Tilings of Sphere by Congruent Pentagons I: Edge Combinations a 2 b 2 c and a 3 bc
aa r X i v : . [ m a t h . M G ] M a r Tilings of Sphere by Congruent Pentagons I:Edge Combinations a b c and a bc Erxiao Wang, Min Yan ∗ Hong Kong University of Science and TechnologyMarch 21, 2019
Abstract
We develop the basic tools for classifying edge-to-edge tilings ofsphere by congruent pentagons. Then we prove such tilings for edgecombination a b c are three families of pentagonal subdivisions of theplatonic solids, with 12, 24 and 60 tiles. We also prove that suchtilings for edge combination a bc are two unique double pentagonalsubdivisions, with 48 and 120 tiles. Keywords : Spherical tiling, Pentagon, Classification.
Mathematicians have studied tilings for more than 100 years. A lot is knownabout tilings of the plane or the Euclidean space. However, results abouttilings of the sphere are relatively rare. A major achievement in this regard isthe complete classification of edge-to-edge tilings of the sphere by congruenttriangles. The classification was started by Sommerville [4] in 1923 andcompleted by Ueno and Agaoka [5] in 2002. For tilings of the sphere bycongruent pentagons, we know the classification for the minimal case of 12tiles [1, 3].Spherical tilings are relatively easier to study than planar tilings, simplybecause the former involve only finitely many tiles. The classifications in ∗ Research was supported by Hong Kong RGC General Research Fund 16303515.
13, 5] not only give the complete list of tiles, but also the ways the tilesare fit together. It is not surprising that such kind of classifications for theplaner tilings are only possible under various symmetry conditions, becausethe quotients of the plane by the symmetries often become compact.Like the earlier works, we restrict ourselves to edge-to-edge tilings ofthe sphere by congruent polygons, such that all vertices have degree ≥ a b c, a bc, a b , a b, a . Here a b c means the five edge lengths are a, a, b, b, c , with a, b, c, distinct. Thefollowing is the classification for this edge combination. Theorem.
Edge-to-edge tilings of the sphere by congruent pentagons withedge combination a b c ( a, b, c distinct) are the following.1. Pentagonal subdivision of tetrahedron, with tiles.2. Pentagonal subdivision of octahedron (or cube), with tiles.3. Pentagonal subdivision of icosahedron (or dodecahedron), with tiles. Pentagonal subdivision is introduced in Section 3.1. The operation canbe applied to any tiling on oriented surface, and dual tilings give the samepentagonal subdivision tiling. Therefore the five platonic solids give threepentagonal subdivisions. Note that we already proved in [1, 3] that edge-to-edge tilings of sphere by (the minimal number of) 12 congruent pentagonsis the deformed dodecahedron. This is exactly the pentagonal subdivisionof tetrahedron in the theorem. We also note that each tiling in the theoremallows two free parameters.The following is the classification when the five edges have lengths a, a, a, b, c . Theorem.
Edge-to-edge tilings of the sphere by congruent pentagons withedge combination a bc ( a, b, c distinct) are the following.1. Double pentagonal subdivision of octahedron (or cube), with tiles. . Double pentagonal subdivision of icosahedron (or dodecahedron), with tiles. Double pentagonal subdivision is introduced in Section 3.2. Unlike pen-tagonal subdivision, each tiling in the theorem allows only one specific pen-tagon, and we provide the exact values in Section 3.2. In fact, we may alsoget a tiling (with 24 tiles) by applying the double pentagonal subdivision totetrahedron. However, the specific pentagon has b = c and therefore the tilinghas edge combination a b . Moreover, the tiling is a degenerate case ( b = c in a b c ) of the pentagonal subdivision of octahedron (or cube). Thereforethe tiling is not listed in the theorem.Our next paper [7] classifies tilings with edge combination a b , and ismuch more complicated. The third in our series [2] classifies tilings withedge combination a , i.e., by congruent equilateral pentagons. This requirescompletely different technique, because we can no longer rely on edge lengthinformation. The remaining case is the edge combination a b , which we callalmost equilateral. The case is much more challenging and will be the subjectof another series.This paper is organized as follows. Section 2 is the basic facts for tilingsof the sphere by congruent pentagons. Section 3 introduces pentagonal anddouble pentagonal subdivisions. We also calculate the specific pentagonsused in the double pentagonal subdivision tilings. Sections 4 and 5 prove thetwo classification theorems.We would like to thank Ka-yue Cheuk and Ho-man Cheung. Some oftheir initial work on the pentagonal subdivision are included in this and thenext paper. Consider an edge-to-edge tiling of the sphere by pentagons, such that allvertices have degree ≥
3. Let v, e, f be the numbers of vertices, edges, and3iles. Let v k be the number of vertices of degree k . We have2 = v − e + f, e = 5 f = ∞ X k =3 kv k = 3 v + 4 v + 5 v + · · · ,v = ∞ X k =3 v k = v + v + v + · · · . Then it is easy to derive 2 v = 3 f + 4 and f − X k ≥ ( k − v k = v + 2 v + 3 v + · · · , (2.1) v = 20 + X k ≥ (3 k − v k = 20 + 2 v + 5 v + 8 v + · · · . (2.2)By (2.1), f is an even integer ≥
12. Since tilings by 12 congruent pentagonshave been classified by [1, 3], we may assume f >
12. We also note that by(2.1), f = 14 implies v = 1 and v i = 0 for i >
4. By [8, Theorem 1], this isimpossible. Therefore we will always assume that f is an even integer ≥ > high degree vertices. Lemma 1.
Any pentagonal tiling of the sphere has a tile, such that fourvertices have degree and the fifth vertex has degree , or . We call three types of tiles in the lemma 3 -tile, 3 H (of degree 4 or 5)by dot. The fourth of Figure 1 is shared by the three neighborhoods, and wecall it the partial neighborhood . We will always label the tiles in the partialneighborhood as in the picture. Proof.
If a pentagonal tiling of the sphere has no tile described in the lemma,then any tile either has at least one vertex of degree ≥
6, or has at least twovertices of degree 4 or 5. Since a degree k vertex is shared by at most k tiles,the number of tiles of first kind is ≤ P k ≥ kv k , and the number of tiles ofthe second kind is ≤ (4 v + 5 v ). Therefore we have f ≤ v + 52 v + X k ≥ kv k . H Figure 1: Neighborhood and partial neighborhood of a special tile.On the other hand, by (2.1), we have f − v + 52 v + X k ≥ kv k ! = 12 + 12 v + X k ≥ ( k − v k > . We get a contradiction.
Lemma 2.
If a pentagonal tiling of the sphere has no -tile, then f ≥ .Moreover, if f = 24 , then each tile is a -tile.Proof. If there is no 3 -tile, then any tile has at least one vertex of degree ≥
4. This implies f ≤ P k ≥ kv k . By (2.1), we get f = 2 f − f ≥
24 + X k ≥ k − v k − X k ≥ kv k = 24 + X k ≥ k − v k ≥ . Moreover, the equality happens if and only if v i = 0 for i > f = 4 v .Since there is no 3 -tile, this means that each tile is a 3 Lemma 3.
If a pentagonal tiling of the sphere has no -tile and -tile,then f ≥ . Moreover, if f = 60 , then each tile is a -tile.Proof. If there is no 3 -tile and 3 ≥
5. This implies f ≤ v + P k ≥ kv k . By (2.1), we get f = 5 f − f ≥
60 + X k ≥ k − v k − v − X k ≥ kv k = 60 + 2 v + X k ≥ k − v k ≥ . v = v = v = · · · = 0 and f = 5 v . Since there is no 3 -tile and 3 The most basic property about angles is that the sum of all angles ( anglesum ) at a vertex is 2 π . Another basic property is the sum of all angles inpentagon. Lemma 4.
If all tiles in a tiling of sphere by f pentagons have the same fiveangles α, β, γ, δ, ǫ , then α + β + γ + δ + ǫ = 3 π + 4 f π. Proof.
