Journal of Combinatorial Theory, Series A

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1 Journl of Comintoril Theory, Series A 0 (0) Contents lists ville t SiVerse SieneDiret Journl of Comintoril Theory, Series A Spheril tiling y ongruent pentgons Hongho Go, Nn Shi, Min Yn Hong Kong University of Siene nd Tehnology, Hong Kong rtile info strt Artile history: Reeived Septemer 00 Aville online Jnury 0 Keywords: Spheril tiling Pentgonl tile Clssifition The tilings of the -dimensionl sphere y ongruent tringles hve een extensively studied, nd the edge-to-edge tilings hve een ompletely lssified. However, not muh is known out the tilings y other ongruent polygons. In this pper, we lssify the simplest se, whih is the edge-to-edge tilings of the -dimensionl sphere y ongruent pentgons. We find one mjor lss llowing two independent ontinuous prmeters nd four lsses of isolted exmples. The lssifition is done y first seprtely lssifying the omintoril, edge length, nd ngle spets, nd then omining the respetive lssifitions together. 0 Elsevier In. All rights reserved.. Introdution Tilings hve een studied y mthemtiins for more thn 00 yers. Some mjor hievements inlude the solution to Hilert s th prolem [,,], the lssifition of wllpper groups[,] nd rystllogrphi groups [,0,], the lssifition of isohedrl tilings of the plne[], nd the lssifition of edge-to-edge monohedrl tilings of the -sphere y tringles [,,]. We refer the reder for more omplete history of the sujet to the exellent 9 monogrph y Grünum nd Shephrd []. In this pper, we study monohedrl tilings, whih re tilings tht ll tiles re geometrilly ongruent. For monohedrl tilings of the plne y polygons, it is esy to see tht ny tringle or qudrilterl n e the tile. It is lso known tht onvex tiles nnot hve more thn six edges, nd there re extly three lsses of onvex hexgonl tiles [] nd t lest lsses of onvex pentgonl tiles [,0]. Our min result ssumes stright line edges(tully gret rs)ut not the onvexity. Moreover, most of the other results in this pper llow ny resonly nie urves to e the edges. E-mil ddress: mmyn@ust.hk (M. Yn). Reserh ws supported y Hong Kong RGC Generl Reserh Fund 00 nd /$ see front mtter 0 Elsevier In. All rights reserved.

