Tilings of Sphere by Congruent Pentagons I

Transcription

1 rxiv: v5 [mth.mg] 8 Mr 018 Tilings of Sphere by Congruent Pentgons I K Yue Cheuk, Ho Mn Cheung, Min Yn Hong Kong University of Science nd Technology Mrch 9, 018 Abstrct We develop some bsic tools for studying edge-to-edge tilings of the sphere by congruent pentgons. Then we prove tht there is no edge-to-edge tiling of the sphere by more thn 1 congruent pentgons, under the ssumption tht the pentgon hs edge length combintion b c, 3 bc, or 3 b,b,c distinct, nd there is tile with ll vertices hving degree 3. 1 Introduction Mthemticins hve studied tilings for more thn 100 yers. A lot is known bout tilings of the plne or the Eucliden spce. However, results bout tilings of the sphere re reltively rre. A mjor chievement in this regrd is the complete clssifiction of edge-to-edge tilings of the sphere by congruent tringles [4, 5]. For tilings of the sphere by congruent pentgons, we completely clssified the miniml cse of 1 tiles [1, 3]. The sphericl tilings should be esier to study thn the plnr tilings, simply becuse the former involves only finitely mny tiles. The clssifictions in [3, 5] not only give the complete list of tiles, but lso the wys the tiles re fit together. It is not surprising tht such kind of clssifictions for the plner tilings re only possible under vrious symmetry conditions, becuse the quotients of the plne by the symmetries often become compct. Reserch ws supported by Hong Kong RGC Generl Reserch Fund nd

2 Like the erlier works, we restrict ourselves to edge-to-edge tilings of the sphere by congruent polygons, such tht ll vertices hve degree 3. These re mild nd nturl ssumptions tht simplify the discussion. The polygon in such tiling must be tringle, qudrilterl, or pentgon [6]. We believe tht pentgonl tilings should be reltively esier to study thn qudrilterl ones becuse 5 is n extreme mong 3, 4, 5. Indeed, in Section, we find vrious restrictions on pentgonl tilings of the sphere. Similr restrictions for qudrilterl tilings re much weker, which is the primry reson tht the study of qudrilterl tilings is much hrder [6]. Our clssifiction progrm strts with Proposition 1, which sys tht pentgonl tilings of the sphere musthvetile, suchthtfourverticeshvedegree3, ndthefifthvertexhs degree 3, 4 or 5. We cll such specil tile 3 5 -, , or tile. Our strtegy is to first try to tile the neighborhood of this specil tile. The neighborhood tiling gives us lot of informtion on edges nd ngles of the pentgon. Then we use the informtion to determine the nglewise vertex combintion ll the possible ngle combintions t vertices, bbrevited s AVC. The AVC guides us to further construct the tiling beyond the neighborhood, nd we eventully construct the tiling of the sphere. Proposition 1 nd the subsequent Propositions nd 3 give only the combintoril structure of the specil tile. Further informtion on edge lengths is provided by [3, Proposition 7] nd its extension Proposition 9 in this pper, which sys tht the pentgon cn only hve 5 possible edge length combintions,b,c re distinct: 5 equilterl, 4 b lmost equilterl, 3 b, 3 bc, b c vrible edge length. The min result of this pper is the clssifiction of the cse of vrible edge length, under the dditionl ssumption tht there is 3 5 -tile. Theorem. There is no edge-to-edge sphericl tiling by more thn 1 congruent pentgons, such tht there is tile with ll vertices hving degree 3, nd the pentgon hs edge length combintion b c, 3 bc, or 3 b,b,c distinct. This is the first of the following series of ppers on the clssifiction of edge-to-edge tilings the sphere by congruent pentgons. O. The miniml cse of tilings by 1 congruent pentgons. [3] I. Edge length combintions 3 b, 3 bc, b c, nd there is 3 5 -tile. [this pper]

3 II. Edge length combintions 3 bc, b c, nd there is no 3 5 -tile. [7] III. Edge length combintion 3 b, nd there is no 3 5 -tile. [8] IV. Equilterl length combintion 5. [] After the 0-th pper [3], we my lwys ssume more thn 1 tiles. After the five ppers, the remining cse is the lmost equilterl length combintion 4 b. The cse is the most complicted, nd llows mny tilings not ppering in the other cses. Sections, 3, nd 4 re generl results. Section is the bsic results on the distribution of vertices, ngles, nd edges. Section 3 gives crude yet very effective geometricl constrint on pentgon with two pirs of equl edges. Section 4 gives precise constrints on sphericl pentgons in terms of sphericl trigonometry. The constrints re used to eliminte two cses in this pper. Sections 5 nd 6 clssify the specific tilings of this pper, which ssumes vrible edge length nd existence of 3 5 -tile. Section 5 studies the neighborhood tiling of 3 5 -tile. In Theorems 13 nd 14, the informtion from the neighborhood tiling quickly implies tht the edge length combintions b c nd 3 bc do not dmit tilings by more thn 1 congruent pentgons. For the edge length combintion 3 b, Proposition 15 shows tht there re 4 possible neighborhood tilings. In Section 6, we further prove tht none of the 4 possible neighborhood tilings led to tilings of the sphere. Vertex, Angle nd Edge Vertex Consider n edge-to-edge tiling of the sphere by pentgons, such tht ll vertices hve degree 3. Let v,e,f be the numbers of vertices, edges, nd tiles. Let v k be the number of vertices of degree k. We hve = v e+f, e = 5f = kv k = 3v 3 +4v 4 +5v 5 +, v = k=3 v k = v 3 +v 4 +v 5 +. k=3 3

4 Then it is esy to derive v = 3f +4 nd f 6 = k 3v k = v 4 +v 5 +3v 6 +,.1 k 4 v 3 = k 43k 10v k = v 4 +5v 5 +8v By.1, f is n even integer 1. Since tilings by 1 congruent pentgons hve been clssified by [1, 3], we my ssume f > 1. We lso note tht by.1, f = 14 implies v 4 = 1 nd v 5 = v 6 = = 0. By [9, Theorem 1], this is impossible. Therefore we will lwys ssume f is n even integer 16. By., most vertices hve degree 3. We cll vertices of degree > 3 high degree vertices. Proposition 1. Any pentgonl tiling of the sphere hs tile, such tht four vertices hve degree 3 nd the fifth vertex hs degree 3, 4 or 5. We cll three types of tiles in the proposition 3 5 -tile, tile, nd tile. Proof. Suppose the tile described in the proposition does not exist. Then ny tile either hs t lest one vertex of degree 6, or hs t lest two vertices of degree 4 or 5. Since degree k vertex is shred by t most k tiles, the number of tiles of first kind is k 6 kv k, nd the number of tiles of the second kind is 1 4v 4 +5v 5. Therefore we hve f v v 5 + k 6kv k. On the other hnd, by.1, we hve f v v 5 + kv k = 1+ 1 v 5 + 6v k > 0. k 6 k 4k We get contrdiction. Proposition. If pentgonl tiling of the sphere hs no 3 5 -tile, then f 4. Moreover, if f = 4, then ech tile is tile. 4