Since the angle sum at each vertex is 2 π , the total sum of all anglesis 2 πv . Moreover, the sum of five angles in each tile is Σ = α + β + γ + δ + ǫ ,the same for all the tiles. Therefore the total sum of all angles is also Σ f .We get 2 πv = Σ f , and by 3 f = 2 v −
4, we further getΣ = 2 π vf = 3 π + 4 f π. The lemma does not require that the angles are arranged in the sameway in all tiles, and does not require that the edges are straight (i.e., greatarcs). However, if we additionally know that all edges are straight, then alltiles have the same area Σ − π , and the equality in the lemma follows fromthe fact that the total area (Σ − π ) f is the area 4 π of the sphere.The angles in the lemma refer to the values, and some angles among thefive may have the same value. For example, if the five values are α, α, α, β, β ,with α = β (different values), then we say the pentagon has angle combina-tion α β . The following is about the distribution of angle values. Lemma 5.
If an angle appears at every degree vertex in a tiling of sphereby pentagons with the same angle combination, then the angle appears at least times in the pentagon.Proof. If an angle θ appears only once in the pentagon, then the total numberof times θ appears in the whole tiling is f , and the total number of non- θ f . If we also know that θ appears at every degree 3 vertex, then f ≥ v , and non- θ angles appear ≤ v times at degree 3 vertices. Moreover,non- θ angles appear ≤ P k ≥ kv k times at high degree vertices. Therefore4 v ≤ f ≤ v + X k ≥ kv k . Then by (2.2), we have0 ≥ v − v + X k ≥ kv k ! = X k ≥ (2(3 k − − k ) v k = X k ≥ k − v k . This implies v i = 0 for i > f = v = 2 v , and contradicts (2.1).Unlike Lemma 4, which is explicitly about the values of angles, Lemma5 only counts the number of angles. The key in counting is to distinguishangles. We may use the value as the criterion for two angles to be the“same”. We may also use the edge lengths bounding angles as the criterion.The observation will be used in the proof of Lemma 9. The observation alsoapplies to the subsequent Lemmas 6, 7, 8. Lemma 6.
If an angle appears at least twice at every degree vertex in atiling of sphere by pentagons with the same angle combination, then the angleappears at least times in the pentagon.Proof. If an angle θ appears only once in the pentagon, then by Lemma 5, itcannot appear at every degree 3 vertex. If it appears twice in the pentagon,then the total number of θ in the whole tiling is 2 f , and the total numberof non- θ angles is 3 f . If we also know that θ appears at least twice at everydegree 3 vertex, then 2 f ≥ v , and non- θ angles appear ≤ v times at degree3 vertices. Moreover, non- θ angles appear ≤ P k ≥ kv k times at high degreevertices. Therefore 3 v ≤ f ≤ v + X k ≥ kv k . This leads to the same contradiction as in the proof of Lemma 5.The proof of Lemma 6 can be easily modified to get the following.
Lemma 7.
If two angles together appear at least twice at every degree vertex in a tiling of sphere by pentagons with the same angle combination,then the two angles together appear at least times in the pentagon. Lemma 8.
Suppose an angle θ does not appear at degree vertices in a tilingof sphere by pentagons with the same angle combination.1. There can be at most one such angle θ .2. The angle θ appears only once in the pentagon.3. v + v ≥ .4. One of αθ , θ , θ is a vertex, where α = θ . The first statement implies that the angle α in the fourth statement mustappear at a degree 3 vertex. Proof.
Suppose two angles θ and θ do not appear at degree 3 vertices. Thenthe total number of times these two angles appear is at least 2 f , and is atmost the total number P k ≥ kv k of angles at high degree vertices. Thereforewe have 2 f ≤ P k ≥ kv k . On the other hand, by (2.1), we have2 f − X k ≥ kv k = 24 + X k ≥ k − v k > . The contradiction proves the first statement.The argument above also applies to the case θ = θ , which means thesame angle appearing at least twice in the pentagon. This proves the secondstatement.The first two statements imply that θ appears exactly f times. Since thisshould be no more than the total number P k ≥ kv k of angles at high degreevertices, by (2.1), we have0 ≥ f − X k ≥ kv k = 12 − v − v + X k ≥ ( k − v k . This implies the third statement.For the last statement, we assume that αθ , θ , θ are not vertices. Thismeans that θ appears at most twice at any degree 4 vertex, and at mostfour times at any degree 5 vertex. Since θ also does not appear at degree3 vertices, the total number of times θ appears is ≤ v + 4 v + P k ≥ kv k .8owever, the number of times θ appears should also be f . Therefore f ≤ v + 4 v + P k ≥ kv k . On the other hand, by (2.1), we have f − v + 4 v + X k ≥ kv k ! = 12 + X k ≥ ( k − v k > . We get a contradiction.
Two pentagons have the same edge combination if the five edge lengths areequal. For example, the first two pentagons in Figure 2 have the same edgecombination a b c . If we further have the same edge length arrangement in the two pentagons, then the two pentagons are edge congruent . For ex-ample, the first two pentagons in Figure 2 have the respective edge lengtharrangements a, b, a, b, c and a, a, b, b, c , and are therefore not edge congruent. Lemma 9.
In an edge-to-edge tiling of the sphere by edge congruent pen-tagons, the edge lengths of the pentagon is arranged in one of the six waysin Figure 2, with distinct edge lengths a, b, c . Moreover, the first of Figure 2cannot be a -tile, and, in case the pentagon is a - or -tile, the fifthvertex of degree or is opposite to the c -edge. abc a b c a b c a bc a b a b a Figure 2: Edges in pentagon suitable for tiling, a, b, c distinct.
Proof.
The lemma is an extension of [3, Proposition 7]. We follow the ar-gument in the earlier paper. By purely numerical consideration, there areseven possible edge combinations ( a, b, c, d, e are distinct) abcde, a bcd, a b c, a bc, a b , a b, a . For abcde , without loss of generality, we may assume that the edges arearranged as in the first of Figure 3. By Lemma 1, we may further assume9hat the vertex shared by c, d has degree 3. Let x be the third edge at thevertex. Then x, c are adjacent in a tile, and x, d are adjacent in another tile.Since there is no edge in the pentagon that is adjacent to both c and d , weget a contradiction. eab c dx aab c dx bca d ax bba c ax Figure 3: Not suitable for tiling.The combination a bcd has two possible arrangements, given (without lossof generality) by the second (adjacent a ) and third (separated a ) of Figure3. By Lemma 1, we may further assume a degree 3 vertex with the thirdedge x . In the second of Figure 3, the edge x is adjacent to both c and d , acontradiction. In the third of Figure 3, the edge x is adjacent to both a and d , again a contradiction.The combination a b c has three possible arrangements, given by the firstand second of Figure 2 and fourth of Figure 3. In the fourth of Figure 3,we may assume (by Lemma 1) a degree 3 vertex with the third edge x . Theedge x is adjacent to both a and c , a contradiction.The combination a bc has two possible arrangements, given by the thirdof Figure 2 ( b, c adjacent) and the left of Figure 4 ( b, c separated). In case b, c are separated, the pentagon has one a -angle that we denote by α , two ab -angles, and two ac -angles. Since there are no b -angle, c -angle, bc -angle,any degree 3 vertex must be one of three on the left of Figure 4. This impliesthat the a -angle