2 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) The tilings we study re lso edge-to-edge, whihmensthtnovertexoftileliesintheinterior of n edge of nother tile. The ssumption simplifies the lssifition of tiling ptterns, or ll the possile wys the tiles fit together to over the whole sphere. It is esy to see tht ny tringle nd qudrilterl n e the tile of n edge-to-edge monohedrl tiling of the plne. Moreover, of the onvex pentgon lsses n e used for edge-to-edge tilings []. On the other hnd,there is no generl lssifition of plne tiling ptterns, even for edge-to-edge nd monohedrl tilings y onvex polygons. The known lssifitions often ssume ertin symmetry [], or ertin speil geometri property of the tiles [,]. Thediffiultywithlssifyingtheplnetilingptternsisthtthenumer of tiles is infinite. By restriting to spheril tilings, the prolem eomes finite nd more mngele. Our lssifition does not ssume ny symmetry or speil geometri property. We lso restrit the study to the tilings suh tht ll verties hve degree. This voids some trivil exmples otined y rtifiilly dding extr verties to the edges, or getting new tilings y wiggling modifitions of the edges. The ssumption further simplifies the lssifition. Now we ome to the sujet of this pper, the edge-to-edge monohedrl tilings of the sphere y polygons, sujet to degree vertexondition.itisnothrdtoseethtthetilenonlyetringle, qudrilterl, or pentgon [, Proposition ]. The first work in this diretion is Sommerville s 9 prtil lssifition of edge-to-edge monohedrl spheril tilings y tringles []. Dvies[] outlined ompletelssifitionin9,ndtheompleteproofwsfinllygivenyuenondagok[] in 00. See [9] for the further study of monohedrl non-edge-to-edge spheril tilings y tringles. As for qudrilterl spheril tilings, Ueno nd Agok [] showed tht the lssifition n e very omplited. For works on the spheril tilings y speil types of qudrilterls, see [,]. As fr s we know, there hve een virtully no study of tilings of the sphere y ongruent pentgons. However, we re of the opinion tht the spheril pentgon tilings should e esier to study thn the qudrilterl ones, euse mong tringle, qudrilterl nd pentgon, pentgon is the other extreme. We feel tht the spheril pentgon should e ompred to the plnr hexgon, nd the spheril qudrilterl should e ompred to the plnr pentgon. To test our onvition, we lssify the miniml se of the edge-to-edge monohedrl tiling of the sphere y pentgons. Min theorem. Any edge-to-edge spheril tiling y ongruent pentgons must elong to one of the five lsses in Fig., sujet to the ondition tht the sum of ngles t ny vertex is π. Here is how to red the tilings in Fig. : Comintorilly,thetilingsrethedodehedron,nd re illustrted on the flt plne fter punhing hole in one tile. The enter tile shows the ngles in tile.the ngles in the other tiles re determined y the lotion of nd the ngle rrngement orienttion. As for the edge lengths, the usul solid lines hve the sme edge length, thethiklines hve the sme length, ndthedottedlineshvethesmelength. Although the lssifition ontins five lsses, only the fifth lss is the essentil one, euse the other four lsses only provide isolted exmples (nd perhps the regulr dodehedron only). Fig. is this essentil lss in its full glory. The tile in the essentil lss T is illustrted on the left of Fig.. Thenglesumonditionis = + + = π. This implies tht the re of the pentgonl tile is π,oronetwelfthofthereofthesphere. It is esy to see tht the tile in T llows two free prmeters. Divide the tile s in the middle of Fig.. Sine = π is lredy fixed, the re A() of the tringle ABC is ompletely determined y. ThefuntionA() is stritly inresing, nd A() is the re of ADE.WhenA() + A() π nd A() + A() is suffiiently lose to π, we n lwys find suitle ngle CAD (equivlent to finding suitle = BAE)sothtthereofthepentgonis π.thisshowsthtwenfreely hoose nd within ertin rnge to get tile tht fits into the tiling, nd the tile is ompletely determined y the hoie. Suppose we put the ue in Fig. inside sphere, so tht the ue nd the sphere hve the sme enter. If we put light soure t the enter, then the projetion of the ue to the sphere is T lss tiling with + = = π.

3 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) T T T T T Fig.. All the possile spheril tilings y ongruent pentgons. For the tiling lss T,thenglesumonditionis = + = + = + + = π.this mens = π, = π, = π +, = π, nd should led to only isolted exmples. The reson is illustrted on the right of Fig.. Suppose nd re given. Fix point A on the sphere nd diretion for the edge AB.StrtingfromA nd trveling long the given diretion y distne, werrivetb. Turningnngle t B nd trveling y distne, werrivetc. Turningnngle t C nd trveling y distne, wer- t D nd trveling y distne, werrivete. Turning rive t D. Turningnngle = π

4 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) 9 0 Fig.. The generl lss T of spheril tilings y ongruent pentgons. C B A D E A B C D E A Fig.. The pentgonl tile. n ngle = π + t E nd trveling long, we get line (tully gret r) EA.Thenwe wnt A to lie on the line EA,whihimposesonereltionetween nd, ndreduesthenumer of free prmeters to one. One A lies on EA,wefurtherneedthepentgontohvere π, whih is the sme s EAB = = π.thisimposesonemorereltionndwereleftwithno free prmeter. Therefore the pentgonl tiles fitting into the tiling lss T must pper in isolted wy. Similr disussion n e mde for T nd T.ForT,wehveevenonelessdegreeoffreedom. The prolem is rystlized elow (the ngles re renmed in four ses in order to get the sme expressions for the ngle sum equtions). Prolem. Find spheril pentgons in Fig. stisfying = + = + = + + = π.are there ny exmples esides the pentgon in the regulr dodehedron?

