Figure 3. With the labelling of this figure, we have necessarily θ 1 = β, θ 1 = δ or θ 1 = γ. These three distinct cases will be now analyzed separately in the following propositions. Proposition 2.1 With the above terminology, if θ 1 = β, then Ω (Q, T ) = ∅. Proof. Suppose that we have two cells in adjacent positions as illustrated in Figure 4. the electronic journal of combinatorics 16 (2009), #R87 3 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 q 1 Figure 3: Local configuration 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 q 2 q 1 b Figure 4: Local configuration As α 1 > α 2 > β > γ > δ and β + γ + δ > π, we have necessarily α 1 + β = π. With the labelling used in Figure 4, we have θ 2 = γ or θ 2 = δ. If θ 2 = γ (Figure 5(a)), then tile 3 comes in a unique way and we must have θ 3 = δ. In fact, since v has valency greater than four (δ + ρ < π, ∀ρ ∈ {α 1 , α 2 , β, γ, δ}), we have α 2 + α 2 + ρ > α 2 + γ + ρ > β + γ + δ > π, ∀ρ ∈ {α 1 , α 2 , β, γ, δ}. 2 3 b 1 1’ q a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 3 g g d q 2 d q 1 b v (a) 2 3 b 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g d d q 3 d g b 4 g b d 5 q 4 q 5 g q 2 q 1 b d (b) Figure 5: Local configurations Therefore, θ 3 = δ and also α 2 + kδ = π, for some k ≥ 2, and we get the local configuration illustrated in Figure 5(b). Now, we have θ 4 = α 2 and θ 5 = α 1 . Note that θ 4 cannot be δ, since θ 4 = δ implies θ 5 = β and β + β < π < β + β + ρ, ∀ρ ∈ {α 1 , α 2 , β, γ, δ}. Ta king into account the angles and edge lengths, θ 6 = β or θ 6 = γ (Figure 6(a)). But α 2 + θ 6 < π < α 2 + θ 6 + ρ, ∀ρ ∈ {α 1 , α 2 , β, γ, δ}, which is a contradiction. If θ 2 = δ (Figure 4), then we obtain the configuration illustrated in Figure 6(b). Note that, using similar argumentation to the one used before, we must have θ 3 = δ. Now, taking into account the relation between angles and edge lengths, θ 4 must be δ, a nd so we get α 2 + k δ = π = γ + kδ, for some k ≥ 2, which is a contra diction (note that α 2 > γ).  Proposition 2.2 If θ 1 = δ and α 1 + δ < π, then Ω (Q, T ) is composed by a single tiling, denoted by N , where α 1 = 3π 5 , α 2 = β = π 2 , γ = π 3 and δ = π 5 . The angles around vertices are po sitioned as illustrated in Figure 7. For a planar representation see Figure 15. Its 3D representation is given in Figure 16. the electronic journal of combinatorics 16 (2009), #R87 4 2 3 b 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g d d q 3 d g b 4 g b d 5 d q 4 a 2 a 1 a 2 6 7 q 6 d g q 2 q 1 b q 5 a 1 (a) 2 3 b 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g g d q 3 d g b 4 d 5 q 4 b g d q 2 q 1 b (b) Figure 6: Local configurations d d b b b dd d d b b b a 2 a 2 d d a 1 a 1 d d g g g g g g d d d d Figure 7: Distinct classes of congruent vertices Proof. Suppose that we have two cells in adjacent positions as illustrated in Figure 3 and consider θ 1 = δ, with α 1 + δ < π. With the labelling of Fig ure 8(a), we have θ 2 = γ or θ 2 = β. 1. Suppose firstly that θ 2 = γ. We consider separately the cases α 1 < β and α 1 ≥ β. 