ARS MATHEMATICA CONTEMPORANEA Also available at http://amc-journal.eu ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 11 (2016) 59-78 Spherical folding tessellations by kites and isosceles triangles IV Catarina P. Avelino , Altino F. Santos Pólo CMAT-UTAD, Centro de Matemática da Universidade do Minho Universidade de Tras-os-Montes e Alto Douro, UTAD, Quinta de Prados, 5000-801 Vila Real, Portugal Received 13 July 2014, accepted 28 September 2014, published online 18 September 2015 The classification of the dihedral folding tessellations of the sphere and the plane whose prototiles are a kite and an equilateral triangle were obtained in [1]. Recently, this classification was extended to isosceles triangles so that the classification of spherical folding tesselations by kites and isosceles triangles in three cases of adjacency was presented in [2, 3, 4]. In this paper we finalize this classification presenting all the dihedral folding tessellations of the sphere by kites and isosceles triangles in the remaining three cases of adjacency, that consists of five sporadic isolated tilings. A list containing these tilings including its combinatorial structure is presented in Table 1. Keywords: Dihedral f-tilings, combinatorial properties, spherical trigonometry, symmetry groups. Math. Subj. Class.: 52C20, 05B45, 52B05 1 Introduction By a folding tessellation or folding tiling of the Euclidean sphere S2 we mean an edge-to-edge pattern of spherical geodesic polygons that fills the whole sphere with no gaps and no overlaps, and such that the "underlying graph" has even valency at any vertex and the sums of alternate angles around each vertex are n. Folding tilings (f-tiling, for short) are strongly related to the theory of isometric foldings on Riemannian manifolds. In fact, the set of singularities of any isometric folding corresponds to a folding tiling, see [13] for the foundations of this subject. The study of this special class of tessellations was initiated in [5] with a complete classification of all spherical monohedral folding tilings. Ten years latter Ueno and Agaoka E-mail addresses: cavelino@utad.pt (Catarina P. Avelino), afolgado@utad.pt (Altino F. Santos) Abstract ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ 60 Ars Math. Contemp. 11 (2016) 35-47 [14] have established the complete classification of all triangular spherical monohedral tilings (without any restriction on angles). Dawson has also been interested in special classes of spherical tilings, see [10], [11] and [12], for instance. The complete classification of all spherical folding tilings by rhombi and triangles was obtained in 2005 [8]. A detailed study of the triangular spherical folding tilings by equilateral and isosceles triangles is presented in [9]. Spherical f-filings by two noncongruent classes of isosceles triangles in a particular case of adjacency were recently obtained [6]. Here we discuss dihedral folding tessellations by spherical kites and isosceles spherical triangles. A spherical kite K (Figure 1(a)) is a spherical quadrangle with two congruent pairs of adjacent sides, but distinct from each other. Let us denote by (a1,a2,a1, a3), a2 > a3, the internal angles of K in cyclic order. The length sides are denoted by a and b, with a < b. From now on T denotes a spherical isosceles triangle with internal angles ft and y (Y = ft), and length sides c and d, see Figure 1(b). We shall denote by Q(K, T) the set, up