5 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) π Fig.. Aspherilpentgontiling. Fig.. Must these spheril pentgons e regulr? To prove the min theorem, we first seprte nd lssify different spets of tilings. At the most rude level, we study tilings y omintorilly ongruent tiles in Setion, whih mens tht ll tiles hve the sme numer of edges, nd edge lengths nd ngles re ignored. Proposition sys tht the dodehedron is the unique spheril tiling y omintorilly ongruent pentgons. Comintoril tilings hve een extensively studied. See [9] for reent survey. Our study is tully topologil one, in the sense tht the lines do not even need to e stright. See [] for our further work for the non-miniml se. In Setions nd, we study tilings y edge ongruent tiles, whih mens tht ll tiles hve the sme edge length omintion nd rrngement. Propositions 9, 0,, (plus the se ll edges hve the sme length) ompletely lssify spheril tilings y edge ongruent pentgons. Our further work in the non-miniml se n e found in []. In Setions nd, westudytilingsyngleongruenttiles,whihmensthtlltileshve the sme ngle omintion nd rrngement. Proposition lssifies the numeris in spheril tilings y ngle ongruent pentgons. Propositions, 9, 0 then further lssify the lotions of the ngles in three mjor ses. The omplete lssifition would involve nother mjor ut rther omplited se. Sine this se will not e needed in the proof of the min lssifition theorem, we only present some exmples insted of the whole proof. In Setion, weominetheedgeongruentndngleongruentlssifitionstoprovethemin theorem. While our edge ongruent nd ngle ongruent results do not require geometry, in tht the edges re not neessrily gret rs, in omining the two, we use geometry to ut down the possile numer of edge ngle omintions. So our min theorem does require the edges to e gret rs. For the non-miniml se, see [] for further prtil results. Finlly, we would like to thnk the referees for the reful reding of the pper nd mny helpful suggestions.. Comintoril onditions In this setion, we study the omintoril spet of tilings. This mens tht we ignore the edge lengths nd the ngles, nd only require the tiles to hve the sme numer of edges. Most results of the setion re well known.

6 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) 9 Fig.. Tiles. Fig.. Not tiles. The results ertinly pply to the edge-to-edge tilings s defined in []. However,some results n e pplied to more generl tilings. Speifilly, we onsider onneted grph niely emedded in the sphere. We ssume tht ll verties in the grph hve degree. The interior of tile is onneted omponent of the omplement of the grph. A tile is the losure of its interior. The oundry of tile is wht is etween the tile nd its interior, nd is union of some edges in the grph. The exlusion of verties of degree mens tht we do not view n n-gon s n (n + k)-gon y rtifiilly hoosing nother k points on the oundry s extr verties. Therefore ll tiles hve nturlly defined verties nd edges tht re lredy given in the grph, nd the tiling is nturlly edge-to-edge. Fig. shows some possile tiles. The first one does not hve ny oundry points identified. The others hve some oundry points (even oundry intervls) identified. In se some oundry intervls re identified, the interior of the tile is different from the topologil version of the interior. Fig. shows some exmples tht re not tiles. They re unions of severl tiles t finitely mny oundry points. Lemm. In ny (nturlly generted) spheril tiling, there is t lest one tile for whih no oundry points re identified. Proof. We ll losed suset of the sphere simple region if it is enlosed y niely emedded simple losed urve. The oundry points of simple region re not identified. The lemm silly sys tht some tile is simple region. It is topologil ft tht if R is the whole sphere or simple region, nd P is tile in R with some oundry points identified, then the losure of some onneted omponent of R P is simple region. Let P e tile, suh tht some oundry points re identified. Then the omplement S P hs onnetedomponentc,suhthtthelosure C is simple region, nd is tully union of tiles. If C is tile, then we re done. Otherwise, C ontins tile P s proper suset. If P is simple region, then we re lso done. If P is not simple region, then some oundry points of P re identified, nd the omplement C P hs onneted omponent C,suhthtthelosure C is simple region. Keep going. As long s we do not enounter tiles tht re simple regions long the wy, we find stritly deresing sequene of simple regions C C C.Sinetherere only finitely mny tiles in the tiling, the sequene must stop. This mens tht some tile found in the proess must e simple region. An immedite onsequene of the lemm is tht in se ll tiles re ongruent (full ongruene, inluding edge lengths nd ngles), the oundry points of ny tile re never identified. Sine the property is needed in the susequent disussion, we inlude it in the following definition.