5 Proof. Suppose there is no 3 5 -tile. Then ny tile hs t lest one vertex of degree 4. Therefore we hve f k 4 kv k. By.1, we further hve f = f f 4+ k 4 4k 3v k k 4kv k = 4+ k 43k 4v k 4. Moreover, the equlity hppens if nd only if v 5 = v 6 = = 0 nd f = 4v 4. Since there is no 3 5 -tile, this mens tht ech tile is tile. Proposition 3. If pentgonl tiling of the sphere hs no 3 5 -tile nd tile, then f 60. Moreover, if f = 60, then ech tile is tile. Proof. Suppose there is no 3 5 -tile nd tile. Then ny tile either hs t lest one vertex of degree 5, or hs t lest two vertices of degree 4. Therefore we hve f 1 4v 4 + k 5 kv k. By.1, we further hve f = 5f 4f 60+ k 4 5k 3v k 8v 4 k 54kv k = 60+v 4 + k 55k 6v k 60. Moreover, the equlity hppens if nd only if v 4 = v 6 = = 0 nd f = 5v 4. Sincethereisno3 5 -tilend3 4 4-tile, thismens thtechtileis3 4 5-tile. Angle The most bsic property bout ngles is tht the sum of ll ngles ngle sum t vertex is π. Another bsic property is the sum of ll ngles in the pentgon. Proposition 4. If ll tiles in tiling of sphere by f pentgons hve the sme five ngles,β,γ,δ,ǫ, then +β +γ +δ +ǫ = 3π + 4 π..3 f Proof. Since the ngle sum t ech vertex is π, the totl sum of ll ngles is πv. Since ll tiles re ngle congruent, the sum Σ of five ngles is the 5

6 sme for ll the tiles, nd the totl sum of ll ngles is lso fσ. Therefore we hve πv = fσ. Then by 3f = v 4, we get Σ = π v f = 3π + 4 f π. The proposition does not require tht the ngles re rrnged in the sme wy in ll tiles. Moreover, if we dditionlly know tht ll edges re stright i.e., gret rcs, then ll tiles hve the sme re Σ 3π, nd.3 follows from the fct tht the totl re Σ 3πf is the re 4π of the sphere. The proposition does not require tht the edges re stright. ThenglesinProposition4refertothevlues, ndsomenglesmongthe five my hve the sme vlue. For exmple, if the five vlues re,,,β,β, with β different vlues, then we sy the pentgon hs ngle combintion 3 β. The following is bout the distribution of ngle vlues. Proposition 5. If n ngle ppers t every degree 3 vertex in tiling of sphere by pentgons with the sme ngle combintions, then the ngle ppers t lest times in the pentgon. Proof. Ifnngleθ ppersonlyonceinthepentgon, thenthetotlnumber of times θ ppers in the whole tiling is f, nd the totl number of non-θ ngles is 4f. If we lso know tht θ ppers t every degree 3 vertex, then f v 3, nd non-θ ngles pper v 3 times t degree 3 vertices. Moreover, non-θ ngles pper k 4 kv k times t high degree vertices. Therefore 4v 3 4f v 3 + k 4kv k. On the other hnd, by., we hve 4v 3 v 3 + kv k = [3k 10 k]v k = 4v k 0. k 4 k 4 k 45k Then we get v 5 = v 6 = = 0 nd f = v 3 = v 4. This contrdicts.1. Unlike Proposition 4, which is explicitly bout the vlues of ngles, Proposition 5 only counts the number of ngles. The key in counting is to distinguish ngles. We my use the vlue s the criterion for the two ngles to be the sme. We my lso use the edge lengths bounding the ngles s the criterion. The observtion will be used in the proof of Proposition 9. The observtion lso pplies to the subsequent Propositions 6, 7, 8. 6

7 Proposition 6. If n ngle ppers t lest twice t every degree 3 vertex in tiling of sphere by pentgons with the sme ngle combintion, then the ngle ppers t lest 3 times in the pentgon. Proof. If n ngle θ ppers only once in the pentgon, then by Proposition 5, it cnnot pper t every degree 3 vertex. If n ngle θ ppers twice in the pentgon, then the totl number of times θ ppers in the whole tiling is f, nd the totl number of non-θ ngles is 3f. If we lso know tht θ ppers t lest twice t every degree 3 vertex, then f v 3, nd non-θ ngles pper v 3 times t degree 3 vertices. Moreover, non-θ ngles pper k 4 kv k times t high degree vertices. Therefore 3v 3 3f v 3 + k 4kv k. This leds to the sme contrdiction s in the proof of Proposition 5. The proof of Proposition 6 cn be esily modified to get the following. Proposition 7. If two ngles together pper t lest twice t every degree 3 vertex in tiling of sphere by pentgons with the sme ngle combintion, then the two ngles together pper t lest 3 times in the pentgon. The following is bout ngles not ppering t degree 3 vertices. Proposition 8. Suppose n ngle θ does not pper t degree 3 vertices in tiling of sphere by pentgons with the sme ngle combintion. 1. There cn be t most one such ngle θ.. The ngle θ ppers only once in the pentgon. 3. v 4 +v One of θ 3, θ 4, θ 5 is vertex, where θ. The first sttement implies tht the ngle in the fourth sttement must pper t degree 3 vertex. Proof. Suppose two nglesθ 1 ndθ donot ppertdegree3vertices. Then the totl number of times these two ngles pper is t lest f, nd is t 7

8 most the totl number k 4 kv k of ngles t high degree vertices. Therefore we hve f k 4 kv k. On the other hnd, by.1, we hve f k 4 kv k = 4+ k 43k 4v k > 0. The contrdiction proves the first sttement. The rgument bove lso pplies to the cse θ 1 = θ, which mens the sme ngle ppering t lest twice in the pentgon. This proves the second sttement. The first two sttements imply tht θ ppers exctly f times. Since this should be no more thn the totl number k 4 kv k of ngles t high degree vertices, by.1, we hve 0 f k 4 kv k = 1 v 4 v 5 + k 6k 6v k. This implies the third sttement. For the lst sttement, we ssume tht θ 3,θ 4,θ 5 re not vertices. This mens tht θ ppers t most twice t ny degree 4 vertex, nd t most four times t ny degree 5 vertex. Since θ lso does not pper t degree 3 vertices, the totl number of times θ ppers is v 4 + 4v 5 + k 6 kv k. However, the number of times θ ppers should lso be f. Therefore f v 4 +4v 5 + k 6 kv k. On the other hnd, by.1, we hve f v 4 +4v 5 + kv k = 1+ 6v k > 0. k 6 k 6k We get contrdiction. Edge Two pentgons hve the sme edge length combintion if the five edge lengths re equl. For exmple, the first two pentgons in Figure 1 hve the sme edge length combintion b c. If we further hve the sme edge length rrngement in the two pentgons, then the two pentgons re edge congruent. For exmple, the first two pentgons in Figure 1 hve the respective edge length rrngements,b,,b,c nd,,b,b,c, nd therefore they re not edge congruent. The subsequent discussion does not hve ny requirement on the vlues of ngles. 8

9 Proposition 9. In n edge-to-edge tiling of the sphere by edge congruent pentgons, the edge lengths of the pentgon is rrnged in one of the six wys in Figure 1. b c b c b c 3 bc 3 b 4 b 5 Figure 1: Edges in the pentgon suitble for tiling,,b,c,d,e distinct. Proof. The proposition is n extension of [3, Proposition 7]. We first follow the rgument in the erlier pper. By purely numericl considertion, there re seven possible edge length combintions bcde, bcd, b c, 3 bc, 3 b, 4 b, 5. For bcde, without loss of generlity, we my ssume tht the edges re rrnged s in the first of Figure. By Proposition 1, one of the two bottom vertices shred by b,c, nd shred by c,d hs degree 3. Without loss of generlity, we my ssume tht the vertex shred by c,d hs degree 3. Let x be the third edge t the vertex. Then x,c re djcent in tile, nd x,d re djcent in nother tile. Since there is no edge in the pentgon tht is djcent to both c nd d, we get contrdiction. e c b b b b c d x b c d x d x c x Figure : Not suitble for tiling. The combintion bcd hs two possible rrngements, illustrted without loss of generlity in the second djcent nd third seprted of Figure. Similr to the cse of bcde, we my ssume degree 3 vertex with the third edge x. In the second of Figure, the edge x is djcent to both c nd d, contrdiction. In the third of Figure, the edge x is djcent to both nd d, contrdiction. 9

