Spherical f-tilings by two noncongruent classes of isosceles triangles-II

In this work, we give a complete classification of spherical dihedral f-tilings when the prototiles are two noncongruent isosceles triangles with certain adjacency pattern. As it will be shown, this class is composed by two discrete families denoted by {ieɛ}m, m ≥ 2, m ∈ ℕ, Fk, k ≥ 4, k ∈ ℕ and two sporadic tilings denoted by G and H.


Introduction
Spherical folding tilings, or f -tilings for short, are edge-to-edge tilings of the Euclidean sphere S 2 , by geodesic polygons, such that all vertices are of even valency and the sum of alternate angles around each vertex is π. Let us denote by Ω(X, Y ) the set, up to isomorphism, of all dihedral f -tilings of S 2 whose cells are congruent to X or Y . In [9], it was shown that any τ ∈ Ω(X, Y ) necessarily has vertices of valency four.
Some monohedral f -tilings of the sphere by triangles such as the spherical octahedron have been known for a very long time. The tilings have been explicitly studied since 1992, see [5]. Ueno and Agaoka, in their work on spherical tilings [1], provided several examples of spherical tilings by congruent quadrangles, see [2]. In [3], the study of all spherical f -tilings by triangles and r-sided regular polygons, for any r ≥ 5 was given. A list on spherical f -tilings by triangles and isosceles trapezoids, by the same authors was given in [4].
The classification of all dihedral spherical folding tilings by triangles is not yet completed. The cases studied so far correspond to the following prototiles: -an equilateral triangle and an isosceles triangle, [7]; -an equilateral triangle and a scalene triangle, [8].
-two non congruent (nonequilateral) isosceles triangles in a particular type of adjacency, [6]. Let T 1 be an isosceles (nonequilateral) spherical triangle with angles α, α, β, paired sides a (opposite to α) and base b (opposite to β); and T 2 another such triangle with angles γ, γ, δ, paired sides c, and base d. The sphere is connected, both prototiles are used, and the tiles meet edge-to-edge; so it is necessary that somewhere in the configuration a prototile of one type shares an edge with one of the other type. "Type I adjacency", in which b = d, was considered in [6]. In this paper we will consider "type II adjacency", with a = d; the remaining case, "type III adjacency", in which one tile's base matches the paired edges of the other, will be dealt with in an upcoming paper. The three cases are illustrated in Figure 1; note that in principle, a pair of tiles could exhibit both type I and type II adjacency, or type III adjacency in two distinct ways. Note that type III adjacency is not compatible with either of the other types, as this would force one triangle or the other to be equilateral. Spherical trigonometry allows us to solve for any edge of a triangle in terms of its angles; doing this for the edge lengths a and d and equating, we obtain the following equation which will be useful in the next section: cos α(1 + cos β) sin α sin β = cos a = cos δ + cos 2 γ sin 2 γ . (1.1) To determine whether a pair of triangles (T 1 , T 2 ) permits a dihedral f -tiling, we often find useful to start by considering local configurations in which the sphere is only partially tiled. We will usually start with a pair of triangles exhibiting type II adjacency, and label these 1 and 2. We will then deduce the location of some tile i from the locations of tiles {1, 2, . . . , i − 1} and the hypothesis that the configuration extends to a complete f -tiling, backtracking when necessary, until either a complete f -tiling or a contradiction is reached.

Triangular Dihedral f -Tilings by Isosceles Triangles with Adjacency of Type II
If τ ∈ Ω(T 1 , T 2 ) has an adjacency of type II , then we may start its configuration with two adjacent cells congruent respectively to T 1 and T 2 , as shown in Figure 2. Lemma 1.1 forces triangle 3 as shown. In order to have the angle folding relation fulfilled, the sum containing the alternate angles α and γ must be of the form α + γ ≤ π. We shall separate cases α + γ = π and α + γ < π.

Lemma 2.1 (Elimination Lemma I)
There are no f -tilings with adjacency type II and α + γ = π.
Proof Suppose that α + γ = π. We shall show that α is acute. Suppose not; then clearly γ ≤ π 2 ; and by the equation (1.1), cos b < 0, which implies that δ > π 2 . An f -tiling cannot have three obtuse angles at a vertex, so tiles 5 and 5 must be as shown in Figure 3-I. As δ + 2γ > π, tiles 6 and 6 must be as shown. Tile 7 is immediate, and as δ + 2γ > π, tile 8 must be as shown; but then we have three α angles at v 1 . This would imply α = γ = π 2 , which by the equation (1.1) gives also δ = π 2 , a contradiction.