7 0 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Fig.. Every tile hs vertex of degree >. Definition. A tiling is omintorilly monohedrl if ll tiles hve the sme numer of edges nd every tile hs no oundry points identified. The tiles in suh tiling re omintorilly ongruent. Let f, e, v e the numers of tiles (or fes), edges nd verties in omintorilly monohedrl tiling. Let n e the numer of edges in tile. Then we hve the Euler eqution nd the Dehn Sommerville eqution v e + f =, e = nf. Note tht the Dehn Sommerville eqution ssumes tht eh edge is shred y two different fes, whih follows from the ondition tht no oundry points of tile re identified. (Atully we only need tht no oundry intervls re identified.) Moreover, if v i is the numer of verties of degree i, then v = v + v + v +, e = v + v + v +. It is simple onsequene of ll these equtions tht n =,,. See Proposition of []. Now we onentrte on the se n =. The equtions eome e = f, v = f +, v 0 = v + v + v +. (.) Lemm. Any spheril tiling y omintorilly ongruent pentgons ontins tile in whih t lest four verties hve degree. Fig. gives omintoril pentgon tiling of the sphere in whih eh tile hs degree vertex. Proof. Let t j e the numer of degree verties in the jth tile. Then v = t j.ontheotherhnd, we hve f = v = v + v + v + v + v + v + v + v + = v + (v 0) < v = t j. This implies tht some t i >. Lemm. A spheril tiling y omintorilly ongruent pentgons hs t lest tiles. Moreover, the minimum is rehed if nd only if ll verties hve degree,ndifndonlyifthenumerofvertiesis0. Proof. The third eqution in (.) impliesthtv v 0, nd v = 0 if nd only if ll verties hve degree. On the other hnd, y the seond eqution, v 0 is the sme s f, nd v = 0 is the sme s f =. Proposition. The spheril tiling y omintorilly ongruent pentgons is uniquely given y Fig. 9.

8 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) 9 0 Fig. 9. Comintoril struture of the miniml pentgon tiling. x A z B x y u C z Fig. 0. Neighorhood of degree vertex. Proof. By Lemm, llvertieshvedegree. We strt with tile P.EhofthefiveedgesofP is shred y P nd extly nother tile. We denote these tiles y P, P, P, P, P nd ssume tht they re rrnged in ounterlokwise order. Consider the vertex A of P on the left of Fig. 0. It is lso vertex of P nd P. Sine A hs degree, esides the two edges of P onneted to A, thereisextlyonemoreedgex lso onneted to A. ThethreeedgesdividetheneighorhoodofA into three orners, eh elonging to one tile. This lol piture implies tht x is ommon edge of P nd P.Thesimilrsitution hppens etween P nd P, etween P nd P, etween P nd P, nd etween P nd P. Therefore the first six tiles re glued together in the wy desried in Fig. 9. Ofoursethisdoesnot yet exlude the possiility tht dditionl identifition my hppen mong these tiles. But t lest the reltions desried in Fig. 9 lredy exist etween these tiles. The edge x hs nother end vertex B esides A. ThethreeedgesonnetedtoB re x, y, z, where y is n edge of P djent to x nd z is n edge of P djent to x. SeemiddleofFig.0. Nowy is shred y P nd nother tile P,ndz is shred y P nd nother tile P.Thethreeedgesx, y, z divide the neighorhood of B into three orners, eh elonging to one tile. Sine two orners lredy elong to P nd P,thethirdornerelongstothesmetile.ThisimpliesthtP = P. Thus the th tile is estlished nd is relted to P nd P s in Fig. 9. Similrly,P n e defined from P nd P, P 9 n e defined from P nd P, P 0 n e defined from P nd P,ndP n e defined from P nd P. The edge z hs nother end vertex C esides B. ItislsovertexofP nd P.Thererethree edges onneted to C. TworeedgesofP nd the third one is u on the right of Fig. 0. Thesitution is the sme s the left of Fig. 0, withp, P, P, C, u in ple of P, P, P, A, x. Wegetthesimilr onlusion tht the edge u is shred y P nd P. We n dedue similr reltions etween P nd P 9,et.ThenthetilesP through P re glued together s in Fig. 9. After onstruting the first eleven tiles, we find tht P, P, P 9, P 0, P hve one free edge eh. The free edge of P is shred y P nd nother tile P.ThefreeedgeofP is shred y P nd nother tile P.ThesitutionissimilrtothereltionetweenP, P, P, P s desried in the middle of Fig. 0. WegetthesimilronlusionthtP = P.Thereforethefivepossileth tiles re ll the sme. Now we find twelve tiles tht re relted s in Fig. 9. Ifthereremoretiles,thenthedditionl tiles hve to e glued to the existing twelve long some verties or edges. This will either inrese the degree of some vertex to e >, or use some edge to e shred y more thn two tiles. Therefore the twelve re ll the tiles in the tiling. Moreover, if there re dditionl identifitions