10 The combintion b c hs three possible rrngements, illustrted in the first nd second of Figure 1 nd fourth of Figure. In the fourth of Figure, we my ssume degree 3 vertex with the third edge x. The edge x is djcent to both nd c, contrdiction. The combintion 3 bc hs two possible rrngements, illustrted in the third of Figure 1 b,c djcent nd left of Figure 3 b,c seprted. In cse b,c re seprted, the pentgon hs one -ngle denote by, two b-ngles, nd two c-ngles. Since there re no b -ngle, c -ngle, bc-ngle, ny degree 3 vertex must be one of three on the left of Figure 3. This implies tht the -ngle ppers t every degree 3 vertex. By Proposition 5 nd the remrk fter the proof of the proposition, the pentgon should hve t lest two -ngles, contrdiction. Figure 3: Also not suitble for tiling. The combintion 3 b hs two possible rrngements, illustrted in the fourth of Figure 1 b djcent nd right of Figure 3 b seprted. In cse the two b-edges re seprted, the pentgon hs one -ngle nd four b-ngles. Since there is no b -ngle, ny degree 3 vertex must be one of two on the right of Figure 3. This implies tht the -ngle ppers t every degree 3 vertex. By Proposition 5, the pentgon should hve t lest two -ngles, contrdiction. The following gives further detils bout the specil tile in Proposition 1 in cse of the first edge length rrngement in Figure 1. Proposition 10. An edge-to-edge tiling of the sphere by pentgons of edge length rrngement in the first of Figure 1 cnnot be 3 5 -tile. Moreover, if the pentgon is or tile, then the fifth vertex of degree 4 or 5 is opposite to the c-edge. Proof. The edge length rrngement in the first of Figure 1 hs three bvertices shred by -edge nd b-edge. If ny such b-vertex hs degree 3, then the third edge t the vertex is djcent to both nd b. The only such edge in the pentgon is c. 10

11 If two b-vertices of degree 3 re djcent, then we hve two c-edges t the two vertices, nd these two c-edges belong to the sme pentgon, contrdiction. Therefore we cnnot hve djcent b-vertices of degree 3. The observtion implies the conclusion of the proposition. 3 Non-symmetric Pentgon Consider the pentgon in Figure 4, with two b-edges nd two -edges s indicted. By [3, Lemm 1], we hve β = γ if nd only if δ = ǫ. Of course the equlities men tht the pentgon is symmetric. We try to prove the following stronger version of the result. Lemm 11. If the sphericl pentgon in Figure 4 is simple nd hs two pirs of equl edges nd b, then β > γ is equivlent to δ < ǫ. b β δ b γ ǫ Figure 4: Geometricl constrint for pentgon. By simple polygon, we men the boundry does not intersect itself. By [3, Lemm 1], in n edge-to-edge tiling of the sphere by congruent pentgons, the pentgon must be simple. To simplify the discussion, we will lwys use strict inequlities in this section. For exmple, this mens tht the opposite of > π will be < π. The specil cse of equlities cn be esily nlysed. The concept of theinterior of simple polygonisreltive onsphere. Once we designte inside, then we hve the outside. Moreover, the inside is the outside of the outside. We note tht the reformultion of Lemm 11 in terms of the outside is equivlent to the originl formultion in terms of the inside. We will lso use the sine lw nd the following well known result in the sphericl trigonometry: If sphericl tringle hs the ngles,β,γ < π nd hs the edges,b,c opposite to the ngles, then > β if nd only if > b. Let A,B,C,D,E be the vertices t thengles,β,γ,δ,ǫ. Among the two gretrcsconnecting B ndc, tke theonewith length<πto formtheedge 11

12 BC. The edge BC is indicted by the dshed line in Figure 5. Since AB nd AC intersect only t one point A nd hve the sme length b, we get b < π. Since ll three edges AB,AC,BC hve lengths < π, by the sine lw, mong the two tringles bounded by the three edges, one hs ll three ngles < π. We denote this tringle by ABC. The ngle BAC of ABC is either of the pentgon, or its complement π. Since the inside nd outside versions of the lemm re equivlent, we will lwys ssume = BAC < π in the subsequent discussion. B A b b G ǫ E γ β δ F D C B A b b β γ δ ǫ D E C B A b b δ ǫ β δ ǫ γ β γ C Figure 5: Constrint for pentgon nd for qudrilterl. The pentgon is obtined by choosing D,E, nd then connecting B to D, C to E, nd D to E by gret rcs. Since BC < π, we find tht BC does not intersect BD nd CE, nd the intersection of BC nd DE is t most one point. If BC nd DE intersect t one point F, then one of D,E is inside ABC nd one is outside we omit the equlity cse of D or E is on BC. The first of Figure 5 shows the cse D is outside nd E is inside. Since BC < π, the interiors of BD nd CE do not intersect BC. This implies tht β > ABC = ACB > γ. On the other hnd, since AC = b < π nd BC < π, the prolongtion of CE intersects the boundry of ABC t point G on AB. Using AB < π, < π, γ < ACB < π ndpplying thesinelwto ACG, we get CG < π, so tht = CE < CG < π. Using < π, BF < BC < π, CF < BC < π, BFD = CFE < π nd pplying the sine lw to BDF nd CEF, we get BDF < π nd CEF < π. Therefore δ = BDF < π < π CEF = ǫ. This proves tht D outside nd E inside imply β > γ nd δ < ǫ. Similrly, D inside nd E outside imply β < γ nd δ > ǫ. 1

13 If BC nd DE re disjoint, then either both B,C re outside ABC, or both B,C re inside ABC. The cse both B,C re outside ABC is the second of Figure 5. Since ABC is n isosceles tringle, we get β β = γ γ. Therefore β > γ if nd only if β > γ. The cse both B,C re inside ABC is the third of Figure 5. Since ABC is n isosceles tringle, we get β+β = γ+γ. Therefore β > γ if nd only if β < γ. Since δ +δ = γ +γ, we lso hve δ < ǫ if nd only if δ < ǫ. In either cse, the proof of Lemm 11 is reduced to the proof of the following similr result for qudrilterls. Lemm 1. If the sphericl qudrilterl in Figure 6 is simple nd hs pir of equl edges, then β > γ is equivlent to δ < ǫ. B β δ D γ C ǫ E Figure 6: Geometricl constrint for qudrilterl. Similrtopentgon, wenotethtthereformultionoflemm1interms of the outside is equivlent to the originl formultion in terms of the inside. ToproveLemm1, weusetheconformllyccurtewyofdrwinggret circles on the sphere. Let the circle Γ be the stereogrphic projection from the north pole to the tngent spce of the south pole of the equtor. The ntipodl points on the equtor re then projected to the ntipodl points on Γ. We denote the ntipodl point of P by P. Since the intersection of ny gret rc with the equtor is pir of ntipodl points on the equtor, the gret circles of the sphere re in one-to-one correspondence with the circles ndstright lines ontheplne thtintersect Γ tpir ofntipodl points. Proof. Suppose > π. In Figure 7, we drw gret circles BPP nd CPP contining the two -edges. They intersect t pir of ntipodl points P,P nd divide the sphere into four -gons. Since > π nd the boundry of the qudrilterl is simple, P,P lie in different -edges. Up to symmetry, therefore, there re two wys the four vertices B,C,D,E of the two -edges cn be locted, described by the two pictures in Figure 7. Moreover, since > π, the ntipodl point B of B lies in the -edge BP D. 13