to isomorphism, of all dihedral folding tilings of S2 whose prototiles are K and T. (a) (b) d Figure 1: A spherical kite and a spherical isosceles triangle Taking into account the area of the prototiles K and T we have 2«i + a2 + a3 > 2n and ft + 27 > n. As a2 > a3 we also have ai + a2 > n. We begin by pointing out that any element of Q (K, T) has at least two cells congruent to K and T, respectively, such that they are in adjacent positions and in one and only one of the situations illustrated in Figure 2. After certain initial assumptions are made, it is usually possible to deduce sequentially the nature and orientation of most of the other tiles. Eventually, either a complete tiling or an impossible configuration proving that the hypothetical tiling fails to exist is reached. In the diagrams that follow, the order in which these deductions can be made is indicated by the numbering of the tiles. For j > 2, the location of tiling j can be deduced directly from the configurations of tiles (1, 2,... ,j - 1) and from the hypothesis that the configuration is part of a complete tiling, except where otherwise indicated. C. P. Avelino and A. F. Santos: Spherical folding tessellations by kites and isosceles triangles IV 61 The cases of adjacency I, II and III have already been analyzed in [2, 3, 4]. In this paper we consider the remaining cases of adjacency IV, V and VI. 2 Case of Adjacency IV Suppose that any element of Q (K, T) has at least two cells congruent, respectively, to K and T, such that they are in adjacent positions as illustrated in Figure 2-IV. As b = d, using trigonometric formulas, we obtain cos y(1 + cos 8) cos a + cos ai cos 03 -^-:—ô— =--■-—a—-. (2.1) sin y sin p sin ai sin Concerning the angles of the triangle T we have necessarily one of the following situations: Y > 8 or y < 8. In the following subsections we consider separately each one of these cases. 2.1 Y > fl In this case we have y > f and a,c a2 > a3. In fact, if a2 > a1 and (a) (b) Figure 3: Local configurations (i) a1 + y = n (Figure 3(b)), we reach a contradiction at vertex v2, as a2 + p > n, for all p G {ai, «2}; (ii) a1 + y < n (Figure 4(a)), at vertex v1 we have necessarily a2 + y + ka3 = n, k > 1. But (a1 + a1 + y) + (a2 + y + a3) > (2a1 + a2 + a3) > 2n, which is impossible. At vertex v1 we have ai + y = n or ai + 7 < n. 1. Suppose firstly that a1 + 7 = n (Figure 4(b)). At vertex v3 we have ka2 = n, with k > 2. As (a1 + y) + (a2 + a2 + a2) > (a1 + a2) + (y + ft + ft) > 2n, it follows that k = 2 (a2 = 2). With the labeling of Figure 4(b), if (i) = a3 (Figure 5(a)), then at vertex v3 we have necessarily a1 + kft = n, k > 2, and so a1 > n = a2 > y > a3 > ft (note that a1 + ft + a3 > n). But then tile 9 must be a triangle, which is impossible; (ii) = ft (Figure 5(b)), then at vertex v2 it follows that a1 + kft = n, k > 2 (note that we must have a3 > ft). But at vertex v3 we have y + Y < n and y + Y + p > n, for all p. (iii) = y, we get the configuration illustrated in Figure 6(a). Now, if C. P. Avelino and A. F. Santos: Spherical folding tessellations by kites and isosceles triangles IV 63 «3 «1 «1 «2 5 «3 «1 «2 «1 1 «1 «3 V2 e «3 «1 7 «1 «2 «2 y p (a) case 1(i) «1 «1 «1 «2 «2 «1 «2 5 «3 «1 «2 «1 «2 «1 «3 9 8 "TO -Y "TO 61 P Y P 11 10 Y YY 12 P (b) case 1 (ii) 4 6 3 2 Figure 5: Local configurations Figure 6: Local configurations (a) 02 = «1 (Figure 6(b)), we have necessarily « + ka3 = n, with k > 2, and ai > f- = a2 > 7 > fi > «3 (a1 + fi + > n). But then, the other sum of alternate angles at vertex v3 must be greater or equal than a1 + fi + fi > n, which is a contradiction (3n > (a1 + 7) + (a2 + a2) + (a1 + fi + fi) > (2«i + «2 + «3) + (fi + 7 + 7) > 3n). (b) 02 = fi (Figure 7(a)), at vertex v3 we have 7 + 7 + ka3 = n, k > 1, and a contradiction arises at vertex v4 as a1 + p > n, for all p G {a1, a2}. (c) 02 = 7, at vertex v3 we have 03 G {fi, a3}. In the first case, illustrated in Figure 7(b), we reach a contradiction at vertex v5. On the other hand, if 03 = a3, due to the angles involved in the sums of alternate angles at vertex v3, we must have a3 = fi. Taking into account the triangle and the kite's areas, it follows that 7 + fi + fi = 7 + a3 + a3 = n (Figure 8). At vertex v6 we have a1 + fi < n and a1 + fi + p > n, for all p G {a1, a2, a3, fi, 7}. (d) 02 = a3, taking into account the analysis of the previous cases, at vertex v3 we have ka3 = kfi = n, k > 3. Due to the kite's area, it follows that 7 -4 < f and consequently cos f < cos (7 - 4). Using equation (2.1), we conclude that fi > f, and so k = 3. The last configuration is then extended to the one illustrated in Figure 9(a). We shall denote this f-tiling by L. Its 3D 64 Ars Math. Contemp. 11 (2016) 35-47 Figure 7: Local configurations representation is given in Figure 9(b). 2. Suppose now that a1 + 7 < n (Figure 3(a)). Again, due to the analysis made in [4] (case 2.1), we use the fact that a side of length c of each triangle must be adjacent to a side of length c of other triangle. At vertex v1 we must have a1 + 7 + ka3 = n, with k > 1. Nevertheless, we reach a contradiction at vertex v2 (Figure 10) since there is no way to satisfy the angle-folding relation around this vertex. □ 2.2 7 < fl In this case we have P > f and a < b = d < c. Proposition 2.2. Under the conditions assumed in this section, there is a single folding tiling, J, such that a2 = |, a1 + 7 = n, 7 = 3 and P + P + a3 = n. Planar and 3D representations of J are given in Figure 12. C. P. Avelino and A. F. Santos: Spherical folding tessellations by kites and isosceles triangles IV 65 r/x ^ \ \ t\ 1 V\ // (a) planar representation (b) 3D representation Figure 9: f-tiling L Figure 10: Local configuration occurring in case 2 Proof. Suppose that any element of Q(K, T) has at least two cells congruent, respectively, to K and T, such that they are in adjacent positions as illustrated in Figure 2—IV. As a = c, we get the configuration illustrated in Figure 11(a), and, at vertex v\, we have a, a. 1 y 1 +2 y P a a3 (a) (b) case 1 Figure 11: Local configurations 66 Ars Math. Contemp. 11 (2016) 35-47 ai + y = n or ai + 7 < n. 1. Suppose firstly that ai + 7 = n (Figure 11(b)). Note that the conditions a2 > ai > a3 and a2 > a3 > ai lead to a contradiction at vertex v2, as a2 + p > n, for all p G {ai, a2}. Therefore ai > a2 > a3. Now, if (i) a2 + a2 = n, then P + P + ka3 = n, k > 1, and so ai > a2 = ^ > P > 7 > a3. Consequently, 7 = f (as P < 2, we have 7 > 4 ). Then, the last configuration is extended to the one illustrated in Figure 12(a). We shall denote this f-tiling by J. Its 3D representation is given in Figure 12(b). (a) planar representation (b) 3D representation 70 30 71 45 61 Figure 12: f-tiling J (ii) a2 + a2 < n, then ka2 = n, k > 3, P + P + ka3 = n, k > 1, and so « > ^ > P > a2 > 7 > a3. As 7 > 4, we have necessarily k = 3 (Figure 13(a)). Now, if at vertex v2 we have k > 1 (Figure 13(b)), one of the angles 02, 03 or must be a3. But then we reach a contradiction at vertex v3, v4 or v5, respectively, as a1 + p > n, for all p € { a 1, a2}. On the other hand, if k = 1, we get the configuration illustrated in Figure 14(a) (note that at vertex v3 we cannot have 7+7+7 = n, as 3 = a2 > 7). At vertex v4 we reach a similar contradiction as in the case k > 1. 