9 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) d d B d A x B A x e A x Fig.. Impossile edge length rrngements. mong these twelve tiles, then the identifition of verties will inrese the degree of some vertex to e >, the identifition of edges will use some edge to e shred y more thn two tiles, nd the identifition of tiles will redue the totl numer of tiles to e <. Sine ll these should not hppen, there is no more dditionl identifition. We onlude tht Fig. 9 gives the omplete desription of the tiling.. Edge ongruent tiling We study the distriution of edge lengths in spheril tiling y ongruent pentgons. Sine ngles re ignored, we introdue the following onept. Definition. Two polygons re edge ongruent, ifthereisorrespondeneetweentheedges,suh tht the djenies of the edges re preserved nd the edge lengths re preserved. Two spheril (or plnr) tringles with gret r (or stright line) edges re ongruent if nd only if they re edge ongruent. However, this is no longer true for qudrilterls nd pentgons. Moreover, our results tully do not even require the edges to e gret rs (or stright). Proposition. Let,,,...,denotedistintnumers.Theninspheriltilingyedgeongruentpentgons, the edge lengths of the tile must form one of the following five omintions: ={,,,, }, ={,,,, }, ={,,,, }, ={,,,, }, ={,,,, }. Proof. By purely numeril onsidertion, there re seven possile omintions of five numers. Wht we need to show is tht the omintions d ={,,,, d}, de ={,,, d, e}, re impossile. Up to symmetry, ll the possile edge length rrngements of these two omintions re listed in Fig.. In the first rrngement, y Lemm, one of the verties A or B must hve degree. By symmetry, we my ssume deg A = withoutlossofgenerlity.letx e the only other edge t A esides nd d. ThenwehvetilesP nd P desried in the piture. The edge x is djent to n edge of length in P.SineP is edge ongruent to P,ndtheedgesdjenttotheuniqueedgeof length in P re the edges of lengths nd d. Therefore,wehvex = or x = d s fr s the edge length is onerned. Similrly, the djeny of x nd d in P leds to x = or x = s fr s the edge length is onerned. Sine,,, d re distint, we get ontrdition. We hve een very reful with the wording in the rgument ove. We sy n edge of length insted of the edge eusethisllowsthepossiilitythtthetilemyhveseverledgesofthe sme length. This my hppen when,,,... re not ssumed to e distint. Sine distint letters indeed denote distint vlues for the length in the urrent proposition, there is tully no onfusion out the wording the edge. The wording in the susequent disussion will e more sul. In the seond rrngement, y Lemm nd symmetry, we my still ssume deg A =. Then we my introdue x, P, P s efore. By onsidering x in P nd P,weseethtx is djent to nd. SinethereisnosuhedgeinP,wegetontrdition.