14 In the first of Figure 7, we consider two gret rcs connecting B nd C. One gret rc the solid one is completely contined in the indicted -gon. The other gret rc the dshed one intersects the -edge BP D t the ntipodl point B nd therefore cnnot be n edge of the qudrilterl. We conclude tht the BC edge is the solid one. By the sme reson, the edge DE is lso the solid one, tht is completely contined in the indicted -gon. Then the picture implies β < π < γ, δ > π > ǫ. Similr rgument gives the BC edge nd DE edges of the qudrilterl in the second of Figure 7, nd we get the sme inequlities bove. In ll the subsequent rgument, we my ssume < π. B β P Γ C δ D γ P B E ǫ β B ǫ E D δ γ C D δ β B γ C D ǫ E Figure 7: Lemm 1: > π, nd DE < π in cse δ,γ < π. Next we rgue tht, if δ,ǫ < π, then we my further ssume DE < π. SupposeDE > π. Wedrwthegretcircle DE continingde inthethird of Figure 7. Since δ,ǫ < π, both DB nd EC lie in the sme hemisphere H bounded by the circle. This implies tht both B nd C lie in H. Therefore mong two gret rcs connecting B nd C, one intersects DE t two ntipodl points. By DE > π, one such point lies on DE, nd therefore this rc is not the BC edge. This proves tht the BC edge lso lies in the hemisphere H. Since ll three edges connecting D,B,C,E lie in H, we cn hve the complement of the qudrilterl in H. The complement is still qudrilterl, which differs from the originl one by replcing DE with the other gret rc of length π DE < π connecting D nd E, nd by replcing β,γ,δ,ǫ with π β,π γ,π δ,π ǫ. Moreover, we still hve π δ,π ǫ < π. It is esy to see tht the lemm for the new qudrilterl is equivlent to the lemm for the originl qudrilterl. Nowwe reredyto prove thelemm. Wedivide into two cses. The first is t lest three ngles < π. By considering the complement qudrilterl, 14

15 this lso covers the cse of t lest three ngles > π. The second is the cse of two ngles > π nd two ngles < π. Suppose t lest three ngles < π. Up to symmetry, we my ssume β,δ,ǫ < π. By the erlier rgument, we my further ssume tht < π nd DE < π. In Figure 8, we drw the gret circles DB nd DE contining DB nd DE. The two gret circles divide the sphere into four -gons. By δ < π, we my ssume tht δ is n ngle of the middle -gon. By DB = < π nd DE < π, B nd E lie on the two edges of the middle -gon. By β,ǫ < π, we find tht BC nd EC re inside the middle -gon. We lso prolong the -edge EC to intersect the boundry of middle -gon t T. The two pictures in Figure 8 refer to γ > π nd γ < π. In the first picture, we hve DT < DB = = EC < ET. Since ll ngles of DET re < π, this implies tht ǫ = DET < EDT = δ. We lso hve β < π < γ. In the second picture, the gret circles DB nd EC contining the two -edges intersect t ntipodl points T nd T, nd the two ntipodl points do not lie on the two -edges. This implies BT ++DT = π = CT ++ET. Since ll ngles in BCT nd DET re < π, we then hve β > γ +ET > +DT +CT < +BT δ < ǫ. T B β D C γ δ D E ǫ Bβ T D C γ δ D E ǫ T Figure 8: Lemm 1: At lest three ngles > π. Suppose two ngles > π nd two ngles < π. Then up to symmetry, we need to consider the following three cses. 1. β,γ > π nd δ,ǫ < π.. β,δ > π nd γ,ǫ < π. 3. β,ǫ > π nd γ,δ < π. 15

16 In the first cse, by the erlier rgument, we my dditionlly ssume < π nd DE < π. We drw gret circles BD nd CE contining the two -edges. Since < π nd the two -edges do not intersect, we my ssume tht they re inside the two edges of the middle -gon bounded by BD nd CE, s in the first of Figure 9. Since DE < π, the edge DE lies in the middle -gon. Since δ,ǫ < π, the two ngles lso lies in the middle -gon. Then β,γ > π implies tht the BC edge lies outside the middle - gon. If we replce the BC edge by the other gret rc c connecting the two points, then we get new qudrilterl with ngles β π,γ π,δ,ǫ. All four ngles of the new qudrilterl re < π, nd the erlier rgument shows tht the lemm holds for the new qudrilterl. Then it is esy to see tht the lemm for the new qudrilterl is equivlent to the lemm for the originl qudrilterl. This completes the proof of the cse. β B D δ c γ C ǫ E C E D δ B β ǫ E γ C Figure 9: Lemm 1: Two ngles < π nd two ngles > π. Inthesecond cse, we myginssume < π. Wedrwthecircle CE contining the -edge CE, s in the second of Figure 9. Since γ,ǫ < π, the two ngles lie in the sme hemisphere H 1 bounded by CE. Then the prolongtion of CB nd ED intersect t point inside the hemisphere H 1. Since CB nd ED do not intersect, we must hve either B,C lie in the sme hemisphere H bounded by the circle DEE, or D,E lie in the sme hemisphere bounded by the circle BCC. Without loss of generlity, we my ssume the first scenrio hppens, s in the second of Figure 9. Since < π, we lso know tht both BC nd CE lie in H, s in the second of Figure9. Now BD = < π implies tht the BD edge cnnot intersect CE t two points. Therefore the BD edge lso lies inside H 1. This implies tht δ is the ngle between n edge BD inside H nd the boundry DEE of H. Such ngle is lwys < π. This contrdicts the ssumption tht δ > π. 16

17 Finlly, the third cse is consistent with the conclusion of lemm. 4 Sphericl Trigonometry Lemm 11 is very rough constrint on sphericl pentgons becuse it is bout inequlities. For exmple, for given vlues of, b nd five ngles stisfying the lemm, there is no gurntee tht such pentgon exists. In this section, we give more precise equlities tht must be stisfied by sphericl pentgon in the first of Figure 10 i.e., the third type 3 b in Figure 1. Note tht the results of this section do not require,b to be distinct, nd re therefore lso pplicble to the edge combintion 5. A generl sphericl pentgon llows 7 free prmeters. The sphericl pentgon in the first of Figure 10 stisfies 3 independent edge length equlities. Therefore the pentgon llows 7 3 = 4 free prmeters. This mens tht four ngles β, γ, δ, ǫ, for exmple completely determine the pentgon. b β δ x b γ ǫ θ b x b θ x β θ γ θ δ ǫ β x y φ 1 φ ǫ δ φ γ φ Figure 10: Sphericl trigonometry of pentgon of 3 b type. We divide the pentgon into n isosceles tringle second of Figure 10 nd qudrilterl third of Figure 10. If < π, then Figure 5 gives three possible scenrios the pentgon is divided. If > π, then Figure 5 gives three possible divisions of the outside pentgon. The subsequent discussion is guided by the first of Figure 10 the second of Figure 5, nd further verifiction shows tht the conclusions re still vlid in ll cses. A generl tringle llows 3 free prmeters. The isosceles tringle introduces 1 equlity nd llows 3 1 = free prmeters. Therefore nd θ completely determine the isosceles tringle. The edge length x is given by cosx = cos+cos θ sin θ The edge length b is further given by = cos+1+coscot θ. sinxcosθ = sinbcosb1 cos, sinxsinθ = sinbsin. 17