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 q 2 q 1 d (a) 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 q 1 d q 2 g b g b d 3 v 1 (b) Figure 8: Local configurations 1.1 If α 1 < β, we have β + γ = π or β + kδ = π, for some k ≥ 1 . As we can observe in Figure 8(b), the case β + γ = π leads t o a contradiction since there is no way to avoid an incompatibility at vertex v 1 . If β + δ = π (Figure 9 ( a)), we also reach a contradiction at vertex v 2 . Note that θ 3 must be β, otherwise there is no way to satisfy the angle-folding relation around vertex v 1 . Suppose finally that β +kδ = π, for some k ≥ 2. At vertex v 1 (Figure 9(b)), we obtain β + kδ = π = α 2 +  k−1 i=1 ρ i + β, with ρ i ∈ {α 2 , δ}, i = 1, 2, . . . , k − 1 (see edge lengths), which is a contradiction (note that we have considered b = d, see Figure 1, as the case b = e lies in the previous one). the electronic journal of combinatorics 16 (2009), #R87 5 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 q 1 d q 2 g b g b d 3 4 d g b v 1 v 2 q 3 (a) 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 q 1 d q 2 g b d d v 1 (b) Figure 9: Local configurations 1.2 If α 1 ≥ β, then α 1 + kδ = π, for some k ≥ 2, and we obtain the local configuration given in Figure 10(a). Now, if tile 10 is a triangle, then θ 3 must be β, γ or δ. In all cases we reach a contradiction at vertex v, see Figure 10(b), Figure 11(a) and Figure 11(b), respectively. 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 4 5 q 3 6 g g g g 7 8 9 g g d d d d d d d b b b b b b b b q 1 d q 2 g 10 (a) 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 4 5 6 g g g g 7 8 9 g g d d d d d d d b b b b b b b b q 1 d q 2 g 10 g d v q 3 b (b) Figure 10: Local configurations 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 4 5 6 g g g g 7 8 9 g g d d d d d d d b b b b b b b b q 1 d q 2 g 10 b d v q 3 g (a) 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 4 5 6 g g g g 7 8 9 g g d d d d d d d b b b b b b b b q 1 d q 2 g 10 b g v q 3 d (b) Figure 11: Local configurations Therefore, tile 10 must be a trapezoid and, using analogous arguments, we get the local configuration illustrated in Figure 12(a). Now, at vertex v 1 , we have α 2 + β < π or α 2 + β = π. 1.2.1 Suppose firstly that α 2 + β < π. Therefore α 1 > β > γ > α 2 > δ. If v 1 has valency six, then α 2 + β + α 2 = π (see edge lengths) and we reach a contradiction at vertex v 2 in Figure 12(b). On the other hand, if v 1 has valency greater than six, we the electronic journal of combinatorics 16 (2009), #R87 6 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 4 5 6 g g g g 7 8 9 g g d d d d d d d b b b b b b b b q 1 d q 2 g a 1 a 1 a 2 a 2 10 a 1 a 1 a 2 a 2 11 12 a 1 a 1 a 2 a 2 d d d d v 1 (a) 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 4 5 6 g g g g 7 8 9 g g d d d d d d d b b b b b b b b q 1 d q 2 g a 1 a 1 a 2 a 2 10 a 1 a 1 a 2 a 2 11 12 a 1 a 1 a 2 a 2 16 g 15 14 13 g g d d d b b b b g d a 1 17 18 a 1 a 2 a 2 a 1 a 1 a 2 a 2 a 1 a 1 v 2 (b) Figure 12: Local configurations reach a contradiction at vertex v 3 in Figure 13(a), since there is no way to satisfy the angle-folding relation around this vertex. 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 4 5 6 g g g g 7 8 9 g g d d d d d d d b b b b b b b b q 1 d q 2 g a 1 a 1 a 2 a 2 10 a 1 a 1 a 2 a 2 11 12 a 1 a 1 a 2 a 2 16 g 15 14 13 g g d d d b b b b g d 17 18 a 1 a 2 a 2 a 1 a 1 a 2 a 2 a 1 d a 1 a 1 d d d v 3 (a) 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 4 5 6 g g g g 7 8 9 g g d d d d d d d b b b b b b b b q 1 d q 2 g a 1 a 1 a 2 a 2 10 a 1 a 1 a 2 a 2 11 12 a 1 a 1 a 2 a 2 13 g 14 15 16 g g d d d b b b b g d x (b) Figure 13: Local configurations 1.2.2 Suppose now that α 2 + β = π. Then, the configuration illustrated in Figure 12(a) is extended in a unique way to the one given in Figure 13(b). With the labelling of this figure, we have x = α 2 or x = β. If x = α 2 , we obtain the configuration given in Figure 