10 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Fig.. Edge length rrngements when ll verties hve degree. y x y x x y x Fig.. Impossile edge length rrngements. In the third rrngement, we my gin ssume deg A = nd introdue x, P, P. Then x is djent to oth nd from the viewpoint of P nd P,yetthereisnosuhedgeinP. Proposition. If spheril tiling y edge ongruent pentgons ontins tile with ll verties hving degree,thentheedgelengthsmusterrngedinoneofthefivewysinfig.. Proof. By purely numeril onsidertion, the five possile edge length omintions,,,, in Proposition hve respetively,,,, possile rrngements. Besides the five in Fig., thereminingfourregiveninfig.. Weneedtorguethtthesefourreimpossile. In the first rrngement, x nd y re djent to nd. SinetheonlysuhedgeinP is, we hve x = y =. Then in P,wegetrowxy of three onseutive edges of the sme length. Sine there is no suh row in P,wegetontrdition. In the seond rrngement, x is djent to, ndy is djent to. BythedjenyinP,we hve x = y =. Thentheedgerowxy = ppers in P ut not in P.Wegetontrdition. In the third rrngement, x is djent to nd. Sine there is no suh edge in P, we get ontrdition. In the fourth rrngement, x nd y re djent to nd. By the djeny in P, we hve x = y =. Then ppers twie in P ut only one in P.Wegetontrdition.. Clssifition of edge ongruent tiling The disussion in Setion is lol, euse we ignored the glol omintoril struture given y Propositions. Inthissetion,wedisusshowthelssifitioninProposition n fit the glol struture. We denote the tiles in Fig. 9 y P, P,...,P.WedenoteyE ij the edge shred y the tiles P i, P j,ndyv ijk the vertex shred y P i, P j, P k. We lso nme n edge y its length nd vertex y the length of the edges t the vertex. For exmple, in Fig., theedgee is -edge, nd the edge E is n -edge. Moreover, the vertex V is n -vertex, nd the vertex V is n -vertex. The tiling hs totl of -edges, -edges, -edges, -verties, -verties, nd -verties. We denote the edgewise vertex omintion of the tiling y {,, }. For the edge length omintion (the first pentgon in Fig. ), we silly ssign to ll edges in Fig. 9. Allvertiesre -verties, nd the edgewise vertex omintion is {0 }.Thisn e relized y the regulr dodehedron tiling. Proposition 9. The spheril tilings y edge ongruent pentgons with edge lengths,,,,, where, re given y the tilings in Fig. up to symmetries. The unleled edges hve length.

11 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Fig.. Tilings for the edge length omintion. In proving the min theorem, we will only use the edgewise vertex omintion {, } derived in the proof. Proof. We need to ssign -edges to Fig. 9, suhthtehtilehsextlyone-edge. In suh n ssignment, there is no -vertex nd no -vertex. So the edgewise vertex omintion is {m,n }. On the other hnd, of ll the 0 edges re -edges. Therefore there re N = -edges nd N = -edges. Then from the edgewise vertex omintion, we get m + n = N = nd n = N =. The solution is m = ndn =. Any tile P in Fig. 9 nd the five tiles round P form tiling T P of hlf sphere (with wiggled oundry). Moreover, the remining six tiles lso form tiling of the omplementry hlf sphere. In ft, P hs n ntipodl tile P (the ntipodl of P is P,forexmple),ndthereminingsix tiles form the tiling T P. We sy n ssignment of -edges is splitting, if there is hlf sphere tiling T P, suh tht if -edge is shred y P i nd P j,thenp i T P if nd only if P j T P.ThisimpliesthtP i T P if nd only if P j T P.WelsollthetileP splitting enter. Asplittingssignmentisthentheunionoftwoindependentssignmentsof-edges to T P nd the other -edges to T P.Inthisse,itisesytoseethtthehoieofthe-edge for P ompletely determines the other two -edges in T P.SeeT P in the first three tilings in Fig.. Uptosymmetry, we hve three possile omintions of the hoie of -edges for P nd P, whih give the first three tilings in Fig.. Theyrenotequivlenteusetheyhverespetively,ndsplitting enters. Next we serh for non-splitting ssignments. Assume there re edges onneting two -edges. Without loss of generlity, we my ssume E = E = s in the fourth tiling in Fig.. SineP nd P re not splitting enters, we get E nd E 0. IfE =, thentheonlypossile -edge of P is E =. This further implies tht the -edge of P is E 9 =. ThenP eomes splitting enter. Therefore E, ndtheonlypossile-edge of P is E =. Thisfurther suessively implies tht the -edge of P 0 is E 90,the-edge of P is E,ndE =. Finlly we look for non-splitting ssignments suh tht no edge onnets two -edges. Without loss of generlity, we my ssume E =. SineE, E, E 9, E 0 re onneted to E y one edge, they re not -edges. This implies tht E = E =. ThentheedgesonnetedtoE nd