α appears at every degree 3 vertex. By Lemma 5 (and theremark after the proof of the proposition), the pentagon should have at leasttwo a -angles, a contradiction. α α α α α α α Figure 4: Not suitable for tiling.The combination a b has two possible arrangements, given by the fourthof Figure 2 ( b adjacent) and the right of Figure 4 ( b separated). In case10he two b -edges are separated, the pentagon has one a -angle α and four ab -angles. Since there is no b -angle, any degree 3 vertex must be one oftwo on the right of Figure 4. This implies that the a -angle α appears atevery degree 3 vertex. By Lemma 5, the pentagon should have at least two a -angles, a contradiction.Finally, the first of Figure 2 has three ab -vertices (shared by a -edge and b -edge). If any such ab -vertex has degree 3, then the third edge at the vertexis adjacent to both a and b . The only such edge in the pentagon is c .If two ab -vertices of degree 3 are adjacent, then we have two c -edgesat the two vertices, and these two c -edges belong to the same pentagon, acontradiction. Therefore we cannot have adjacent ab -vertices of degree 3.The observation implies the final statement of the lemma. In this paper, we study tilings by pentagons with edge combinations a b c and a bc . By Lemma 9, there are three possible ways of arranging the edges.Then we denote the angles as in Figure 5. αβ γδ ǫ αβ γδ ǫ αβ γδ ǫ Figure 5: Edges and angles for a b c and a bc , with a, b, c distinct.Lemma 9 only assumes that the edge lengths a, b, c are distinct. As re-marked after Lemma 5, this may cause some ambiguity about angles. Onesource of ambiguity is that angles with distinct notations may have equalvalue. In this case, we may further use the edge lengths to distinguish the an-gles. The simplest case is the second of Figure 5, where the angles α, β, γ, δ, ǫ can be unambiguously described as ab -angle, a -angle, b -angle, ac -angle, bc -angle. In the first of Figure 5, there is no ambiguity for the bc -angle δ and ac -angle ǫ . However, α, β, γ are all ab -angles, and can be distinguished onlyby referring to the pentagon (not adjacent to δ, ǫ , adjacent to δ , and adja-cent to ǫ ). Similarly, in the third of Figure 5, α, β, γ are bc -angle, ab -angle, ac -angle. Then δ is the a -angle adjacent to β , and ǫ is the a -angle adjacentto γ . 11e introduce a notation for angle and edge configurations at vertices.First we note that angles are bounded by edges of certain lengths. For exam-ple, for α , we indicate the bounding edges by denoting α (or α ) for the firstand second of Figure 5, and α (or α ) for the third of Figure 5. Then we maydenote the vertices in Figure 6 by β β γ · · · , α ǫ δ · · · , α α β · · · . Here · · · consists of the additional angles at the vertex, and is called the remain-der . We may also denote α ǫ δ · · · by δ ǫ α · · · , α ǫ δ · · · , α ǫ · · · , δ ǫ · · · ,etc. More briefly, we may omit the edge lengths and denote α ǫ δ · · · by αδǫ · · · , αǫ · · · , αδ · · · , δǫ · · · , etc. Without indicating the edge lengths, thenotation no longer implies order or adjacency, unless we explicitly say consec-utive . For example, αδ · · · simply means that vertex has at least one α andat least one δ , and consecutive αδ means α and δ are adjacent at a vertex. δβ αβδ αǫ γα βα γǫδ γβ δǫ βα γαβ γδ βα Figure 6: Adjacent angle deduction.If the first vertex in Figure 6 is part of a tiling by the first of Figure 5,then we may determine the arrangements for the three tiles and get vertices α α · · · , α δ · · · just off β β γ · · · . The vertices are obtained by the followingprocess. First we denote angles α, δ adjacent to β (in the pentagon) by β → δα (or β → αδ ), and denote angles α, ǫ adjacent to γ by γ → ǫα .We may then combine these into the adjacent angle deduction β β γ → δα αδ ǫα . On the right, we can clearly see the vertices α α · · · and α δ · · · . Similarly, ifthe second and third of Figure 6 are parts of tilings by the second and thirdof Figure 5, and we get adjacent angle deductions α ǫ δ → βγ γδ ǫβ , α α β → βγ γβ αδ . The adjacent angle deduction is an efficient notation that deduce newvertices from existing ones without drawing pictures. But we need to beaware of the possible ambiguities for the angles, because the configurationat a vertex alone may not be enough to uniquely identify some angles. The12eaders are advised to double check adjacent angle deductions by drawingpictures.An adjacent angle deduction may have several outcomes, due to possiblydifferent ways of arranging the tiles in compatible way. For example, consec-utive γδ for the third of Figure 5 has two possible adjacent angle deductions γδ = γ δ → αǫ βγ , αǫ γβ . We denote all the possible outcomes by αǫ ( βγ ) , where ( βγ ) can be either βγ or γβ .The adjacent angle deduction is symmetric with respect to flipping (cor-responding to the change of vertex orientation). For example, the adjacentangle deduction above for γδ is also δ γ → γβ ǫα , βγ ǫα . If all the angles at a vertex are included, then the adjacent angle deductionis circular . For example, a vertex γ δ for the third of Figure 5 is γ γ δ , andis also δ γ γ or γ δ γ . Correspondingly, we have the same adjacent angledeductions γ γ δ → ǫα αǫ ( βγ ) ,δ γ γ → ( βγ ) ǫα αǫ ,γ δ γ → αǫ ( βγ ) ǫα . As an application of the technique of adjacent angle deduction, let usconsider a vertex γ k in a tiling by the second of Figure 5. The vertex is k copies of γ , and adjacent angle deduction at the vertex is γ γ · · · γ γ → ( αǫ ) ( αǫ ) · · · ( αǫ ) ( αǫ ) . Since the deduction is circular, we find that on the right, the number of α α is the same as the number of ǫ ǫ . In particular, if α α appears on the right,then ǫ ǫ also appears.
Given any tiling of an oriented surface, we add two vertices to each edge.If an edge belongs to a tile, then we can use the orientation to label the13wo vertices as the first and the second, with respect to the tile (and theorientation). See the labels for the five edges of the pentagonal tile on theleft of Figure 7. Note that each edge is shared by two tiles, and the labelsfrom the viewpoints of two tiles are different. See the edge shared by thepentagon and the triangle below it.We further add one vertex at the center of each tile, and connect the centervertex to the first vertex of each boundary edge of the tile. This divides an m -gon tile into m pentagons. See the right of Figure 7. The process turns anytiling into a pentagonal tiling. We call the process pentagonal subdivision .Each tile in the pentagonal subdivision has three edge vertices, one oldvertex (from the original tiling), and one center vertex. The edge verticeshave degree 3. The degree of the old vertex remains the same. The degree ofthe center vertex is the “degree” (number of edges) of the original tile. It iseasy to see that the pentagonal subdivisions of a tiling and its dual tiling arecombinatorially the same, with old vertices and center vertices exchanged. Figure 7: Pentagonal subdivision.Suppose we start with a regular tiling, and the distance from the new edgevertices to the nearby original vertices are the same. Then the pentagonalsubdivision is a tiling by congruent pentagons. See Figure 8. In fact, we nolonger require the original edges to be straight. This means that we onlyrequire α + δ + ǫ = 2 π instead of α + δ = ǫ = π . Moreover, if the originalregular tiling is made up of m -gons and each vertex has degree n , then β = πm and γ = πn .The regular tilings