18 For the qudrilterl in the third of Figure 10, we relbel β θ,γ θ by β,γ nd get the fourth of Figure 10. A generl qudrilterl llows 5 free prmeters. Three equl edges introduces equlities nd leves 5 = 3 free prmeters. Therefore its four ngles stisfy one equlity, nd the edge lengths nd x cn be expressed in terms of the four ngles. We dd n rc y to the qudrilterl nd get two tringles. By the sine lw for the tringle lbeled 1 nd by pplying the known formule to the isosceles tringle lbeled, we hve sinsinβ = sinysinγ φ Cnceling sin, we get = sinysinγcosφ sinycosγsinφ = sincossinγ1 cosǫ sincosγsinǫ. sinβ +cosγsinǫ = cossinγ1 cosǫ. 4.1 By the symmetry of exchnging β nd γ, nd exchnging δ nd ǫ, we get sinγ +cosβsinδ = cossinβ1 cosδ. 4. Cnceling cos from 4.1 nd 4., we get n equlity for the four ngles of the qudrilterl sinβ +cosγsinǫsinβ1 cosδ = sinγ +cosβsinδsinγ1 cosǫ. Then is determined by the four ngles through 4.1 or 4.. Moreover, x is determined below cosx = coscosy +sinsinycosδ φ = coscosy +sincosδsinycosφ+sinsinδsinysinφ = coscos +sin cosǫ +sincosδsincos1 cosǫ+sinsinδsinsinǫ = cos sin cos1 cosδ1 cosǫ+sin sinδsinǫ = cos 3 1 cosδ1 cosǫ cos sinδsinǫ +coscosδ +cosǫ cosδcosǫ+sinδsinǫ

19 Bck to the pentgon in the first of Figure 10, we substitute β,γ,δ,ǫ in 4.1 nd 4. by β θ,γ θ,δ,ǫ. After dividing by sinθ. We get sinβ +cosγsinǫcotθ cosβ +sinγsinǫ = cossinγcotθ cosγ1 cosǫ, sinγ +cosβsinδcotθ cosγ +sinβsinδ = cossinβcotθ cosβ1 cosδ. This is system of two equtions for cos nd cotθ. The solutions re qudrtic eqution for cos nd cotθ The coefficients re functions of β,γ,δ,ǫ Lcos +M cos+n = 0, 4.4 P cot θ+qcotθ+r = L = sinβ γ1 cosδ1 cosǫ, M = cosβ γsinǫ sinδ +sinδ ǫ, N = sinδ sinǫ sinβ γ1 sinδsinǫ, P = sinβsinβ +cosγsinǫ1 cosδ sinγsinγ +cosβsinδ1 cosǫ, Q = sinβ +cosβ +γsinǫ1 cosδ +sinγ +cosβ +γsinδ1 cosǫ, R = cosβcosβ sinγsinǫ1 cosδ cosγcosγ sinβsinδ1 cosǫ. Combined with 4.3 tht determines x, we my further determine nd b of the isosceles tringle. 5 Neighborhood Tiling Our strtegy for constructing tilings is to first construct tilings of the neighborhood of the specil tile in Proposition 1. In this pper, we concentrte on the neighborhood of 3 5 -tile. The neighborhood tiling implies some useful informtion bout edges nd ngles. The informtion is lredy enough for dismissing the edge combintions b c nd 3 bc. 19

20 Theorem 13. If n edge-to-edge sphericl tiling by congruent pentgons hs 3 5 -tile nd the pentgon hs edge length combintion b c, with,b,c distinct, then the number of tiles is f = 1. Proof. By Propositions 9 nd 10lso see[3, Proposition 7], the edge lengths of the pentgon re rrnged s the second of Figure 1. A 3 5 -tile hs nother 5 tiles round it, nd its neighborhood is combintorilly given by Figure 11. We denote the tile by P 1 nd its five neighboring tiles by P,...,P 6. The pentgon with edge combintion b c is the lst in Figure 1. Up to the combintoril symmetry of the neighborhood, we my ssume tht the edges of P 1 re given s indicted, nd the edge shred by P,P 3 is or c. The first of Figure 11 is the cse tht the edge shred by P,P 3 is. By the edge length considertion, we my successively determine ll edges of P,P 6,P 5,P 4. Then we find three -edges in P 3, contrdiction. The second of Figure 11 is the cse tht the edge shred by P,P 3 is c. We denote the five ngles of the pentgon by b-ngle, β -ngle, γ b -ngle, δ cngle, ǫ bc-ngle. By edge length considertion, we my determine ll edges nd ngles of P,P 3. Then we my further determine the -ngle β of P 4 nd b -ngle γ of P 6. The ngle sums t the three vertices imply +δ +ǫ = 3β = 3γ = π. This further implies + β + γ + δ + ǫ = 10 π. By the ngle sum eqution 3.3 for pentgon, we conclude f = γ ǫ δ β 3 ǫ δ β γ β β γ 1 γ 4 δ ǫ 6 5 ǫ δ ǫ δ 3 γ γ β β γ β γ 1 β β 4 ǫ δ ǫ δ 6 γ δ ǫ 5 Figure 11: Neighborhood tilings for b c nd 3 bc. Theorem 14. If n edge-to-edge sphericl tiling by congruent pentgons hs 3 5 -tile nd the pentgon hs edge length combintion 3 bc, with,b,c distinct, then the number of tiles is f = 1. 0

21 Proof. By Proposition 9 lso see [3, Proposition 7], the edge lengths of the pentgon re rrnged s the third of Figure 1. Up to symmetry, we my ssume tht the edges of P 1 is given by the third of Figure 11, nd with ll ngles s indicted. Since ech tile hs only one b-edge nd one c-edge, the edge shred by P,P 3 must be. This determines ll edges nd ngles of P,P 3. Then we my further determine P 4,P 6. Given the three known -edges of P 5, there re two wys of rrnging edges nd ngles of P 5. Either wy gives two vertices δ ǫ nd δǫ shred by P 1,P 4,P 5 nd by P 1,P 5,P 6. The ngle sums t the two vertices nd t the vertex βγ imply δ = ǫ = π nd + β + γ = π. Then we get 3 +β +γ +δ +ǫ = 10 π. By the ngle sum eqution.3 for pentgon, we 3 conclude f = 1. Next we turn to the edge length combintion 3 b. By Proposition 9 lso see [3, Proposition 7], the edge lengths of the pentgon re rrnged s the fourth of Figure 1. Up to the combintoril symmetry of neighborhood, we myssume tht theedgesof P 1 regiven byfigure1. Weignorethengles for the moment, nd concentrte on the edge lengths. If the edge shred by P,P 3 is, s in the first of Figure 1, then we my successively determine ll edges of P,P 3,P 4,P 6,P 5. If the edge shred by P,P 3 is b, s in the second nd third of Figure 1, then we my determine ll edges of P,P 3. Since the two b-edges of P 5 re djcent, either the edge shred by P 4,P 5 is, or the edge shred by P 5,P 6 is. Up to the symmetry of horizontl flipping, we my ssume tht P 5,P 6 shre. Then depending whether P 4,P 5 shre or b, we get the second nd third of Figure 1. We lbel the three neighborhood tilings in Figure 1 with ngles to be discussed s Cses 1,, 3. 3 γ β γ 1 β β 4 ǫ δ ǫ δ 6 γ δ ǫ 5 1 δ β γ ǫ 3 ǫ γ β δ γ ǫ β γ 1 ǫ γ 4 δ δ ǫ 6 β ǫ δ δ β γ 5 β 3 β γ 1 4 δ ǫ Figure 1: Neighborhood tilings for 3 b. Now we consider the ngles. We denote the ngles by,β,γ,δ,ǫ s in P 1 in Figure 1. First we rgue tht β γ nd δ ǫ. By Lemm 11 or 1