Figure 3 Local configurations
Thus α must be acute, and γ obtuse. There cannot be a second γ at v 2 , as this would immediately create an alternating sum greater than π. Nor can triangle 5 be as in Figure 3-II, as this would require the vertex to have exactly one more angle, supplementary to β. This angle could only be δ, but neither of the edges adjacent to a δ angle matches the α − α edge that is already present. We are thus forced to the configuration of Figure 3-III, with four α angles coming together. As α < π 2 , the alternating sum must contain at least two more angles, which must be α or γ angles; and as γ > α we have α < π 3 . But at v 2 we have two adjacent β angles and a γ; for this to exist in an f -tiling, β + γ ≤ π, which would imply β ≤ α. But then 2α + β < π, which is impossible.
Proof Suppose the triangles T 1 , T 2 to have adjacency type II and α > β.

Figure 5 Local configurations
The third case leads also to an impossibility, since the possible extensions are the ones illustrated in Figure 5, which have already been ruled out.

Figure 8 Local configurations
If α + γ + pγ + qβ = π, p, q ≥ 1, we are led to a contradiction similar to that in the two previous cases.
In the first and third cases, we would have α = β which is impossible. In the second case, we conclude, by the adjacency condition (1.1), that α ≈ 64.085 • or α ≈ 115.91 • , which are both impossible, since β < α < π 2 .
Assume now that δ > π 2 . The configuration in Figure 7 takes the form illustrated in Figure 9-II.
The vertex partially surrounded by the angles δ, γ, β is of valency greater than four (otherwise β + δ = π and γ = π 2 violating the equality 2α + kγ = π, k ≥ 1). On the other hand, the vertex partially surrounded by the angles γ, δ, δ is also of valency greater than four (otherwise, as β > γ, one has β + δ > γ + δ = π which is impossible). Therefore, the only way to arrange the angles around vertices of valency four is (δ, δ, α, α); but this cannot be realized in a tiling due to incompatible edge lengths.
The first case implies that α = β, which is impossible. In the second case, one has γ = β and so π 4 < γ = β < α < π 2 = δ. By the adjacency condition (1.1), we get γ = β ≈ 52.884 • and α ≈ 74.232 • . Accordingly, we have three type of vertices: vertices of valency four surrounded exclusively by angles δ, vertices of valency six, with alternate angle sets surrounded by alternate angles α, γ, β and α, γ, γ. Thus, we may extend the configuration in Figure 11-II in a unique way, getting a vertex v 5 (Figure 12-I) partially surrounded by an angle sequence (γ, γ, γ, γ, γ). But π − 3γ ≈ 21.348 • which is less than any of the angles, so we cannot complete the tiling at this vertex while satisfying the folding condition.

3.2.1
Consider first the case in which α ≥ π 2 . By the adjacency condition (1.1), δ > π 2 . In case, α > π 2 , we extend the configuration in Figure 7 to the one in Figure 15, where tile 7 is congruent to T 1 or to T 2 . If tile 7 is congruent to T 1 , then the extended configuration ends up at a vertex partially surrounded by the angle sequence (δ, γ, α), which gives an alternate sum of at least δ + α forbidden by the angle folding relation (see Figure 15-I). If tile 7 is congruent to T 2 , then γ + δ = π (Figure 15-II); any other possibility leads to edge incompatibility.
The only remaining possible configuration is that shown in Figure 15-II with γ + δ = π where the vertex, v 7 is partially surrounded by a sequence of angles (δ, β, δ). The alternate sum 2δ > π is too large; so the configuration cannot be extended at this vertex. .

Figure 15 Local configurations
We must then have α = π 2 . As seen before, δ > π 2 . Extending the configuration in Figure 7, tile 7 is congruent to T 1 or T 2 . In the first case, we get the configuration illustrated in Figure 16-I giving rise to a vertex partially surrounded by a sequence (δ, γ, α), with an impossible alternate sum δ + α > π. In the second case, we are led to the configuration in Figure 16-II. Edge matching requires there to be a γ or a δ at v 8 next to triangle 7. But this gives an alternate sum of δ + 2γ > π or 2δ + γ > 2δ > π contradicting the folding condition.
In the first case, we may expand the configuration and get a vertex (v 11 at Figure 19-I) partially surrounded by a sequence (α, α, α). This requires an alternate sum 2α + γ = π; therefore, γ = α > π 3 , contradicting the condition α + 2γ = π. In the second case, for r > 1 the angle arrangement, at vertex v 10 (Figure 19-II) leads to another vertex partially surrounded by a sequence of angles (α, α, α); this leads to a contradiction as in the previous subcase.
For r = 1, the adjacency condition (1.1) leads us to γ ≈ 112.98 • (impossible) or γ ≈ 50.180 • , which implies β ≈ 29.46 • and α ≈ 79.64 • . Proceeding with the extension of the configuration, we must keep in mind that tile 19 has two possible positions, as shown in Figure 20-I.
One of the two possible positions for tile 19 leads to a vertex, v 12 partially surrounded by a sequence (α, α, α), forcing an alternate sum with two α angles; but the remaining angle is π − 2α ≈ 20.72 • , less than any of the four angles (see Figure 20-II). The other position for tile 19 gives rise to a similar contradiction at the vertex v 12 (see Figure 20-III).
The vertex v 15 is thus of valency six or eight, and surrounded only by α angles. Valency six would imply α = π 3 and thus α + 2γ > π which is false. If the valency is eight, then α = π 4 , γ = 3π 8 , β = 5π 8 and, by the adjacency condition (1.1), δ ≈ 64, 916 • . We extend the configuration in Figure 25 at vertex v 15 . The α angle of tile 3 and the γ angle of tile 2 form part of an alternating sequence that must be completed by one more γ angle. This must be from a tile positioned as tile 7 in Figure 26-I; if it had the other orientation, edge matching would force a δ next to tile 2 and a β next to tile 1. But this would imply α + β + δ = π, which is inconsistent with the values above.
Once we have settled the orientation of tile 7, the other tiles of Figure 26-I follow, and we find a vertex, v 16 , partially surrounded by a sequence (δ, δ, δ). Since the sum 2δ + μ does not satisfy the angle folding relation, for any μ ∈ {α, β, γ, δ}, the configuration cannot be extended at this vertex.