of the sphere are the five platonic solids. Since dualtilings give the same pentagonal subdivision, the pentagonal subdivision givesthree tilings of the sphere by congruent pentagons.1. Pentagonal subdivision of tetrahedron: f = 12.14 βγδ ǫα βγ δǫ γαβ γ δǫαβ γδ ǫγ aa b bcαβ γδ ǫ Figure 8: Pentagonal subdivision of regular tiling.2. Pentagonal subdivision of cube or octahedron: f = 24.3. Pentagonal subdivision of dodecahedron or icosahedron: f = 60.The first tiling is the deformed dodecahedron T in [3]. Further by [1], thisis the only edge-to-edge tiling of the sphere by 12 congruent pentagons. Thesecond tiling is the first of Figure 9. P QC MVP ′ Figure 9: Pentagonal subdivision of platonic solid.The three tilings are pentagonal subdivisions of three regular triangulartilings (corresponding to m = 3 and n = 3 , , πn = π , π , π at the vertex V . Let C be the center of triangle, and let M be the middle point of an edge oftriangle. The pentagonal subdivision is completely determined by a point P in the following way. The point P determines another point Q , by requiringthat M is the middle point of the great arc P Q . Moreover, rotating P around C by π gives P ′ . Then C, P, Q, V, P ′ are the vertices of a tile in thepentagonal subdivision. Rotating this tile around C by π and π give two15ore tiles, and the three tiles together form the pentagonal subdivision of oneregular triangular tile of the original platonic solid. The whole pentagonalsubdivision tiling is obtained by replacing each triangle in the platonic solidby the second of Figure 9. The description gives a one-to-one correspondencebetween the tiling and the location of P , subject to the only condition thatthe three pentagons in the second of Figure 9 do not overlap. In particular,the pentagonal subdivision tiling allows two free parameters. The study ofthe 2-dimensional moduli of pentagonal subdivision tiling will be the subjectof another paper. Given any tiling of a surface, we have the dual tiling. The original tiling andthe dual tiling give a combined tiling by quadrilaterals. If the surface is alsooriented, then we may use the orientation to further cut each quadrilateralinto two halves in a compatible way. The process turns any tiling into apentagonal tiling. We call the process double pentagonal subdivision .Figure 10: Double pentagonal subdivision.Similar to pentagonal subdivision, we hope that the double pentagonalsubdivision of a platonic solid can be a tiling of the sphere by congruent pen-tagons. Since the dual tiling gives the same double pentagonal subdivision,we only need to consider the double pentagonal subdivision of one triangletile in regular tetrahedron, octahedron, icosahedron (each vertex has degree n = 3 , , α = π, β = (cid:0) − n (cid:1) π, γ = δ = π, ǫ = n π. The number of tiles are respectively f = 24 , , βγ δ ǫ αβγδ ǫ ααγγ δǫǫαβγ δǫαβγ δǫ ααγ γδǫ ǫ αβ γδǫαβ γ δ ǫαα γ γδ ǫ ǫ a a ab cαβ γδ ǫ Figure 11: Double pentagonal subdivision of regular tiling.7 degrees of freedom. On the other hand, the specific values of 5 anglesand 3 equal edges amount to 7 equations. Therefore we expect the doublepentagonal subdivision tilings to be isolated examples. This can be arguedas follows. We connect three edges of length a together at angles δ = π and ǫ = n π . Then at the two ends, we produce the b -edge and c -edge at angles β = (1 − n ) π and γ = π with the existing a -edges. They meet and createa pentagon P ( a ) that is determined by a . In fact, b, c, α are also determinedby a . The area of P ( a ) is A ( a ) = α + β + γ + δ + ǫ − π = α + ( n − ) π, and is a strictly increasing function of a . When a is very small, the pentagonis very small, and therefore α ( a ) = ( − n ) π + A ( a ) is slightly bigger than( − n ) π . When a grows, A ( a ) will reach ( n − ) π , and the angle α will reachthe desired value π , and we get the desired pentagon.Next we calculate the exact size of the pentagons in the double pentag-onal subdivision tilings. For a triangle face in the tetrahedron, octahedron,icosahedron, we form the triangle in the first of Figure 12 by connecting thecenter point δ to the middle point α of the edge and the vertex ǫ n (thepoints are the center and the middle by symmetry). We know the angles ofthe triangle, and denote the edges by x, y, z . The edges are determined bythe anglescos x = cot π cot n π, cos y = cos n π sin π , cos z = cos π sin n π . The first of Figure 11 shows that the x -edge connects the two edges ofthree connected a -edges at alternating angles δ and ǫ . See the second of17 y zα δ ǫ n π π n π a a ax δǫδ ǫ n ya bδ α β Figure 12: Calculate pentagon in double pentagonal subdivision tiling.Figure 12. By [1, Lemma 3], we havecos x = cos a + sin a (sin a cos a cos ǫ + cos a sin a cos δ ) − sin a sin a (sin ǫ sin δ + cos a cos ǫ cos δ )= cos a (1 − cos δ )(1 − cos ǫ ) + cos a sin δ sin ǫ + cos a (cos δ + cos ǫ − cos δ cos ǫ ) − sin δ sin ǫ. For f = 24, this meanscos x = 13 = 94 cos a + 34 cos a −
54 cos a − . The cubic equation has only one real solution (the approximate value means0 . π ≤ a < . π )cos a = 29 q
19 + 3 √
33 + 89 p
19 + 3 √ − , a ≈ . π. The first of Figure 11 also gives the triangle in the third of Figure 12. Thenwe have the cosine lawcos y = cos a cos b + sin a sin b cos β. This can be regarded as a quadratic equation for cos b or sin b . Then we canget a precise formula for cos b and sin b in square and cubic roots, and we get b ≈ . π . Since β = γ = δ = ǫ for f = 24, the pentagon is symmetric,and we have b = c .For f = 48, we get the cubic equation for cos a cos x = 1 √ a + √
32 cos a −
12 cos a − √ . a = 19 q √ √
35 + 43 p √ √ − √ , a ≈ . π. We can also get the precise formula for cos b, sin b, cos c, sin c , and then get b ≈ . π , c ≈ . π .For f = 120, we get the cubic equation for cos a cos x = √ q − √
5) = 18 (cid:18) − √
5) cos a + q √
5) cos a +( − √
5) cos a − q √ (cid:19) . The cubic equation has only one real solution. We have precise formula forcos a, cos b, cos c , and then get a ≈ . π, b ≈ . π, c ≈ . π . a b c By Lemma 9, for an edge-to-edge tiling of the sphere by congruent pentagonswith edge combination a b c , the pentagon is the first or the second of Figure5. Moreover, in case the pentagon is a special tile in Lemma 1, the first mustbe a 3
4- or 3 H of degree 4 or 5 is the vertex where α is. Therefore our classification can be divided into four cases in Figure 13,with the dot vertex being the vertex H of the special tile. The case thatthere is a 3 -tile is part of the second pentagon, with the dot vertex havingdegree 3. The detailed argument is in Section 4.2. This means that there isno 3 -tile in Sections 4.1, 4.3, 4.4, 4.5. In particular, the dot vertex alwayshas degree 4 or 5 in these sections, and by Lemma 2, we may assume f ≥ αβ γδ ǫ αβ γδ ǫ αβ γδ ǫ αβ γδ ǫ a b c .19ur strategy is to first tile the partial neighborhood in the fourth of Figure1. Then we study the configuration around H , and further tile beyond thepartial neighborhood if necessary. We will always start by assuming that thecenter tile P is given by the pentagon in Figure 13.To facilitate discussion, we denote by P i the tile labeled i , by E ij the edgeshared by P i , P j , and by V ijk the vertex shared by P i , P j , P k . We denote by A i,jk the angle of P i at V ijk , and by θ i the angle θ in P i . When we say a tileis determined , we mean that we know all the edges and angles of the tile. ( a b c ) Let the first of Figure 13 be the center tile P