22 [3, Lemm 1], we know tht β γ is equivlent to δ ǫ. Therefore we only need to ssume β = γ nd δ = ǫ. In ech of three pictures in Figure 1, we hve vertex with three nd nother vertex with one nd two b. The ngle sums t the two vertices re + β = 3δ = π. This implies +β +γ +δ+ǫ = +β +δ = 10 π. By the ngle sum eqution.3 for 3 pentgon, we conclude f = 1. Therefore the cse cn be dismissed. The cse of 3 b differs from b c nd 3 bc in tht the ngles cnnot be uniquely described by the two bounding edges with the exception of the b -ngle. After showing β γ nd δ ǫ, the sitution becomes less mbiguous. To fcilitte further discussion, we dopt the nottion of [3]. Denote by V ijk the vertex shred by P i,p j,p k. Denote by A i,jk the ngle of P i t V ijk. Cse 1 In the first of Figure 1, we lredy know ll edges nd ngles of P 1. We lso know the b -ngles of P,P 3. Moreover, the b-ngles A 4,13,A 6,1 re β or γ. By the ngle sums t V 134,V 16 nd β γ, we know A 4,13 = γ, A 6,1 = β. This determines ll edges nd ngles of P 4,P 6. Given we know ll edges of P 5, there re two possible wys of rrnging ngles of P 5. Either wy implies tht one of V 145 nd V 156 is δ ǫ, nd the other is δǫ. The ngle sums t δ ǫ,δǫ nd V 16 = βγ imply δ = ǫ = π 3 nd +β+γ = π. Then we get +β+γ+δ+ǫ = 10 π. By the ngle sum 3 eqution.3 for pentgon, we conclude f = 1. Therefore the cse cn be dismissed. Cse In the second of Figure 1, we lredy know ll edges nd ngles of P 1. The vertices V 145,V 156 hve only -edges nd therefore involve only δ,ǫ. By the nglesumtthetwoverticesndthefctthtδ ǫ, eitherv 145 = V 156 = δ ǫ, or V 145 = V 156 = δǫ. If V 145 = V 156 = δ ǫ, then by the fct tht there is only one δ nd one ǫ in P 5, the distribution of ngles t the two vertices must be s indicted. This further determines ll ngles of P 4,P 5,P 6. If V 145 = V 156 = δǫ, we cn similrly determine P 4,P 5,P 6. Since exchnging β,γ nd exchnging δ,ǫ switches between the two scenrios, we only need to consider the cse in the second of Figure 1. The ngle sums t 3,βγǫ,δ ǫ

23 nd the ngle sum eqution.3 for pentgon imply = π, β +γ = π, δ = π, ǫ = 8 π f 3 f 3 f Cse 3 In the third of Figure 1, we lredy know ll edges. We lso know ll ngles of P 1 nd the ngles of ll tiles. The ngles of P 4 cn be rrnged in two wys. Figure 13 gives the first wy. Figure 14 gives the second wy. In Figure 13, we know the b-ngles A 3,14,A 5,14 re β or γ. Moreover, the ngle sums t V 134,V 145 imply tht A 3,14,A 5,14 hve the sme vlue. The first nd second of Figure 13 re the cse A 3,14 = A 5,14 = β, nd the third picture is the cse A 3,14 = A 5,14 = γ. The ngles A 3,14,A 5,14 determine P 3,P 5. ǫ γ β δ 3 δ β γ ǫ ǫ δ β γ 1 ǫ γ γ 4 β δ ǫ δ 6 β δ β 5 ǫ γ 3.1 ǫ γ γ ǫ 3 δ β βδ ǫ δ β γ 1 ǫ γ γ 4 β δ ǫ δ 6 β δ β 5 ǫ γ 3. δ β β δ 3 ǫ γ γ ǫ ǫ δ β γ 1 ǫ γ γ 4 β δ ǫ 6 γ δ ǫ β 5 δ β 3.3 Figure 13: Neighborhood tilings for Cse 3. The tile P 6 hs two possible rrngements. All three in Figure 13 hve the sme rrngement, nd re lbeled s Cses 3.1, 3., 3.3 for 3 b. Then in the first two pictures, we further consider the two possible rrngements of P. In the first picture, the ngle sums t 3,β δ,γ ǫ,δ ǫ nd the ngle sum eqution.3 for pentgon imply = 3 π, β = f π, γ = δ = f π, ǫ = f π. 5. In the second picture, the ngle sums t 3,β δ,βγǫ,δ ǫ nd the ngle sum eqution.3 for pentgon imply = π, β = 5 π, γ = π, δ = + 4 π, ǫ = 3 6 f 6 f 3 f 4 8 π. 3 f 5.3 In the third picture, we lso consider two possible rrngements of P. If the rrngement is not the sme s the third picture, then the ngle sums 3

24 t V 134 = βγδ nd V 16 = βγǫ imply tht δ = ǫ, contrdiction. Therefore the ngles of P must be rrnged s indicted. Then the ngle sums t 3,βγδ,γ ǫ,δǫ nd the ngle sum eqution.3 for pentgon imply = π, β = f π, γ = 5 6 f π, δ = f π, ǫ = f π. 5.4 We remrk tht by f 16, we hve β < γ nd δ > ǫ for 5. Cse 3.1 nd 5.4 Cse 3.3, nd we lso hve β > γ nd δ < ǫ for 5.3 Cse 3.. Therefore the geometric constrint in Lemm 11 is stisfied for ll three cses. We still need to consider the other rrngement of P 6, given we lredy know P 1,P 3,P 4,P 5. The new rrngement of P 6 mens tht V 16 = βγδ or γ δ, nd V 156 = ǫ 3. In the first two pictures of Figure 13, we compre the ngle sums t V 134 = β δ nd V 16, nd lwys get β = γ, contrdiction. In the third picture, the ngle sums t 3,βγδ,ǫ 3 imply = ǫ = π nd 3 β + γ + δ = π. Then we get + β + γ + δ + ǫ = 10 π. By the ngle 3 sum eqution.3 for pentgon, we conclude f = 1, nd the cse cn be dismissed. Now we turn to Figure 14, where the ngles of P 4 re rrnged differently from Figure 13. Then we consider two possible rrngements of P 5. The first picture shows one rrngement, nd the second nd third pictures show the other rrngement. ǫ γ γ ǫ 3 δ β β δ ǫ β γ δ 1 δ β β 4 γ δ ǫ ǫ 6 β δ γ 5 ǫ γ β δ 3 γ ǫ δ ǫ β γ 1 δ β β 4 γ δ ǫ ǫ 6 γ ǫ γ 5 δ β δ β γ ǫ 3 ǫ γ βδ δ ǫ β γ 1 ǫ γ β 4 γ δ ǫ 6 γ δ ǫ β 5 δ β Figure 14: More neighborhood tilings for Cse 3. In the first of Figure 14, we hve A 3,14 = β or γ. If A 3,14 = γ, then the ngle sums t V 134 = βγǫ nd V 145 = βγδ imply δ = ǫ, contrdiction. Therefore A 3,14 = β, which determines P 3. Now we consider two possible rrngements of P 6. If P 6 is rrnged differently from the first of Figure 14, then V 16 = βγǫ or γ ǫ. Compring the ngle sums t V 134 = β ǫ nd V 16 lwys gives β = γ, contrdiction. This implies tht P 6 must be rrnged s in the picture. Moreover, compring the ngle sums t V 145 = βγδ nd 4