14. At vertex v 4 , one of the alternating angle sums must contain the sequence (. . . , β, γ, γ, . . .). But β + γ + γ > π, which is an impossibility. If x = β, then α 2 = β = π 2 and consequently γ + δ > π 2 . It follows that a vertex surrounded by six angles γ takes place, i.e., γ = π 3 . And so δ > π 6 , since β + γ + δ > π. The construction of the planar configuration is unique (Figure 15) and leads to a vertex surrounded by only angles δ. Therefore δ = π 4 or δ = π 5 . However, if δ = π 4 , then the electronic journal of combinatorics 16 (2009), #R87 7 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 4 5 6 g g g g 7 8 9 g g d d d d d d d b b b b b b b b q 1 d q 2 g a 1 a 1 a 2 a 2 10 a 1 a 1 a 2 a 2 11 12 a 1 a 1 a 2 a 2 13 g 14 15 16 g g d d d b b b b g d 17 a 1 a 2 a 2 a 1 a 1 a 1 a 2 a 2 18 d d d d v 4 Figure 14: Local configurations π = α 1 + kδ ≥ α 1 + 2δ > α 1 + π 2 > π, which is an impossibility. And so δ = π 5 and also k = 2, as indicated in Figure 15. We also have α 1 = 3π 5 . We shall denote such f -tiling by N . Its 3 D representation is shown in Figure 16. 31 29 28 a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 32 33 34 g g d d d b b b b 13 12 11 a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 14 15 16 g g d d d b b b b g d d 49 g 50 51 52 g g d d d b b b b 45 g 46 47 48 g g d d d b b b b g d d g 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 4 5 g g d d d b b b b q 1 d q 2 g 23 10 21 a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 24 25 22 g g d d d b b b b g d 6 g 7 8 9 g g d d d b b b b 43 g 42 41 40 g g d d d b b b b g d d g 37 30 a 1 a 1 a 2 a 2 38 g d b b g d 44 62 g d b b g d g d g g g b b b b g g b b g g b b g g b d d d d d d d d 36 35 20 18 17 19 26 27 39 d g g g b b b b g g b b g g b b g g b d d d d d d d d 61 59 57 56 54 53 55 58 60 d d g g b b b b 63 64 d d g g 66 65 b g d a 1 a 1 d a 2 a 2 b g 67 68 Figure 15: Planar representation of N 2. If θ 2 = β (Figure 8( a)), then we obtain the local configuration illustrated in Figure 17(a). Note that v 1 must be enclosed exclusively by angles γ and δ. Now, we have θ 3 = γ, θ 3 = α 2 or θ 3 = β. 2.1 Suppose firstly that θ 3 = γ. Then β + γ = π. Now, if there is at least one angle γ the electronic journal of combinatorics 16 (2009), #R87 8 Figure 16: 3D representation of N 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 3 g d b b q 1 d q 2 g q 3 v 1 4 (a) 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 5 6 7 g g g g 8 9 10 g g d d d d d d d b b b b b b b b q 1 d q 2 g d b 4 q 4 v 2 g q 3 (b) Figure 17: Local configurations in the alternating angle sum containing α 1 (see vertex v 1 , Figure 17(a)), we get β > α 1 , and so β + α 2 > π, which is not po ssible (see tile 4). Therefore α 1 + kδ = π, for some k ≥ 2, and we get the local configuration illustrated in Figure 17(b). We also obtain the following relation between angles: α 1 > β > π 2 > γ ≥ α 2 > δ. Now, θ 4 must be α 2 or γ. 2.1.1 If θ 4 = α 2 (and consequently γ = α 2 ) a nd the vertex v 2 has valency six, we get the local configuration illustrated in Figure 18(a) and consequently a contradiction at vertex v 3 . If vertex v 2 has valency eight, we reach a contradiction as we can observe at Figure 18(b) (vertex v 4 ). Using the same kind of reasoning when v 2 has valency greater than eight, we also obtain a contradiction. 