in the partial neighborhood inthe first of Figure 14. Since E and E are adjacent to both a and b , weget E = E = c . This determines (all edges and angles of) P , P , P , P .Then we know three edges of P , which further determines P .The angle sums at βδǫ, γδǫ and the angle sum for pentagon (Lemma 4)imply β = γ and α + β = (1 + f ) π . By the edge length consideration, wehave H = α βγ, αβ γ, αβγ , αβγδǫ . By 2 α + β + γ = 2( α + β ) > π and α + β + γ + δ + ǫ = (3 + f ) π > π , we get H = α βγ, αβγδǫ .Note that the partial neighborhood tiling is symmetric with respect tothe exchange a ↔ b, β ↔ γ, δ ↔ ǫ . Therefore up to the symmetry, we mayassume H = αβ γ . By the extra angle sum at H , we get α = ( + f ) π, β = γ = ( − f ) π, δ + ǫ = ( + f ) π. We have adjacent angle deduction at H = α β β γβ γ → αδ αǫ . This gives a vertex α δ · · · = α δ · · · . Since an angle bounded by c -edge is δ or ǫ , the vertex is α δ δ · · · or α δ ǫ · · · . By α + δ + ǫ = (2 + f ) π > π , α δ ǫ · · · is not a vertex. Therefore the vertex is α δ δ · · · , and we haveadjacent angle deduction δ δ → βǫ ǫβ at the vertex. Therefore ǫ ǫ · · · = ǫ ǫ · · · is a vertex. Since there is no a -angle, we get ǫ ǫ · · · = θ ǫ ǫ ρ · · · forat least two angles θ, ρ .Since α δ δ · · · is a vertex, the angle sum at the vertex implies δ ≤ (2 π − α ) = ( − f ) π . Then ǫ = ( + f ) π − δ ≥ ( + f ) π > α > β = γ .This implies that the minimal value among the five angles is either β or δ .Therefore the angle sum at θ ǫ ǫ ρ · · · is ≥ β + 2 ǫ or 2 δ + 2 ǫ . By β + ǫ ≥ ( − f ) π + ( + f ) π > π and δ + ǫ = ( + f ) π > π , we get a contradiction.20 β γδ ǫβ δ ǫγαǫγα βδδǫγ α ββδǫγ α γǫδβ α αβ γδ ǫ αβ γδ ǫ αβ γδ ǫǫγ α βδβδǫ γα αβ δǫγγβ β γδβ α γǫγβ Figure 14: Cases 1 and 2 for a b c . ( a b c ) Let the second of Figure 13 be the center tile P in the partial neighborhoodin the second and third of Figure 14. The two pictures show two possible waysof arranging edges and angles of P . In the first picture, we may determine P , P , and then get b -angle γ and a -angle β . In the second picture, wemay get b -angle γ and a -angle β . Then E = a or c , and either way gives γ . Similarly, we get β .In the second of Figure 14, the angle sums at αδǫ, β , γ and the anglesum for pentagon imply f = 12. By [1, 3], the whole tiling is the deformeddodecahedron, which is the pentagonal subdivision of regular tetrahedron.We note that by the edge length consideration, the tile P in the third ofFigure 14 cannot be a 3 -tile. Therefore the case of edge combination a b c and there is a 3 -tile is included in the second of Figure 14. This meansthat we may assume there is no 3 -tile in the subsequent discussion for edgelength combination a b c .In the third of Figure 14, by the edge length consideration, the vertex H = α βγ, α βγ , α β γ, αβγδǫ . By Lemma 4, αβγδǫ is not a vertex. Moreover,the first three possibilities imply 2 α + β + γ < π . Combined with the anglesums at βδ , γǫ , we get α + β + γ + δ + ǫ = ((2 α + β + γ ) + ( β + 2 δ ) + ( γ + 2 ǫ )) < π, again contradicting Lemma 4. ( a b c ) Let the third of Figure 13 be the center tile P in the partial neighborhoodin Figure 15. We consider two possible arrangements of P . The first picture21hows one arrangement, and the second and third show the other arrange-ment. We label the three partial neighborhood tilings as Cases 3.1, 3.2, 3.3.We also recall that, after Case 2, the degree of H is 4 or 5.In the first picture, we get b -angle γ and a -angle β . Then E = a or c , and either way gives γ and E = b . This determines P .In the second and third pictures, we may determine P , P . Then the twopictures show two possible arrangements of P . In the second picture, we get a -angle β . In the third picture, we may determine P . βδ αǫ γ βδ αǫ γ βδ αǫ γ γ γδβα γ ǫβ β α γǫδ ǫγα β δ αβ δ ǫγ γ ǫ δβαβαγǫ δ β ǫγα β δ αβ δ ǫγ γα βδǫβαγǫ δ β δ ǫγα a b c . Case 3.1
The angle sums at α γ, γǫ , βδ and the angle sum for pentagon imply α = ǫ = π − γ, β = f π, δ = (1 − f ) π. The angle A , = α or δ since it is adjacent to β . By the edge length consid-eration and the fact that α γ, βδ are vertices, we get H = α β , α β , αβ δǫ .Suppose H = αβ δǫ . The extra angle sum at H implies α = ǫ = ( − f ) π, β = f π, γ = (1 + f ) π, δ = (1 − f ) π. By the edge length consideration, we have H = α ǫ · · · , and adjacent anglededuction α ǫ → βγ γδ . Therefore γ γ · · · is a vertex, contradicting γ > π .Suppose H = α β , α β . The extra angle sum at H implies H = α β : α = ǫ = (1 − f ) π, β = f π, γ = f π, δ = (1 − f ) π ; H = α β : α = ǫ = (1 − f ) π, β = f π, γ = f π, δ = (1 − f ) π. By the edge length consideration, we have H = α α · · · , and adjacent an-gle deduction α α → βγ γβ . Therefore γ γ · · · is a vertex. If the re-mainder of γ γ · · · has one α or ǫ , then by the edge length consideration,22he remainder has another α or ǫ . By γ + α > π and γ + ǫ > π , we geta contradiction. Therefore there is no α and ǫ in the remainder. Thenby the edge length consideration, the remainder also has no β , and we get γ γ · · · = γ γ · · · γ = γ k . We have adjacent angle deduction γ γ · · · γ γ → ( αǫ ) ( αǫ ) · · · ( αǫ ) ( αǫ ) . Recall that the initial pair γ γ is obtained from adjacent angle deductionon α α . Therefore we have α α on the right side of the adjacent anglededuction above. This implies that ǫ ǫ also appears. See the example at theend of Section 2.4. Therefore ǫ ǫ · · · = θ ǫ ǫ ρ · · · is a vertex, with θ, ρ = δ or ǫ by the edge length consideration. By δ > ǫ , the angle sum at ǫ ǫ · · · is ≥ ǫ . On the other hand, the angle sum at γ k implies γ ≤ π . Therefore weget f ≥
24 in case H = α β and f ≥
36 in case H = α β . This implies4 ǫ > π in both cases, and we get a contradiction. Case 3.2
The angle sums at α β, αδǫ, γ and the angle sum for pentagon imply α = ( − f ) π, β = ( + f ) π, γ = π, δ + ǫ = ( + f ) π. The angle A , = α or δ . By the edge length consideration and the factthat α β, αδǫ are vertices, we get H = β δ , β δ . The extra angle sum at H implies H = β δ : δ = ( − f ) π, ǫ = ( + f ) π ; H = β δ : δ = ( − f ) π, ǫ = ( + f ) π. Consider the vertex ǫ ǫ · · · shared by P , P in the second of Figure 15. Bythe edge length consideration, the vertex is θ ǫ ǫ ρ · · · , with θ, ρ = δ or ǫ . By δ + ǫ > π and 2 ǫ > π , we get a contradiction. Case 3.3
The angle sums at αδǫ, γ and the angle sum for pentagon imply α + δ + ǫ = 2 π, β = ( + f ) π, γ = π. By the edge length consideration, we have H = α β , β δ , β , β .23uppose H = α β . The extra angle sum at H implies α = ( − f ) π, β = ( + f ) π, γ = π, δ + ǫ = ( + f ) π. By the third of Figure 15, we have adjacent angle deduction β α α β → αδ βγ γβ αδ at H . This gives a vertex β δ · · · = β δ δ · · · or β δ ǫ · · · .The vertex β δ δ · · · implies δ ≤ π − β = ( − f ) π , ǫ = ( + f ) π − δ ≥ ( + f ) π , and adjacent angle deduction δ δ → βǫ ǫβ . Then we have avertex ǫ ǫ · · · , with the remainder having at least one γ or at least two from α, ǫ , by the edge length consideration. By ǫ ≥ ( + f ) π and f ≥