25 V 16 gives A,16 = β. This determines P. Now the neighborhood tiling hs collection of vertices 3,β ǫ,βγδ,δǫ. The collection is obtined from the collection for Cse by exchnging δ nd ǫ. Then the equlities β > γ nd δ < ǫ for Cse become β > γ nd δ > ǫ for the first of Figure 14. This contrdicts the geometric constrint in Lemm 11. In the second nd third pictures of Figure 14, we lredy know P 1,P 4,P 5 ndthenconsider two possiblerrngements ofp 6. Thesecond pictureshows the first rrngement of P 6. By compring the ngle sums t V 145 = γ δ nd V 16, we find A,16 = γ, which determines P. Then we hve two possible rrngements of P 3. One rrngement of P 3 hs vertices 3,β ǫ,γ δ,ǫ 3, nd the other rrngement hs vertices 3,βγǫ,γ δ,ǫ 3. Combined with the ngle sum eqution.3 for pentgon, we get = β = ǫ = 1 π, γ = 4 π, δ = 8π, 3 f f nd = ǫ = π, β = 1 + π, γ = 3 f 5 π, δ = 6 f π. 3 f By f 16, we hve β < γ nd δ < ǫ in both cses, contrdicting the geometric constrint in Lemm 11. In the third of Figure 14, we hve the other rrngement of P 6. By compring the ngle sums t V 16 nd V 134, we find A,16 = β,a 3,14 = γ. This determines P,P 3. Theneighborhoodtilinghsvertices 3,βγǫ,γ δ,δǫ. The nglesumstthevertices implyβ = δ ndγ = ǫ, contrdicting thegeometric constrint in Lemm 11. The following summrises the discussion on tilings of the neighborhood of 3 5 -tile with edge combintion 3 b. Proposition 15. If n edge-to-edge sphericl tiling by more thn 1 congruent pentgons hs edge length combintion 3 b, with,b distinct, then up to symmetry, the neighborhood of 3 5 -tile hs four possible tilings given by the second of Figure 1 nd the three pictures in Figure Tiling for Edge Combintion 3 b We need to strt from the four neighborhood tilings in Proposition 15 nd try to extend the tiling beyond the neighborhood. The following technicl results re useful for constructing the extension. 5

26 Lemm 16. In tiling by congruent pentgons with edge length combintion 3 b,,b distinct, the number of b-ngles t ny vertex is even. Proof. The proposition is consequence of the fct tht the number of b-edges tvertexisthenumber ofb -nglesplushlfofthenumber ofb-ngles. Lemm 17. Suppose β γ nd δ ǫ in the 3 b -pentgon in Figure 10. Suppose tiling by congruent copies of the pentgon hs t most one ǫ t every vertex. Then there cnnot be consecutive γδ δγ t vertex consists of δ only. In cse there is no δ, this mens tht two γ cnnot shre n -edge t vertex. We note tht the lemm is symmetric with respect to the exchnge of β,γ nd exchnge of δ,ǫ. Proof. Suppose wehve consecutive γδ δ t vertex sinfigure15. Then we my determine ll edges nd ngles of P 1. Since δ is djcent to β nd ǫ in the pentgon, P 1,P shre vertex βǫ or ǫ. Since ǫ is not vertex, the vertex shred by P 1,P is βǫ. This determines β of P. Combined with δ of P, we determine P. By pplying the similr rgument to the vertex shred by P,P 3, we lso determine P 3. The process continues. If we hve consecutive γδ δγ, then we my strt from one end γ nd consecutively determine the tiles contining the intermedite δ. When we rech the other end γ, we find tht the two tiles contining consecutive γ nd δ shre vertex ǫ, contrdicting to the ssumption. γ ǫ γ β β ǫ δ 3 δ β ǫ δ γ 1 Figure 15: Cnnot hve consecutive γδ k γ if ǫ is not vertex. 6.1 Cse, = β The neighborhood tiling for Cse is the second of Figure 1, nd the ngles re given by 5.1. By f 16, we hve ǫ > > δ nd β + γ <. By Lemm 11, we lso hve β > γ. Then by β+γ < nd β+γ = δ, we get γ < nd δ β,γ. By β +γ < ǫ, we lso hve ǫ β,γ. We conclude tht 6

27 the only wy for some of the five ngles to be equl is = β. This section studies the cse = β, nd the next section studies the cse β. By = β nd 5.1, we get ll the ngles 4 π, ǫ = 3 8 π. f = β = 3 π, γ = 8 1 f π, δ = f By f 16, we hve π ǫ = + 16 π < ll ngles. This implies tht 3 f ǫ is not vertex, nd therefore Lemm 17 cn be pplied. In the second of Figure 1, P 5,P 6 shre vertex β, with ngles in the reminder being consecutive. Since the ngle sum of the reminder is π β = π = = β < δ,ǫ, by Lemm 16, the vertex β is β, 3 β γ c δ, or β γ c. By Lemm 17, the reminder cnnot be γ c δ or γ c. Therefore the vertex β shred by P 5,P 6 is β. This belongs to tile outside P 5,P 6. Since γ is djcent to in the pentgon, the new tile shres vertex γ with either P 5 or P 6. By Lemm 16, the vertex γ is either βγ or γ. If βγ is vertex, thenthereminderhsnglesum 3 8 f π <,β,δ,ǫ. Therefore the vertex is βγ c δ or βγ c, c 1. If c 3, then by the edge length considertion, we get two γ shring n -edge, contrdiction to Lemm 17. Therefore by Lemm 16, the vertex is βγδ or βγ. By f > 16, we find the ngle sum of βγ to be < π. Therefore βγ cn only be βγδ. Now we turn to the possibility tht γ is γ. Hving discussed βγ, we my ssume tht the reminder of γ does not contin β. Since +γ+ǫ = π, the reminder does not contin ǫ. Therefore the vertex is γ c δ d or γ c δ d, with c. By the edge length considertion, we get contrdiction to Lemm 17. We conclude tht the vertex γ shred by P 5,P 6 is βγδ. Then the ngle sum t βγδ implies f = 36 nd = β = 3 π, γ = 9 π, δ = 4 9 π, ǫ = 10 9 π. Now consider the vertex βγ shred by P 4,P 5 in the second of Figure 1. The reminder hs ngle sum 10 π. By the ngle length considertion nd 9 the ngle sum estimtion, the reminder cnnot contin β which by Lemm 16 contins nother β or γ or ǫ. Therefore the vertex is βγ c δ d or βγ c δ d. Since β nd γ shre -edge, by the edge length considertion, we must hve c 3. However, this contrdicts Lemm 17. 7