2.1.2 If θ 4 = γ, we obtain the configuration illustrated in Figure 19(a) and now we have θ 5 = α 2 or θ 5 = γ. 2.1.2.1 If θ 5 = α 2 (therefore α 2 = γ), the last configuration is extended in a unique way to the one illustrated in Figure 19(b). In this configuration we have obtained two vertices surrounded by the cyclic sequence of angles (α 2 , γ, α 2 , γ, α 2 , γ, . . .), with α 2 = γ = π k , for some k ≥ 3. We have considered k = 3 for convenience. Although a complete planar representation was possible to draw, we may conclude that such a configuration cannot be realized by an f-tiling since there is no spherical trapezoid satisfying the relations that come from Figure 19(b). In fact, we have Q = T ∪ T ′ (Figure 20) and then β + x = α 2 , the electronic journal of combinatorics 16 (2009), #R87 9 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 5 6 7 g g g g 8 9 10 g g d d d d d d d b b b b b b b b q 1 d q 2 g d b 4 b d g 13 q 4 a 2 14 g d a 2 a 1 a 1 11 a 1 a 1 a 2 a 2 12 b d g 15 b d d d d g b b b b g g g 19 18 17 16 a 2 a 2 a 2 a 2 a 1 a 1 a 1 a 1 20 21 v 3 g q 3 a 1 (a) 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 5 6 7 g g g g 8 9 10 g g d d d d d d d b b b b b b b b q 1 d q 2 g d b 4 b d g 13 q 4 a 2 14 g d a 2 a 1 a 1 11 a 1 a 1 a 2 a 2 12 b d g 15 b d d d d g b b b b g g g 19 18 17 16 a 2 a 2 a 2 a 2 a 1 a 1 a 1 a 1 20 21 v 23 22 g g b b d d b d g 24 4 g q 3 (b) Figure 18: Local configurations 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 5 6 7 g g g g 8 9 10 g g d d d d d d d b b b b b b b b q 1 d q 2 g d 4 q 4 b d g b 11 g b d 12 q 5 g q 3 (a) 2 1 1’ a 1 a 1 a 2 a 2 a 1 a 1 a 2 a 2 g 3 5 6 7 g g g g 8 9 10 g g d d d d d d d b b b b b b b b q 1 d q 2 g d 4 q 4 b d g b 11 g b d 12 q 5 a 2 a 2 a 1 a 1 13 d g 14 b 15 a 2 a 2 a 1 a 1 g b d a 2 a 2 a 1 a 1 17 16 g b d 18 19 d b g g d b 20 b 21 g d d g b 22 d b g 23 a 2 d g b 24 d b g 25 26 a 2 a 1 a 1 b d g g 27 g 28 b d b d g q 3 (b) Figure 19: Local configurations for some x > 0, which is a contradiction, as β > α 2 . 2.1.2.2 If θ 5 = γ, then the last configuration is extended to the one illustrated in Fig- ure 21(a). Using similar arguments to the ones used before in 2.1.2.1, we conclude that the case θ 6 = α 2 (Figure 21(b)) leads to a contradiction (by using an argument of symmetry). On the other hand, the case θ 6 = γ leads the the configuration illustrated in Figure 22(a) (by using arguments of symmetry). Now, if tile 18 is a triangle, then it must be set up the electronic journal of combinatorics 16 (2009), #R87 10 [...]... given in Figure 72(b) and 3D repreβγ k sentations, for k = 3, k = 4 and k = 5, are given in Figure 73 The case (iii) leads to a family of f-tilings, M3 , with π < δ < 1 arccos − 1 Planar and δ 6 2 4 3D representations are given in Figure 81(b) and Figure 82, respectively In the last situation, we obtain a sporadic tiling, denoted by G Figure 83 and Figure 84 illustrate planar and 3D representations... α2 + β = π and δ = π , k ≥ 3, k (iii) β = π , γ = 2 π 3 or and α2 + 2δ = π, (iv) β = π , γ = π , δ = 2 3 π 5 and α2 = or 2π 5 In the first case, for each k ≥ 2, we obtain a single tiling, denoted by T k , with γ = sin π arccos 22k A planar representation of T k is given in Figure 66 and 3D representations, for k = 2 and k = 3, are given in Figure 67 The case (ii) leads to a family of f-tilings, Rk... α1 = 3π 5 and δ = π But then α1 + α2 = 5 3π 5 + π 3 = 14π 15 < π, which is a contradiction Proposition 2.3 If θ1 = δ and α1 + δ = π (Figure 3), then Ω(Q, T ) = ∅ iff (i) α2 + γ = π and β = (ii) α2 + 2γ = π, β = π 2 π 2 or and kδ = π, for some k ≥ 4, or (iii) α2 + β = π and kγ = π, for some k ≥ 3 the electronic journal of