24, thisimplies that the angle sum at ǫ ǫ · · · is always > π , a contradiction.The remainder of the vertex β δ ǫ · · · has value ( − f ) π < α, β, γ , andtherefore has only δ, ǫ . By the edge length consideration, the remainder hasat least one δ and at least one ǫ . This contradicts δ + ǫ > ( − f ) π .Suppose H = β δ . The extra angle sum at H implies α + ǫ = ( + f ) π, β = ( + f ) π, γ = π, δ = ( − f ) π. By the third of Figure 15, we have adjacent angle deduction β δ δ β → αδ βǫ ǫβ αδ at H . This gives vertices β α · · · and ǫ ǫ · · · . By the edgelength consideration, we have β α · · · = β α α · · · , β α γ · · · , β α ǫ · · · .We also know that the remainder of ǫ ǫ · · · has at least one γ or at leasttwo from α, ǫ .The angle sum at β α α · · · or β α γ · · · gives an upper bound for α that further implies (using α + ǫ ) ǫ > π . Then α + 2 ǫ = ( α + ǫ ) + ǫ > π ,2 γ + 2 ǫ > π , 4 ǫ > π , so that ǫ ǫ · · · is not a vertex, a contradiction.The remainder of β α ǫ · · · has value ( − f ) π < β, γ, δ , and thereforehas only α, ǫ . By the edge length consideration, the remainder has at leastone α and at least one ǫ . This contradicts α + ǫ > ( − f ) π .We will discuss the remaining cases H = β , β in Section 4.5. ( a b c ) Let the fourth of Figure 13 be the center tile P in the partial neighborhoodin Figure 16. We consider two possible arrangements of P . The first andsecond pictures show one arrangement. The third picture shows the otherarrangement. We label the three partial neighborhood tilings as Cases 4.1,4.2, 4.3. Again, the degree of H is 4 or 5.24n the first and second pictures, we may determine P and b -angle γ .Then the two pictures show two possible arrangements of P . In the firstpicture, we may determine P and a -angle β . In the second picture, we get a -angle β . Then E = b or c , and either way gives β . In the third picture,we get b -angle γ . Then E = a or c , and either way gives γ . This furtherdetermines P and gives a -angle β . δǫ βγ α δǫ βγ α δǫ βγ α αβδ ǫ γ γα β δǫ δ ǫ γαβǫδβα γ β αβδ ǫ γ γǫ δ βαβǫδβα γ β γ γ αγ δǫββδǫγα β a b c .In Case 4.1, the angle sums at αδǫ, β , γ and the angle sum for pentagonimply f = 12. After Case 2, the case can be dismissed.In Case 4.2, since αδǫ is a vertex, the remainder of H = βδǫ · · · has no α . Then by the edge length consideration, we have H = βδ ǫ . Then anglesums at α β, γ and H imply α + β + γ + δ + ǫ = (2 α + β ) + γ + ( β + 2 δ + 2 ǫ ) = π, contradicting the angle sum for pentagon.In Case 4.3, the angle sums at α γ, β , γǫ and the angle sum for pentagonimply α = ǫ = π − γ, β = π, δ = ( + f ) π. Since α γ is a vertex and β + 4 δ > π , by the edge length consideration, wehave H = α βδ , α δ , αδ ǫ, δ .The angle sum at H implies H = α βδ : α = ǫ = ( − f ) π, β = π, γ = ( + f ) π, δ = ( + f ) π ; H = α δ : α = ǫ = ( − f ) π, β = π, γ = ( + f ) π, δ = ( + f ) π ; H = αδ ǫ : α = ǫ = ( − f ) π, β = π, γ = (1 + f ) π, δ = ( + f ) π. By the edge length consideration, H = α βδ , α δ is α α · · · , and H = αδ ǫ is α ǫ · · · . By adjacent angle deductions α α → βγ γβ and α ǫ → γ γδ , we know γ γ · · · is a vertex. For H = α βδ , αδ ǫ , this contradicts γ > π . For H = α δ , the remainder of γ γ · · · has value ( − f ) π <α, β, γ, δ, ǫ . Therefore the remainder is a single δ . However, γ δ is not avertex by edge length consideration. We get a contradiction.We will discuss the remaining case H = δ in Section 4.5. After Sections 4.1, 4.2, 4.3, 4.4, the only remaining cases for the edge com-bination a b c are the following.Case 3.3, H = β : The extra angle sum at H gives α + δ + ǫ = 2 π, β = π, γ = π, f = 24 . Case 3.3, H = β : The extra angle sum at H gives α + δ + ǫ = 2 π, β = π, γ = π, f = 60 . Case 4.3, H = δ : The extra angle sum at H gives α = ǫ = π − γ, β = π, δ = π, f = 24 . The partial neighborhood tiling for Case 3.3 is the third of Figure 15, andfor Case 4.3 is the third of Figure 16. Moreover, we know there is no 3 -tileafter Case 2.By Lemma 2, for f = 24, we know each tile is a 3 H = δ , the four tiles P , P , P , P around a vertex δ must be the first of Figure17. Since every tile is a 3 δ have degree 3. In particular, we have tiles P , P , P as indicated, withtheir edges (those not shared with P , P , P ) and angles to be determined.Up to the symmetry of vertical flipping, we may assume that the edges of P are given as indicated. This determines one b -edge and one c -edge of P , andtherefore determines all edges of P . Then we find that P has 3 b -edges, acontradiction. This proves that Case 4.3, H = δ , allows no tiling.Next consider Case 3.3, H = β . Again every tile is a 3 β . First assume that adjacent tiles always have oppositeorientations. Then we get the second of Figure 17. Since every tile is a 3 δ δδ ββ ββ δ ǫγαβ ǫ γδ αβα βδǫγδǫγ α βǫ γδ αβαβδǫ γ β ǫγδαβββ Figure 17: Case 4.3, H = δ ; and Case 3.3, H = β .is a c -angle, a contradiction. Therefore there are two adjacent tiles around β with the same orientation, say P , P in the third of Figure 17. We mayassume that the edges and angles of P , P are given as indicated. Since everytile is a 3 P , P , . . . , P , with their edges and anglesto be determined.The edges E = c and E = b determine P . The edges E = b and E = a determine P . Since P already has two b -edges, and E = c , wehave E = a . This determines P , P . Then E = a and E = b determine P . This shows that, by starting with P , P , we can derive P . By repeatingwith P , P in place of P , P , we can further derive P . We have shownthat the neighborhood of β is uniquely given by four tiles with the sameorientation.It remains to show that the vertex β · · · shared by P , P is β . If thevertex has degree 3, then by the edge length consideration, it is β , contra-dicting the fact that β is already a vertex. Since every tile is a 3 β · · · has degree 4, and the vertex αǫ · · · shared by P , P has degree3. Therefore we have a tile P outside P , P . Then E = a and E = c determine P . In particular, we know A , = β . Since V has degree 4, bythe edge length consideration, we get V = β .By starting at a vertex V = β , we find another vertex V = β . Wemay repeat the argument that started from V = β by restarting from V = β . After repeating six times, we get the pentagonal subdivision ofcube (or octahedron).The final remaining case is Case 3.3, H = β . After finishing Case 3.3, H = β , and Case 4.3, H = δ , we may further assume that there is no 3 f = 60, we know each tile is a 3 β , there must be two adjacent tiles with the same orientation.Then the argument given by the third of Figure 17 can be carried out, becausethe argument never uses P . The argument shows that all five tiles have thesame orientation, and also derives another degree 5 vertex V = β · · · ,with three tiles P , P , P having the same orientation.By the edge length consideration, we have V = β , α β , β δ . The latertwo cases are given by Figure 18, with P , P (these are P , P in the third ofFigure 17) having the same orientation. In both pictures, we can determine P , P , and then determine P . Since every tile is a 3 ǫ · · · shared by P , P and the vertex αβ · · · shared by P , P have degree 3. Bythe edge length consideration, the vertices are γǫ , α β . Moreover, P , P , P also share a vertex βδ . We also have vertices αδǫ, γ in the third of Figure15. The angle sums at the five degree 3 vertices imply all the angles are π , contradicting β = π . Therefore V = β · · · is β , and the processof deriving new V = β from existing V = β can be repeated. Afterrepeating the process twelve times, we get the pentagonal subdivision ofdodecahedron (or icosahedron). β β δ ǫγαα βδǫγαγǫδβ βδǫγ αδǫγ α β β δ ǫγαδ βαγǫδǫγαβ βδǫγ αδǫγ α Figure 18: Case 3.3, H = β : β · · · must be β . a bc By Lemma 9, for an edge-to-edge tiling of the sphere by congruent pentagonswith edge combination a bc , the pentagon is the third of Figure 5. Again westart with the neighborhood of a special tiling in Lemma 1. In case the tileis 3