2. Suppose now that a1 + 7 < n (Figure 11(a)). If a2 > a1 > a3 or a2 > a3 > a1, at vertex v1 we must have a2 + k7 = n, with k > 2, and consequently at vertex v2 it follows that a1 + a1 < n, and so a1 < f and a2 + a3 > n. But then an incompatibility on the sides arises at vertex v1. If «1 > «2 > «3, and (i) = a3 (Figure 14(b)), then 02 must be P, otherwise we get, at vertex v3, 03 = a1 and a1 + p > n, for all p € {a1, a2}. Nevertheless, an impossibility cannot be avoided at vertex v1 since we obtain P + 7 + p > n, for all p € {a1, a2}. C. P. Avelino and A. F. Santos: Spherical folding tessellations by kites and isosceles triangles IV 67 Figure 14: Local configurations (ii) 01 = 7 and (a) 02 = ft (Figure 15(a)), then 7 < 4 and ft > |. At vertex v4 we must have ft+a2 < n, however 2n > (ai+7+Y) + (ft+a2) = (ft+Y+Y) + (ai + a2) > 2n is impossible. (b) 02 = 7, it follows that a1 + k7 = n, k > 2, as illustrated in Figure 15(b). But any choices for d3 and 94 lead to a contradiction. 3 Case of Adjacency V Proposition 3.1. Q(K, T) is composed by a single folding tiling, M, such that a2 = f, ai + ft = n and 7 = a3 = |. For a planar representation see Figure 20(b). Its 3D representation is given in Figure 21. 68 Ars Math. Contemp. 11 (2016) 35-47 Figure 15: Local configurations occurring in case 2(ii) Proof. Suppose that any element of Q (K, T) has at least two cells congruent, respectively, to K and T, such that they are in adjacent positions as illustrated in Figure 2-V. The case analyzed in [4] (case 2.1) give rise to no f-tilings including two cells in these adjacent positions, and so a side of length c of each triangle must be adjacent to a side of length c of another triangle. 1. If a2 > «1, then a2 > f and we get the configuration of Figure 16(a). If a2 + fi = n (Figure 16(b)), we have a1 = 2 (vertex v1), and so a2 + a3 > n, justifying the choice for 01. But at vertex v2 we obtain a contradiction as «3 + 7 + 7 > n ((ai + «1) + («2 + fi) + («3 + Y + 7) = (2«1 + «2 + «3) + (fi + Y + Y) > 3n). Figure 16: Local configurations occurring in case 1 If a2 + p < n, then a2 + kp = n, with k > 2 (note that a2 + a3 > n). Consequently, Y > p and a3 > p. Observing Figure 17(a), we conclude that there is no way to satisfy the angle-folding relation around vertex v2 (a2 + a2 > a2 + a > n, a2 + a3 > n, a2 + y + P > n, for all p, and = p implies 02 = y and y + Y + p > n, for all p). 2. Suppose now that a1 > a2 (Figure 17(b)). It follows that a1 > f > a2 > p and y > n. 2.1 If p > y, then at vertex v1 we must have a1 + p + ka3 = n, with k > 1, or a1 + p = n. In the first case we have a1 > 2 > a2 > p > y > a3 (Figure 18(a)). As or 02 must be a3, we get an impossibility at vertex v2 or v3, respectively. C. P. Avelino and A. F. Santos: Spherical folding tessellations by kites and isosceles triangles IV 69 (a) case 1 (b) case 2 Figure 17: Local configurations (a) (b) Figure 18: Local configurations occurring in case 2.1 Therefore a1 = n. At vertex v1 we cannot have a1 = n = a 1+a3, as illustrated in Figure 18(b), otherwise at vertex v2 we get a1 + 7 + ka3 = n, k > 1, and a contradiction arises at vertex v3. Consequently, we get the configuration illustrated in Figure 19(a). Note Figure 19: Local configurations occurring in case 2.1 that at vertex v2 we cannot have 7 + 7 + ka3 = n, k > 1, nor 7 + 7 + 7 + ka3 = n, k > 1, 70 Ars Math. Contemp. 