combinatorics 16 (2009), #R87 15 The first case leads to a family of f-tilings, ... Local configuration a1 a d a2 a1 T ’’ T’ b d e d a d T a2 g b b f f Figure 34: Q = T ′ ∪ T ′′ In Figure 34, the spherical trapezoid is divided in two triangles, T ′ and T ′′ Since T and T ′ have two sides in common and the angle formed by these sides is equal (α2 = γ), we conclude that T and T ′ are congruent Now, as Q = T ′ ∪ T ′′ , we have α2 > β, which is an incongruity Thus, we conclude that the... = π, i.e., γ = π , and so α1 = 2π , 3 3 α2 = β = π , γ = π and δ = π We get the configuration illustrated in Figure 69(b), 2 3 4 and consequently the decomposition in Figure 70 As T and T ′ have one side in common p 3 p p 3 T’ T 2 2p 3 p p 4 4 Figure 70: Q = T ∪ T ′ and the adjacent angles are equal, we conclude that T and T ′ are congruent, which is an impossibility On the other hand, if θ5 = α2 ,... vertices v1 and v2 in Figure 66 are in antipodal positions since there exist two distinct geodesics of the same length joining them, and so b = e = π We have α1 = π −γ, 3 π α2 = β = π and δ = 2k , for some k ≥ 2 Using these relations, it is a straightforward 2 sin π exercise to show that γ = arccos 22k We shall denote such family of dihedral f-tilings by T k , k ≥ 2 3D representations of T 2 and T 3... 52: Planar representation of Rk δβ Figure 52 As α1 + α2 > π and by an area argument, we may conclude that β + δ ∈ (k−1)π π , k + arccos − cos2 π (β = arccos − cos2 π , otherwise T would be isosceles) k k k The corresponding 3D representations for 3 ≤ k ≤ 5 are illustrated in Figure 53 We denote such family of f-tilings by Rk , with k ≥ 3 and β+δ ∈ (k−1)π , π + arccos − cos2 π , δβ k k k δ < π < β k... by R2 , with δ, γ ∈ 0, π A planar δγ 2 representation is given in Figure 38(a) For its 3D representation see Figure 38(b) The case (ii) leads to a family of f-tilings, denoted by Mk , with γ π (k − 2)π 1 , arccos − cos2 2k 2 k γ∈ In Figure 41 is given the corresponding planar representation 3D representations for k = 4 and k = 5 are given in Figure 42 In the last situation, there is a family of f-tilings, ... β = π, i.e., β = π , and hence γ + δ > π Now, we 2 2 consider separately the cases α2 + γ = π and α2 + γ < π (i) If α2 + γ = π, then α1 > α2 > β = π > γ > δ and the refereed local configuration 2 is extended in a unique way to the one given in Figure 38(a), where δ, γ ∈ 0, π The 2 corresponding 3D representation is illustrated in Figure 38(b) We denote such family of f-tilings by R2 , with δ, γ ∈ 0,... π and kδ = π, for some k ≥ 4 The f-tiling with such a planar representation will be denoted by Mk , k ≥ γ g g a2 4 b 5 18 g g a2 g a2 g 16 17 2 a1 q1 d q2 b b a1 d b b d a1 d a1 b 19 b b dd dd a2 g g 1’ b b b b 15 1 a2 g q3 g a2 q6 g q4 g 12 d a1 d a1 14 7 3 a1 d a1 d 13 g g a2 8 6 d d d d 9 q5 b b b b11 10 a2 g g Figure 41: Planar representation of Mk γ 4 Using spherical trigonometry formulas and by . discuss dihedral f-tilings by spherical scalene triangles, T , and spherical isosceles trapezoids, Q, with a certain adjacency pattern. We present in Table 1 a list containing all the f-tilings obtained. presenting the study of all dihedral spherical f-tilings by scalene triangles and isosceles trapezoids in the remaining case of adjacency. A list containing all the f-tilings obtained in this paper. all dihedral f-tilings by spherical parallelograms and spherical triangles was obtained. Later on, in [5], the classification of all dihedral f-tilings of the sphere by tri- angles and r-sided