4- or 3 H of degree 4 or5. The classification is then divided into three cases in Figure 19. We willinclude the discussion of the neighborhood of 3 -tile in the first case. Thismeans we will assume there is no 3 -tile in the second and third cases.28 β γδ ǫ αβ γδ ǫ αβ γδ ǫ a bc . ( a bc ) Let the first of Figure 19 be the center tile P in the partial neighborhood inFigure 20. We consider whether E = a or E = a . If both are not a -edges,then we get the first picture. If one is a -edge and the other is not, by thesymmetry of exchanging b and c , we may assume that E = b and E = a .This is the second picture. If both are a -edges, then we get the third picture.We label the three partial neighborhood tilings as Cases 1.1, 1.2, 1.3. αβ γδ ǫ αβ γδ ǫ αβ γδ ǫ γ α βδǫβδ ǫ γααγǫδβα βαγǫ δ γ α βδǫβδ ǫ γααβδǫ γ α γ ǫδβαβδǫ γ a bc .In Case 1.1, since V = αβγ is a vertex, the vertex V = αβγ · · · hasempty remainder, so that P is a 3 -tile. We also note that, by the edgelength consideration, the tile P cannot be a 3 -tile in Cases 1.2 and 1.3.Therefore Case 1.1 is equal to the existence of 3 -tile for a bc . This meansthat, after Case 1.1, we may assume there is no 3 -tile. By Lemma 1, thisimplies f ≥ Case 1.1
No matter how we arrange angles of P , we always get vertices δ ǫ and δǫ . The angle sums at αβγ, δ ǫ, δǫ and the angle sum for pentagon imply f = 12. By [1, 3], the tiling is the deformed dodecahedron, and cannot haveedge combination a bc . 29 ase 1.2 By the edge length consideration, we have H = α βγ, α γ · · · . Since αβγ isa vertex, we have H = α βγ . We will further show that α + γ > π , so that H = α γ · · · .Consider the edge E . If E = a , then V and V are combinations of δ, ǫ . If the combinations are different, then the angle sums at the two verticesimply δ = ǫ = π . Combined with the angle sum at αβγ and the angle sumfor pentagon, we get f = 12, contradicting f ≥
24. The requirement of thesame combination of δ, ǫ implies A , = ǫ , A , = δ , A , = δ . This furtherdetermines P , P , and we get the first of Figure 21. If E = b or c , then wemay determine P , P and get the second and third of Figure 21. The anglesums at the four degree 3 vertices and the angle sum for pentagon imply E = a : α + γ = (1 + f ) π, β = δ = (1 − f ) π, ǫ = f π ; E = b : α + γ = ( − f ) π, β = ( + f ) π, δ = (1 − f ) π, ǫ = f π ; E = c : α = β = ( + f ) π, γ = ǫ = (1 − f ) π, δ = f π. In all cases, we have α + γ > π . αβ γδ ǫγ α βδǫβδ ǫ γααβδǫ γ αβ γδ ǫγ α βδǫβδ ǫ γααβδǫ γ αβ γδ ǫγ α βδǫβδ ǫ γααβδǫ γ δǫγ α βǫδβαγ δβα γ ǫδβαγǫ ǫγα β δǫγαβδ Figure 21: Case 1.2 for a bc . Case 1.3
By the edge length consideration, we have H = α , α βγ . In the third ofFigure 20, we also note that, since b and c are adjacent, one of E and E is a . By the symmetry of exchanging b and c , we may further assume that E = a . Then we consider three possibilities for E .The case E = a is illustrated by Figure 22. The vertices V and V are combinations of δ, ǫ . If the combinations are different, then the30ngle sums at the two vertices imply δ = ǫ = π . The angle sums at V and V further imply β = γ = π . By [3, Lemma 21], we get b = c , acontradiction. Therefore V and V are the same combinations of δ, ǫ .This implies A , = δ, A , = ǫ, A , = A , . Then we determine P andget two pictures according to A , = A , = δ or ǫ . αβ γδ ǫα γ ǫδβδǫγ α βαβδǫ γ αβ γδ ǫα γ ǫδβδǫγ α βαβδǫ γ ǫδ β αγǫδβαγ δǫ γ αβδǫγαβ Figure 22: Case 1.3 for a bc , E = a , E = a .In the first of Figure 22, the angle sums at β ǫ, γ ǫ, δ ǫ and the angle sumfor pentagon imply α = f π + ǫ, β = γ = δ = π − ǫ. By 3 α + β + γ > π , we get H = α βγ . On the other hand, if H = α , thenthe extra angle sum at H implies α = π, β = γ = δ = ( + f ) π, ǫ = (1 − f ) π. By the edge length consideration, the remainder of the vertex β β · · · sharedby P , P has two angles from α, β . This implies that the angle sum at thevertex is > π , a contradiction. By exchanging b ↔ c, β ↔ γ, δ ↔ ǫ , the sameargument shows that the second of Figure 22 also leads to contradiction.The case E = b is illustrated by Figure 23. This determines P , P .Then two possible ways of arranging P give two pictures.In the first of Figure 23, the angle sums at β δ, β ǫ, γ δ, δ ǫ imply β = γ = δ = ǫ . By [3, Lemma 21], this implies b = c , a contradiction. In thesecond of Figure 23, if H = α βγ , then the angle sums at β ǫ, γ δ, δ , H andthe angle sum for pentagon imply α = ( + f ) π, β = ( − f ) π, δ = γ = π, ǫ = ( + f ) π. Then the remainder of the vertex ǫ ǫ · · · shared by P , P has value ( − f ) π < β, γ, δ, ǫ . Since α is a bc -angle, we get a contradiction.31 β γδ ǫα γ ǫδβδβα γǫβδǫ γ ααβδǫ γ αβ γδ ǫα γ ǫδβδβα γǫβδǫ γ ααβδǫ γ δǫγαβ ǫδβαγ Figure 23: Case 1.3 for a bc , E = b , E = a .We will discuss the remaining case H = α (for second of Figure 23) inSection 5.4.The case E = c is illustrated by Figure 24. Then two possible ways ofarranging P give two pictures. αβ γδ ǫα γ ǫδβǫγα βδγǫδ β ααβδǫ γ αβ γδ ǫα γ ǫδβǫγα βδγǫδ β ααβδǫ γ δǫγαβ αβ γδ ǫγα ǫβδ ααα δ ǫδβαγ Figure 24: Case 1.3 for a bc , E = c , E = a .In the first of Figure 24, if H = α βγ , then the angle sums at β δ, γ ǫ, δǫ , H and the angle sum for pentagon imply α = ( + f ) π, β = ǫ = ( − f ) π, γ = ( + f ) π, δ = (1 + f ) π. The vertex γ δ · · · shared by P , P is α γ δ · · · or γ γ δ · · · . By 2 γ + δ >α + γ + δ > π , we get a contradiction. Therefore H = α , and the anglesums at β δ, γ ǫ, δǫ , H and the angle sum for pentagon imply α = π, β = ǫ = (1 − f ) π, γ = ( + f ) π, δ = f π. By α + β + γ > π , a vertex α · · · is not αβγ · · · . Then by the edge lengthconsideration, we have α · · · = α α · · · , α α · · · . Moreover, the remainderof α α · · · has at least two angles from α, γ , and the remainder of α α · · · α, β . By α = π , α + β > π , α + γ > π ,we conclude that α · · · = α . This gives P , P and determine all the edgesand angles of the two tiles. Note that f = 24 implies β = γ = δ = ǫ . By[3, Lemma 21], this implies b = c , a contradiction. Therefore f >
24, andthe remainder of the vertex β β · · · shared by P , P has value δ < α, β, γ, ǫ .This implies the vertex β β · · · is β δ . Now we have a vertex γ δθ · · · sharedby P , P , P , with θ adjacent to δ . Then θ = β or ǫ , and 2 γ + δ + θ > π , acontradiction.In the second of Figure 24, the angle sums at β ǫ, γ ǫ, δ ǫ and the anglesum for pentagon imply α = f π + ǫ, β = γ = δ = π − ǫ. By 3 α + β + γ > π , we have H = α βγ . Therefore H = α , and the extraangle sum at H implies α = π, β = γ = δ = ( + f ) π, ǫ = (1 − f ) π. By the edge length consideration, the vertex β δ · · · shared by P , P is α β δ · · · or β β δ · · · . The remainder of α β δ · · · has value ( − f ) π ,which is nonzero and strictly less than all the angles. We get a contradiction.The remainder of β δ · · · , has value ( − f ) π , which is strictly less thanall the angles. Therefore the remainder is zero. This implies f = 24 and β = γ = δ = ǫ , which by [3, Lemma 21] further implies b = c , again acontradiction. ( a bc ) Let the second of Figure 19 be the center tile P in the partial neighborhoodin the first of Figure 25. The b -edge of P and the c -edge of P imply E = a .This determines P , P . In particular, αβγ is a vertex. Then H = α β · · · has no γ . Therefore H = α α β · · · = α β , α β δ, α β ǫ . The angle sum at H implies α + β ≤ π . Then the angle sum at αβγ implies γ ≥ π . Therefore γ · · · is not a vertex. On the other hand, we have adjacent