1.2.2.2
If vertex v 14 satisfies 2α + γ = π, then π 4 < α < π 3 ; thus a vertex cannot be entirely surrounded by α angles. If a vertex has an α angle and no γ angle, edge matching prevents the presence of δ angles. Thus if one of the alternate sets of angles at a vertex contains two α angles it must be completed by a γ angle. Moreover, matching bases, the other alternate set also contains two α angles. Consequently vertex v 15 (and all other such vertices) must have four α angles and two (adjacent) γ angles.
Similar arguments show that if one of the alternate sets of angles at a vertex contains an α angle and a γ angle, it is completed by an α. The only alternative would be α+γ +tδ = π, t ≥ 1, which forces vertices surrounded by two alternate angles γ, whose sum 2γ + μ violates the angle folding relation, for any μ ∈ {β, γ, δ} and 2γ + α = π implies the impossibility α = γ > π 3 , as illustrated in Figure 26-II.
On the other hand, vertices with even one β angle must be of valency four with two β angles and two γ angles. Any other possibilities led to the edge incompatibility or a contradiction arises from the adjacency condition (1.1).
We may continue extending the configuration in Figure 25 and taking into account the two previous observations, vertices surrounded by several angles δ must satisfy mδ = π, m ≥ 2. Every such configuration may be completed, so the local configuration extends to a complete tiling, denoted by E m , m ≥ 2. For each m ∈ N, m ≥ 2, E m is composed of vertices with two β and two γ angles, vertices with four α and two γ angles, and vertices with 2m δ angles. The number of triangular faces of E m is 8m; the triangles are equally distributed among two congruence classes. The next figures shows a 2D and 3D representation of E 2 . E 2 is composed of eight triangles congruent to T 1 and eight triangles congruent to T 2 , where α = γ = 60 • , β = 120 • , and δ = 90 • .
This family of tilings can be obtained from the family of dihedral tilings R k described in [9], where the prototiles are an isosceles triangle and a spherical square, by splitting each square into two isosceles triangles.
We study each of these possibilities.