11 (2016) 35-47 otherwise we obtain a similar contradiction as before (in fact we cannot have two angles a3 adjacent). Observe also that we have necessarily a2 + a2 = n, as a2 + a2 + a2 > n. Now, 9\ = a3, 6\ = 7 or 6\ = ft. 2.1.1 If 6\ = a3 (Figure 19(b)), at vertex v3 we must have ai + a3 = n, which implies a3 = ft. Nevertheless, a contradiction arises at vertex v4 since we get a1 + 7 + ka3 > n, for all k > 1. 2.1.2 If 61 = 7 (Figure 20(a)), at vertex v4 we obtain ft + 7 + ka3 = n. But at vertex v3 we get a1+7+ka3 = n, which is not possible as 3n > (a1 +7+a3) + (a1+ft) + (a2 + a2) > (2ai + a2 + a3) + (ft + 7 + 7) > 3n. (a) (b) f-tiling M Figure 20: Local configurations occurring in case 2.1 2.1.3 If = ft, the last configuration is extended to the one illustrated in Figure 20(b). We shall denote this f-tiling by M. Its 3D representation is given in Figure 21. Figure 21: f-tiling M C. P. Avelino and A. F. Santos: Spherical folding tessellations by kites and isosceles triangles IV 71 2.2 Suppose now that 3 < 7 (Figure 22(a)). In this case we have 7 > 3 and 91 = 3 or 0\ = «3. a, 03 4 02 a, a, 02 5 a2 a, 1 a a3 (a) (b) Figure 22: Local configurations occurring in case 2.2 If 61 = fi (ai + fi < n, see Figure 22(b)), then at vertex we must have 7 + 7 + ka3 = n, with k > 0. As we seen before, as two angles a3 in adjacent positions lead to a contradiction, we must have 7 + 7 = n. Moreover, 62 cannot be a3, otherwise we would obtain 63 = a1 and, at vertex v3, a1 + 7 > n. The case 62 = fi also leads to a contradiction as 7 + 7 = n and vertex v3 cannot have valency four. Finally, if 61 = a3, we obtain the configuration illustrated in Figure 23. At vertex v1 we reach a contradiction as (a1 + fi + a3) + (a1 + 7) + (a2 + a2) > (2a1 + a2 + a3) + (fi + Y + Y) > 3n. □ Figure 23: Local configuration occurring in case 2.2 7 4 2 2 5 4 Case of Adjacency VI Suppose that any element of Q (K, T) has at least two cells congruent, respectively, to K and T, such that they are in adjacent positions as illustrated in Figure 2-VI. As b = c, using trigonometric formulas, we obtain cos 3 + cos2 7 cos or + cos «1 cos or -~2-- = —^-—a—-. (4.1) sin2 y sin ai sin a- 72 Ars Math. Contemp. 11 (2016) 35-47 Remark 4.1. The cases analyzed in [2] and [3] give rise to no f-tilings including two cells in these adjacent positions, and so a side of length c of each triangle must be adjacent to a side of length c of another triangle. Proposition 4.2. T) = 0 iff (i) ai + y = n, a2 = f, y + 7 + a3 = n and ft = |, or (ii) ai + y = n, a2 = ft = f and 7 + 7 + a3 = n. In the first case, there is a single f-tiling denoted by N. A planar representation is given in Figure 26(b) and a 3D representation is given in Figure 27. In the second case, there is a single f-tiling, P. The corresponding planar and 3D representations are given in Figure 29(b) and Figure 30, respectively. Proof. Concerning the angles of the triangle T we have necessarily one of the following situations: 7 > ft or y < ft. We consider separately each one of these cases. 