angle deduction α α → βγ γβ . Therefore γ · · · is a vertex, a contradiction. ( a bc ) Let the third of Figure 19 be the center tile P in the partial neighborhoodin the second of Figure 25. The b -edge of P and the c -edge of P imply33 δ αǫ γα β δǫγγαβ δǫ δǫ βγ αǫ γ αβδαβδ ǫγγαβ δ ǫδβαγ ǫ Figure 25: Cases 2 and 3 for a bc . E = a . This determines P , P , and further determines P , P . The anglesum at αβγ and the angle sum for pentagon imply α + β + γ = 2 π, δ + ǫ = (1 + f ) π. We claim that the remainder of H = δǫ · · · has no β β . If this is the case,then the angle sum at H implies β ≤ π − ( δ + ǫ ) = ( − f ) π, α + γ ≥ π − β ≥ ( + f ) π. Moreover, we have adjacent angle deduction β β → δα αδ at H . Therefore α α · · · = θ α α ρ · · · , with θ, ρ = α or γ , is a vertex. By α + γ > π , oneof θ, ρ is α . Then α α · · · = α α α · · · , and the angle sum at the verteximplies 4 α ≤ π or 3 α + γ ≤ π . Combined with α + γ ≥ ( + f ) π , wealways get γ > π . On the other hand, we have adjacent angle deduction α α → βγ γβ . Therefore γ · · · is a vertex, and we get a contradiction.This proves our claim that H = δǫ · · · has no β β . By the same reason, H has no γ γ . Moreover, by δ + ǫ > π , we have H = δ ǫ · · · . Since H hasdegree 4 or 5, by the edge length consideration, we get H = δ ǫ, δǫ , δ ǫ, δǫ .We will continue studying the case in Section 5.4. After Sections 5.1, 5.2, 5.3, the only remaining cases for the edge combination a bc are the following.Case 1.3, H = α , E = b , E = a : The extra angle sum at H gives α = π, β = ( − f ) π, γ = δ = π, ǫ = ( + f ) π. H = δ ǫ, δǫ , δ ǫ, δǫ : The extra angle sum at H gives α + β + γ = 2 π, δ + ǫ = (1 + f ) π. The partial neighborhood tiling for Case 1.3 is the second of Figure 23, andfor Case 3 is the second of Figure 25. Moreover, we know there is no 3 -tile,which by Lemma 2 implies f ≥ H = α , if f = 24, then β = γ = δ = ǫ . By [3, Lemma 21],this implies b = c , a contradiction. Therefore we have f >
24. If α a β b γ c δ d ǫ e is a vertex, then we have a + ( − f ) b + ( c + d ) + ( + f ) e = 2 . This implies (3 a + 5 b + 4 c + 4 d + 2 e − f = 24( b − e ) . By f >
24, we have β > π . We also have ǫ > π and specific valuesfor α, γ, δ . This implies a ≤ , b ≤ , c ≤ , d ≤ , e ≤
5. We substitutethe finitely many combinations of exponents satisfying the bounds into theequation above and solve for f . Those combinations yielding even f >
24 aregiven in Table 1. In the table, “ f = all” means that the angle combinationscan be vertices for any f . Further consideration by edge length shows thatthe vertices in the column “not vertex” are impossible.vertex not vertex f β ǫβ ǫ, γ δ, δ , α γδ , γ all αβ , ǫ α ǫ, α ǫ , αǫ π πδǫ γǫ π παγǫ , αδǫ ǫ βǫ π παǫ H = α : α = π, δ = γ = π .The table tells us that the all the possible vertices ( anglewise vertex com-bination ) for f = 48 isAVC = { αβ , β ǫ, γ δ, δ , α , ǫ } .
35e construct the tiling based on the AVC and the fact that there is a vertex δ , such that the tiles around the vertex are arranged as P , P , P in thesecond of Figure 23. Figure 26 illustrates such a vertex δ , with three tiles P , P ′ , P ′′ around the vertex as assumed. We will use P ′ n , P ′′ n to denote thetwo rotations of P n , and subsequent conclusions remain valid after rotations.By the AVC, the vertex βǫ · · · shared by P , P ′ is β ǫ . This determines P (and its two other rotations). The vertex α · · · shared by P , P is α (wewill omit mentioning “by the AVC”). This determines P , P . We also knowthat P ′ , P , P ′ (and their rotations) share a vertex γ δ . The vertex γ · · · shared by P , P is γ δ . This gives P , δ . The vertex ǫ · · · shared by P , P ′ is ǫ . This gives ǫ . Then δ , ǫ determine P . The vertex βǫ · · · sharedby P , P is β ǫ . This determines P . The vertex δ · · · shared by P , P is δ . This gives P , δ . By the edge length consideration, the vertex β β · · · shared by P , P cannot be αβ and is therefore β ǫ . This gives ǫ . Then δ , ǫ determine P . We find that P , P , P around a vertex δ just like P , P ′ , P ′′ around the starting vertex δ . The argument for the tiling can therefore berepeated by starting from the new δ . δβǫ α γβαδ γǫ αγ βδǫ α γβ δǫδǫ αγ ββα δγ ǫ δǫβ γαδβ ǫαγ βα δγ ǫαγβδ ǫ αγβδ ǫ δ ǫα γββαδγǫ δ ǫβ γα δ βǫ αγ βαδ γǫα γβ δ ǫαγ βδǫ δǫ αγββ αδ γǫδǫ βγ α ′ ′′ ′ Figure 26: Construct double pentagonal subdivision.The tiles P , P and their rotations form a local tiling that is exactlythe same as the tiling of triangle in Figure 11. The repeated constructionaround the new δ creates another such tiling of triangle next to the existingtiling of triangle. Therefore further repetition gives the double pentagonalsubdivision tiling.The argument for f = 48 also applies to f = 120. By Table 1, the AVC36or f = 120 is { β ǫ, γ δ, δ , α , ǫ } . The argument still starts with the sameassumption on the neighborhood of one δ vertex. The only place ǫ is usedfor f = 48 is the vertex shared by P , P , P ′ . In fact, the only fact we useis that all angles at ǫ · · · = ǫ are ǫ . Therefore the argument still works,as all angles at ǫ · · · = ǫ are ǫ . At the end, we still get double pentagonalsubdivision tiling.For f = 72, the AVC is { β ǫ, γ δ, δ , α , δǫ } . The argument for f = 48works until the vertex ǫ · · · shared by P , P ′ . By the AVC, we have ǫ · · · = δǫ . However, the circular adjacent angle deduction at δǫ δ ǫ ǫ ǫ → βǫ ( δγ ) ( δγ ) ( δγ )shows that one of β γ · · · , γ γ · · · , γ ǫ · · · should be a vertex. By the AVC,we find β γ · · · and γ ǫ · · · are not vertices. Moreover, the only vertex γ · · · in the AVC is γ δ = γ γ δ = γ γ · · · . The contradiction shows that ǫ · · · isnot a vertex. Therefore there is no tiling for f = 72.For general f (i.e., f = 48 , , { β ǫ, γ δ, δ , α } . Againwe repeat the argument for f = 48 until the vertex ǫ · · · shared by P , P ′ .Since ǫ · · · is not in the AVC, there is no tiling for general f .This completes the discussion for Case 1.3, H = α . The conclusion isthe double pentagonal subdivisions for f = 48 and 120.Now we turn to Case 3, with H = δ ǫ, δǫ , δ ǫ, δǫ . In the second partialneighborhood tiling in Figure 25, if H = δ ǫ , then A , = δ . This implies A , = β or ǫ . By the edge length consideration, we get A , = ǫ . Therefore δǫ is a vertex. Then the angle sums at δ ǫ, δǫ and δ + ǫ = (1 + f ) π imply f = 20, a contradiction. We get the same contradiction for H = δǫ .The final case we need to consider is Case 3, H = δ ǫ, δǫ . After finishingall the other cases for a bc , we may assume that there is no 3 -tile and no3 f ≥
60. On the other hand, for H = δ ǫ ,the argument for H = δ ǫ can be used to show that δǫ is a vertex. Thenthe angle sums at δ ǫ, δǫ and δ + ǫ = (1 + f ) π imply f = 28, contradicting f = 60. We get the same contradiction for H = δǫ . References [1] Y. Akama, M. Yan. On deformed dodecahedron tiling. preprint ,arXiv:1403.6907, 2014. 372] Y. Akama, E. X. Wang, M. Yan. Tilings of sphere by congruent pentagonsIII: edge combination a . preprint , 2019.[3] H. H. Gao, N. Shi, M. Yan. Spherical tiling by 12 congruent pentagons. J. Combinatorial Theory Ser. A , 120(4):744–776, 2013.[4] D. M. Y. Sommerville. Division of space by congruent triangles andtetrahedra.
Proc. Royal Soc. Edinburgh , 43:85–116, 1922-3.[5] Y. Ueno, Y. Agaoka. Classification of tilings of the 2-dimensional sphereby congruent triangles.
Hiroshima Math. J. , 32(3):463–540, 2002.[6] Y. Ueno, Y. Agaoka. Examples of spherical tilings by congruent quadran-gles.