28 6. Cse, β We continue the study of Cse, under the dditionl ssumption tht ll five ngles re distinct. From the discussion t the beginning of Section 6.1, we lso know β > γ nd δ < ǫ. Suppose ǫ 8 is vertex. Then 0 > π ǫ = 1 π. This implies f 3 f < 4. By.1, we get 5 v 4 +v 5 +3v 6 +. By [9, Theorems 1 nd 6], we get v >6 = 0 for f =, v >5 = 0 for f = 0, nd v >4 = 0 for f = 18 or By f 16, we hve π ǫ < + 4 π,β,δ,ǫ. Therefore the vertex 3 f ǫ must be γ k ǫ. By Lemm 16 nd the fct tht ll vertices hve degree 6, we get ǫ = γ ǫ or γ 4 ǫ. If γ 4 ǫ is vertex, then v 6 1. By 5 v 4 +v 5 +3v 6 nd [9, Theorems 1 nd 6], we get v 4 =,v 5 = 0,v 6 = 1. By v >6 = 0 nd.1, we get f =. By f = nd the ngle sum t γ 4 ǫ, we my clculte ll the ngles = 67 π, β = 3 66 π, γ = π, δ = π, ǫ = π. Since no four ngles from bove repetition llowed cn dd up to π, we get contrdiction to v 4 =. If γ ǫ is vertex, then the ngle sum t γ ǫ gives β = π nd γ = 8 f 1 3 π. For ech f = 16,18,0,, we know the vlues of ll five ngles nd cn further find ll possible ngle combintions t vertices. We lso tke into ccount of Lemm 16 nd the constrint v >6 = 0 for f =, v >5 = 0 for f = 0, nd v >4 = 0 for f = 18 or 16. Then we get the following nglewise vertex combintions ll possible ngle combintions t vertices AVC = { 3, βγǫ, δ ǫ, γ ǫ } for f = 16,18,; AVC = { 3, βγǫ, δ ǫ, γ ǫ, γ δ} for f = 0. By the AVC, the vertex γδ shred by P,P 6 in the second of Figure 1 is γ δ, nd we must hve f = 0. It is esy to see tht γ δ must be configured s in Figure 16. This determines P 1,P 4. Then P 1,P shre vertex β or γ. Since β is not in the AVC, we get the γ ngle of P, which further determines P. By the sme rgument, we determine P 3. Then P,P 3 shre vertex β, contrdicting the AVC. This completes the proof tht ǫ is not vertex. Therefore Lemm 17 holds. 8

29 δ δ ǫ 3 β β ǫ γ γ β γ γ 4 ǫ 1 β ǫ δ δ δ Figure 16: Impossible vertex for Cse. Next we find vertices β b γ c δ d ǫ with one ǫ. Let b = min{b,c}. Then the ngle sum t such vertex gives b+ f d+ f 3 8 f = b+d+4+ 4 f b+d. 6.1 If b+d > 0, then + b+d <. This implies = b = 0 nd d = 0 or 1, nd further implies b + d 0, contrdicting to the ssumption. Therefore b+d. If b+d =, then 6.1 implies = 0. By b + d =, we get vertices β b γǫ b 1, βγ c ǫ c 1, β b δ ǫ, γ c δ ǫ. In the second of Figure 1, we see tht βγǫ nd δ ǫ re lredy vertices. Then the ngle sum implies tht the four possibilities β b γǫ, βγ c ǫ, β b δ ǫ, γ c δ ǫ cn only be the existing vertices βγǫ nd δ ǫ. If b+d = 1, then 6.1 nd f 16 imply = 0. By b+d = 1, we get vertices β b δǫ b 1, γ c δǫ c 1. By β+δ+ǫ f f π f π+ π > π, we cn only hve γ c δǫ. By Propositions 16 nd 17, we further find tht the only possibility is γ δǫ. If b+d = 0, then6.1ndf 16imply = 0or1. By b+d = 0, weget vertices β b ǫb 1, γ c ǫc 1, β b ǫb, γ c ǫc. By+β+ǫ > π nd β +ǫ > π re > π, we cn only hve γ c ǫ,γ c ǫ. By Propositions 16 nd 17, we further find tht the only possibilities re γ ǫ,γ ǫ. The ngle sum t γ ǫ implies β = γ, contrdiction. The ngle sum t γ ǫ implies β = + 4 π nd γ = 4 π. Note tht the vertex β shred by P 3 f f 5,P 6 in the second of Figure 1 hs the remining ngle π β = 8 π, which 3 f is strictly less thn,β,ǫ. Therefore the vertex is β γ c δ d. The second of 9

30 Figure 1 further shows tht γ c δ d is bounded by two b-edges, contrdicting Lemm 17. Itreminstoconsidervertex β b γ c δ d withoutǫ. Tovoidcontrdicting Lemm 17, ech γ needs to be combined with one β into chin βδ δγ bordered by two b-edges. This implies b c. All the possible vertices we found so fr give the following complete list of possible ngle combintions t vertices AVC = { 3, βγǫ, δ ǫ, γ δǫ, β b γ c δ d b c}. 6. Suppose γ δǫ is not vertex. Since both numbers of β nd γ in the whole tiling should be equl to f, to blnce the equl totl number, we must hve b = c in every vertex β b γ c δ d. The ngle sum t β b γ b δ d is = f b f d = f b+d. If the vertex is not 3, this implies + b + d 5. By trying ll,b,d stisfying the inequlity nd using f 16, we get ll the solutions = 0, b+d = 4, f = 4: βγδ, β γ, δ 4 ; = 1, b+d = 3, f = 36: βγδ, δ 3 ; = 0, b+d = 5, f = 60: β γ δ, βγδ 3, δ 5. The vertex β shred by P 5,P 6 in the second of Figure1 does not pper in the list for f = 36, is β γ for f = 4, nd is β γ δ for f = 60. The second offigure1 further shows thtγ forf = 4 orγ δ forf = 60is bounded by two b-edges, contrdicting Lemm 17. We conclude tht γ δǫ must be vertex. The ngle sum t γ δǫ implies β = is f π nd γ = f f π. Then the ngle sum t β b γ c δ d b c 3b+c+d =. If 3b+c+d = 0, then the vertex is 3. If 3b+c+d > 0, then the eqution implies 4+3b+c+d < 1. Substituting those,3b+c+d stisfying the inequlity nd tht 3b+c+d is positive nd even by Lemm 16 into the eqution nd solve for f. Those combintions,3b+c+d yielding even f 16 re 0,8 for f = 4, 1,6 for f = 36, nd 0,10 for f =

31 Forf = 4, we consider thevertex β shredby P 5,P 6 inthesecond of Figure 1. By the AVC 6. nd the discussion bove, the vertex is β b γ c δ d, stisfying b, b c, 3b+c+d = 8, nd b,c hve the sme prity. The solution shows tht the vertex β is β γ or β δ. The second of Figure 1 further shows tht γ or δ is bounded by two b-edges. By Lemm 17, we cnnot hve γ bounded by two b-edges. Moreover, δ is not b -ngle. For f = 36, we consider the vertex βγ shred by P 4,P 5 in the second of Figure 1. By the AVC 6. nd the discussion bove, the vertex is βγǫ or β b γ c δ d, stisfying b 1, c 1, b c, 3b+c+d = 6, nd b,c hve the sme prity. The solution shows tht the vertex βγ is βγǫ or βγδ. The second of Figure 1 further shows tht ǫ or δ is bounded by two b-edges. Both re contrdictions. For f = 60, we hve = 3 π, β = 3 5 π, γ = 1 5 π, δ = 5 π, ǫ = 6 5 π. Substituting the ngles into 4.5, we get P = 0, Q =, R = Therefore cotθ = R Q = Since < π, we hve θ < π. Therefore θ = 3 π = β, nd the pentgon is 5 given by Figure 17, in which D lies in the gret rc connecting B nd C. A B β δ D ρ ǫ E γ ρ C Figure 17: Impossible sphericl pentgon. Since CDE nd ABC re isosceles tringles, we hve ρ = EDC = ECD = β γ = π. Then δ +ρ = 4 π π, contrdicting to the fct tht 5 5 D lies on the gret rc connecting B nd C. 31