1. Suppose firstly that 7 > ft. If a2 > ai, then a2 > | and we get the configuration of Figure 24(a). Due to the edge Figure 24: Local configurations occurring in case 1 lengths and also Remark 4.1, v1 cannot have valency four and so a2 + 7 + ka3 = n, k > 1. Therefore, analyzing vertices v1 and v2 we conclude that a2 + a3 < n and a1 < 2, which is impossible taking into account the kite's area. Thus, a1 > a2 > a3 (Figure 24(b)) and = ft or = 7. In the first case, v1 cannot have valency four and there is no way to satisfy the angle-folding relation around this vertex. Consequently, = 7 and (i) if a1 + 7 < n, then a1 + 7 + ka3 = n, k > 1 (Figure 25(a)). At vertex v2 we reach a contradiction as a1 + p > n, for all p G {a1, a2}. (ii) if a1 + 7 = n, then the last configuration extends to the one illustrated in Figure 25(b). Now, if 02 = ft (Figure 26(a)), we obtain a contradiction at vertex v2. On the other hand, if 02 = 7 a global planar representation is achieved as illustrated in Figure 26(b). We denote such f-tiling by N. The corresponding 3D representation is given in Figure 27. C. P. Avelino and A. F. Santos: Spherical folding tessellations by kites and isosceles triangles IV 73 ai "3 a! a2 3 6 V2 y-h— "1" "3 "3 7 ai "2 a2 "1 e, y y 5 4 a3 a! 1 "3 2 y P P a2 ai y a a2 13 (a) case 1(i) (b) case 1(ii) Figure 25: Local configurations occurring in case 1 22 a, a3 y ee 24 y 16 y y e e y 15 12 y y y y a1 a2 13 . y y y 23 e 21 e y a3 a1 11 4 a2 a1 \y " y y 14 3 e 0 y e e y e 17 19 y yy a1 a2 a2 a1 y ee 2 10 y y y al a3 18 \y y y 7 20 p y P p y 11 y / yy 6 e9 ey 8 a1 a2 y 0 e 25 y 26 (a) (b) f-tiling N Figure 26: Local configurations occurring in case 1(ii) 8 1 2. Suppose now that 7 < ft. If a2 > ai, then a2 > f and we get the configuration of Figure 28(a). Due to the edge lengths and also Remark 4.1, vi cannot have valency four and so ai + ai + k7 = n, k > 1. But then the other sum of alternate angles must contain a2 + a3 > n, which is not possible. 74 Ars Math. Contemp. 11 (2016) 35-47 Figure 27: f-tiling N a2 > a3 and ka2 = n, k > 2, and we have a\ +7 = n or a\ +7 < n. 2.1 If a1 + 7 = n, with the labeling of Figure 28(b), we have 6\ = 7 or 61 = p. 2.1.1 If 61 = 7, the last configuration is extended to the one illustrated in Figure 29(a). a a3 Y Y 3 4 7 a2 a1 Y P v a1 a2 a2 a1 Y 2 P P 5 1 6 a3 a a1 a3 Y Y Y Y Y Y Y V2 9 8 e2 \ P P (a) (b) f-tiling P Figure 29: Local configurations occurring in case 2.1 5 v C. P. Avelino and A. F. Santos: Spherical folding tessellations by kites and isosceles triangles IV 75 2.1.1.1 If 62 = 7, at vertex v2 we have a3 + 7 + 7 = n or a3 + 7 + 7 + 7 = n. Note that we cannot have more angles a3 around v2, as two angles of this type in adjacent positions lead to an impossibility, as seen before. The condition a3 + 7 + 7 = n implies a2 + a2 = 2, and we get the configuration illustrated in Figure 29(b). We denote this f-tiling by P, whose 3D representation is given in Figure 30. ( / / v \ \ v \ 1 \ \ \x a // 1 J Figure 30: f-tiling P On the other hand, if a3 + 7 + 7 + 7 = n (Figure 31(a)), the angles 63 and 94 cannot be a3 otherwise we reach a contradiction at vertices v3 and v4, respectively. But this implies that at vertex v5 we have two angles a3 in adjacent positions, which is not also possible. Figure 31: Local configurations occurring in case 2.1 2.1.1.2 If 62 = ft, then at vertex v3 we have ft + 7 + ka3 = n, k > 1, which leads to a contradiction as illustrated in Figure 31(b) (see vertex v4). 2.1.2 If 61 = ft, we obtain a similar impossibility as in the previous case. 2.2 If a1 + 7 < n (Figure 32(a)), then 61 G {ft, 7}. 