12 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Fig.. Tilings for the edge length omintion. E y one edge nnot e -edges. This is enough for us to determine the remining -edges nd get the fifth tiling in Fig.. Proposition 0. The spheril tiling y edge ongruent pentgons with edge lengths,,,,, where, is given y the left of Fig. up to symmetries. Proof. By Proposition, wemystrtwithp,withtheedgesrrngedsthethirdinfig.. See Fig.. ThelengthofE is either or. The se E = is desried on the left of Fig.. FromE = E =, weseethttheother three edges of P hve length. Bythesmereson,wegetlltheedgelengthsofP.Thenmong E nd E,oneis nd the other is. Bysymmetry,wemyssumeE = nd E =. Then E =. FromE =, E =, wegetlltheedgesofp.frome = E = E =, wegetll the edges of P.FromE 0 =, E 0 =, wegetlltheedgesofp 0.FromE = E 0 =, weget ll the edges of P.FromE = E = E =, wegetlltheedgesofp.frome 9 =, E 9 =, we get ll the edges of P 9.FromE =, E =, wegetlltheedgesofp.thisonludeslltheedge lengths. The se E = is the right of Fig.. WenimmeditelygetlltheedgesofP, P.Afterwrds, we get ll the edges of P, P.NowweknowthreeedgesofP to e, sothttheothertwoedges re. ThenwegetlltheedgesofP 9,ndfindthtP hs one -edge djent to two -edges, ontrdition. Proposition. The spheril tiling y edge ongruent pentgons nnot hve edge lengths,,,,, for distint,,. Proof. By Proposition, we my strt with P,withtheedgesrrngedsthefourthinFig.. See Fig.. Sine E is djent to nd, wegete =. ThenwesuessivelygetlltheedgesofP, P, P, P, P, P. Now we hve four -edges in P,ontrdition. Proposition. The spheril tiling y edge ongruent pentgons with edge lengths,,,,, where,, redistint,isgivenytheleftoffig. up to symmetries. Proof. By Proposition, wemystrtwithp,withtheedgesrrngedsthefifthinfig.. See Fig.. Sine E is djent to, itslengthiseither or. The se E = is the left of Fig.. FromE =, E =, E =, wegetlltheedgesof P, P.ThenwefurthergetlltheedgesofP.Sinewennothvethree-edges or three -edges in tile, we must hve E = nd then get ll the edges of P, P.Bythesmereson,weget

13 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Fig.. Tilings for the edge length omintion Fig.. Tilings for the edge length omintion. E 90 = nd ll the edges of P 9, P 0.ThenwemygetlltheedgesofP, P, P.Thisonludes ll the edge lengths. The se E = is the right of Fig.. WeimmeditelygetlltheedgesofP.Thenwehve E = or in P.EitherseimpliesE = in P.ThenwegetlltheedgesofP nd find three -edges in P,ontrdition.. Angle ongruent tiling Now we turn to the distriution of ngles in spheril tiling y ongruent pentgons. Sine edge lengths re ignored, we introdue the following onept. Definition. Two polygons re ngle ongruent, ifthereisorrespondeneetweentheedges,suh tht the djenies of edges re preserved nd the ngles etween the djent edges re preserved. Two spheril tringles with gret r edges re ongruent if nd only if they re ngle ongruent. Two plnr tringles with stright line edges re similr if nd only if they re ngle ongruent. However, this is no longer true for qudrilterls nd pentgons. In the regulr dodehedron tiling, ll the ngles re = π. In wht follows, we will lwys reserve for this speil ngle, nd let,,... e ngles not equl to. Inthissetion,wemyosionllyllowsomefrom,,... to e equl. In the lter setions, we will ssume,,... to e distint.