1.2.2.4
Assume that vertex v 14 satisfies α + γ + tδ = π for t ≥ 1. Then, vertex v 15 is surrounded by six, eight or ten α angles, or by two γ angles and four or six α angles. We study these five possible cases.
As in Case 1.2.2.4, it is impossible to complete the tiling for t ≥ 2; and for t = 1, we would have β < 0.
1.4 If θ 8 = δ and β + θ 8 = π, the angle arrangement forces 2γ = π. As δ < γ, we must have δ < π 2 < β. Extending the configuration in Figure 30, tile 6 has two possible positions, as shown in the next figure. Either of these configurations can be extended in a unique way leading to a vertex surrounded in part by a sequence of adjacent angles β, β, α (on the bottom edge of the diagrams in Figure 33). This vertex cannot be of valency four, which would require both α and β to be right angles. However, since γ = π 2 , β > π 2 , β + δ = π and 2α + β > π, then α + β + μ > π, for any μ ∈ {α, β, γ, δ}, so neither of these partial tilings can be completed as an f -tiling. Assume that β > α and δ > γ. Starting from the configuration illustrated in Figure 24, we have two cases to consider, depending on whether θ 8 is γ or δ.
1.1 If β + θ 8 = π with θ 8 = γ, extending the configuration leads us to a vertex surrounded by three angles α as shown in Figure 34-I. If α = π 2 , then from the adjacency condition, we conclude that δ > π 2 . As β > α, then β > π 2 and so γ < π 2 . Therefore, at vertex v 14 , the sum containing the alternate angles α and γ must be of the form α + tγ = π, t ≥ 2. However, at the vertex with three angles δ, there would have to be an alternating sum containing two δ angles, which is impossible (Figure 34-II).
1.1.2 If θ 9 = δ and 2δ = π, then γ > π 4 . Taking into account that 2α > π − β = γ > π 4 and the compatibility of the edges, we conclude that α > π 8 ; and at a vertex surrounded (in part) by four angles α, the alternating sum containing two angles α is either of the form kα = π for k ∈ {3, . . . , 7}, or pα + qγ = π, where p ≥ 2, q ≥ 1, and p + q ≤ 6. It is impossible to extend the configuration for cases k = 4, 5, 6, 7, since at vertex v 14 , the sum α + γ + μ violates the angle folding relation, for any μ ∈ {α, β, γ, δ}. For the case k = 3, one has α = γ = π 3 , β = 2π 3 and consequently vertex v 14 is of valency six satisfying 2γ + α = π. The configuration in Figure 35 extends to a tiling of the entire sphere, which is denoted by G. Figure 37 presents a 2D and 3D representation of G, which is composed of six copies of T 1 and twelve copies of T 2 . This tiling can be obtained from the spherical cube by dividing the three faces that meet at one vertex into two equilateral triangles and the others into four. Bisecting the larger triangles yields the tetrakis cube, the dual of the truncated octahedron. However, parity considerations make it impossible to combine pairs of triangles and obtain one of the (triangle, well-centered quadrangle) f -tilings of [9].

1.1.2.2
In the second case, suppose first that q = 1. Then, p = 2, 3, 4, or 5. For p = 2, 2α + γ = π, β + γ = π, and δ = π 2 . We can extend the configuration of Figure 35 to the complete tiling E 2 shown in Figure 27. The angles here are the same as in G, but the tiling has eight tiles of each type.
Like G, E 2 can be obtained as a subdivision of the spherical cube or by combining faces of the spherical tetrakis cube. Combining the larger faces along their hypotenuses yields the f -tiling R 2 described in [9].
In Figure 40-I, we reach a vertex with contiguous angles β, γ, γ, α. But as β + γ = π the vertex would have to be of valency four and α + γ = π, in violation of the folding condition. With the other choice for tile 8, tile 12 can be as in Figure 40-II or as in Figure 41. These lead to similar impossibilities. If θ 9 = δ but 2δ < π, it is impossible to extend the configuration in Figure 35, since there is always incompatibility of the edges or violation of the angle folding relation at the vertex partly surrounded by three angles δ.

1.2.2
We will consider the case β + kγ + α = π and we shall study the cases k = 1 and k > 1 separately. If the angle θ 10 was δ, then vertex v 14 would be of valency greater than four. However, as γ < α < β, γ < δ and 2γ + δ > π, the sum containing the angles α and δ would not satisfy the angle folding relation. Therefore, θ 10 = γ and δ ≤ π 2 . Extending the configuration further, the tile numbered 6 has two possible positions, as shown in Figure 44-I and II.
By the adjacency condition (1.1), β ≈ 71.774 • and γ ≈ 48.226 • . Extending the configuration, we get a tiling denoted by H. 2D and 3D representations of H are shown in Figure 45.
It uses forty-eight triangles congruent to T 1 and twenty-four triangles congruent to T 2 . Combinatorially, it is a tetrakis-hexakis truncated octahedron, but it has the symmetry group of a snub cuboctahedron.  Figure 44-II, we may add some cells to the configuration and end up at a vertex partially surrounded by the sequence of angles α, α, γ, γ, α (Figure 46). This vertex would have to have exactly one more angle, with 2α + γ = π. But β + γ + α = π, so we have a contradiction. If δ < π 2 , the vertex v 20 with three consecutive δ angles cannot be completed while satisfying the angle folding relation, since γ < α < β, γ < δ and 2γ + δ > π.
In the first case, α > π 4 and from β + kγ + α = π < 2γ + δ, we conclude that δ > β + α > π 2 . Therefore, adding some cells to the configuration in Figure 30, we are led to a vertex with contiguous angles α, γ, δ. But α + δ = π, nor can we add any other angle or combination of angles give us a sum of π. Thus this configuration cannot appear in an f -tiling.
Extending the configuration in Figure 47-II, we are led to vertex v 22 , surrounded in part by three angles β. But 2β < π while π − 2β is less than any of the four angles (Figure 48-I.) Extending the configuration in Figure 47-III, we are lead to a vertex v 23 (see Figure 48-II) partly surrounded by a sequence of contiguous angles (α, γ, γ, β, β) arises. But α + γ + β > π, so this is impossible in an f -tiling.