76 Ars Math. Contemp. 11 (2016) 35-47 ai a3 3 a2 a1 e2 0i a1 a3 3 a2 a, 5 Y a3 1 4 6, P Y a2 a1 1 ai a3 Y P 2 Y a2 a, 1 a1 a3 Y P 2 Y (a) (b) Figure 32: Local configurations occurring in case 2.2 If = ft (Figure 32(b)), then a + ft + ka3 = n, k > 1. It follows that the other sum of alternate angles at vertex v1 must be greater or equal to a1 + 7 + 7 > n, which is an impossibility. If = 7 and (i) = 7 (Figure 33(a)), then ft > a1 > 2, which implies tile 6. At vertex v2 we obtain ft + 7 + ka3 = n, k > 1, giving rise to two angles a3 in adjacent positions, which leads to a contradiction, as seen previously. Figure 33: Local configurations occurring in case 2.2 (ii) = «3 (Figure 33(b)), vertex v1 has valency six or greater than six. In the first case, we obtain two angles a3 in adjacent positions, which is not possible. In the last case, we have necessarily 03 = 7, and so ft > a1 > 2. Consequently, a contradiction arises at vertex v2 or v3. C. P. Avelino and A. F. Santos: Spherical folding tessellations by kites and isosceles triangles IV 77 Concerning to the combinatorial structure of each tiling obtained, we have that (i) the symmetries of the f-tilings L, J, M and N that fix a vertex v of valency four and surrounded by (a2,a2, a2,a2) are generated by a reflection and by the rotation through an angle 2 around the axis by ±v. On the other hand, for any vertices vi and v2 of this type, there is a symmetry sending v1 into v2. It follows that the symmetry group has exactly 48 = 6 x 8 elements and it forms the group of all symmetries of the cube - the octahedral group, sometimes referred as C2 x S4. (ii) the f-tiling P has only two vertices surrounded by (a2,a2, a2,a2), say the north and south poles. The symmetries of P that fix the north pole are generated by a reflection and by the rotation through an angle | around the zz axis, giving rise to a subgroup isomorphic to D4 (the dihedral group of order 8). Now, the reflection on the equator is also a symmetry of P, and so it follows that the symmetry group of P is isomorphic to C2 x D4. 5 Summary In Table 1 is shown a list of the spherical dihedral f-tilings whose prototiles are a spherical kite and an isosceles spherical triangle, K and T, of internal angles (a1,a2,a1, a3), and (ft, y, y), respectively, in cases of adjacency IV, V and VI. Our notation is as follows: 7i is the solution of equation (2.1), with a2 = n, a1 = n - 71 and a3 = ft = 3; ft1 is the solution of equation (2.1), with a2 = f, a1 = n - 7 and a3 = n - 2ft1; ft2 is the solution of equation (2.1), with a2 = 2, a1 = n - ft2 and a3 = 7 = | ; y2 is the solution of equation (4.1), with a2 = 2, ft = |, a1 = n - y2 and a3 = n - 2y2; 73 is the solution of equation (4.1), with a2 = ft = 2, a1 = and a3 = n - 273. \V | is the number of distinct classes of congruent vertices; N1 is the number of triangles congruent T and N2 is the number of kites congruent to K (used in the dihedral f-tilings); G(t) is the symmetry group of each tiling t g Q (K, T). f-tiling ai a2 a3 P Y |v | Ni n2 G(T ) L n - yi n n 2 yi 3 24 24 C2 x s4 J 2n 3 2 n - 2 p pi 3 4 48 24 c2 x s4 M n - ^2 2 3 p2 3 4 48 24 c2 x s4 N n - y2 2 n - 2y2 3 y2 3 48 24 c2 x s4 P n - y3 2 n - 2y3 2 y3 3 16 8 c2 x D4 Table 1: Combinatorial structure of dihedral f-tilings of S2 by spherical kites and isosceles triangles in cases of adjacency IV, V and VI References [1] C. 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