14 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Lemm. In spheril tiling y ngle ongruent pentgons, the sum of five ngles in tile is 0π =. In se the edges re gret rs, the lemm is onsequene of the formul for the re of spheril pentgon. However, the lemm holds even if the edges re not gret rs. Proof. The sum of ngles t eh vertex is π.bylemm, therere0verties.thereforethetotl sum of ll the ngles in the tiling is 0π.Ontheotherhnd,thesumΣ of five ngles in tile is the sme for ll tiles euse of the ngle ongruene. Sine there re tiles, we get 0π = Σ. This implies Σ = 0π. In Setion, wenmedvertexythelengthsofedgestthevertex.similrly,wenlsonme vertexythenglestthevertex.forexmple,infig., V is -vertex, nd V is n -vertex. The tiling hs -verties nd -verties, nd hs the nglewise vertex omintion {, }. Welsosytht nd re verties in the tiling, nd the other omintions re not verties. We sy vertex is of -type, ifthevertexisn -vertex, where nd my or my not e nd.thusn -type vertex n e n -vertex, or n -vertex, or n -vertex, et. From purely symolil viewpoint (rell tht the symol is speil), degree vertex must e one of the following types:,,,,,,. Lemm. Avertexofdegree must e one of the following types:... with.. with.. with,, distint. Proof. We need to show tht, if, then,, nnot e verties. If is vertex, then + = π =, whih implies =, ontrdition.if is vertex, then + = π =, whih gin implies =. If is vertex, then = π =, whihstillimplies =. Lemm. Suppose,,, re distint ngles.. If is vertex, then,, re not verties.. If is vertex, then, re not verties. The lemm tully llows some ngles to e. Moreover,wegetmoreexlusionsysymmetry. For exmple, if is vertex, then the following nnot e verties:,,,,,,,,. Proof. If oth nd re verties, then + = nd + =. Thisimplies =, ontrdition. If oth nd re verties, then + = nd + =. Thisimplies =,ontrdition. If oth nd re verties, then + = nd + + =. Thisimplies =, gin ontrdition. The proof of the seond sttement is similr. Proposition. Let = π nd let,,,... denote distint ngles. Then spheril tiling y ngle ongruent pentgons must hve one of the following (unordered) ngle omintions:

15 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0). : The ngles in the pentgon re,,,,.thenglewisevertexomintionis{0 }.. : The ngles in the pentgon re,,,,,stisfying + =.Thenglewisevertexomintion is {, }.. : The ngles in the pentgon re,,,,,stisfying + =. Thenglewisevertex omintion is {, }.. : The ngles in the pentgon re,,,,,stisfying + + =. Thenglewisevertex omintion is {, }.. : The ngles in the pentgon re,,,,, stisfying =, =, =. There re three possile nglewise vertex omintions: {,,, }, {,,,, }, {,,, }. The fourth se degenertes into the seond (when = ) ndthethird(when = ) ses,whih further degenerte into the first se. The tiling T in Fig. nd Fig. is the fourth se nd its degenerte ses. The tilings T, T, T, T in Fig. elong to the third nglewise vertex omintion in the fifth se. Proof. We divide the proof y onsidering how mny times the ngle ppers in the pentgon. Cse. If the ngles in the pentgon re,,,,,thenlemm tells us + =, so tht the ngles re relly,,,,. Thisisthefirstseintheproposition. Cse. The ngles in the pentgon re,,,,,with,. Notetht, re not yet ssumed to e distint. Lemm eomes + =. If is vertex, then + = π =, ndweget = =, ontrdition.bythesimilrreson, is not vertex. By Lemm, theonlypossileverties re nd,nd, must e distint. Let {m,n } e the nglewise vertex omintion. Sine tiles with ngles,,,, in eh tile hve totl of ngle, ngle nd ngle, we get m + n = nd n =. Therefore m = ndn =. This is the seond se of the proposition. Cse. The ngles in the pentgon re,,,,,with,,. Notetht,, re not yet ssumed to e distint. Lemm eomes + + =. If is vertex, then + + =, ndweget =, ontrdition.thergumenttullyshowsthtthereisno -type vertex. If is vertex, then + = nd = =. Thusthe -vertex is relly -vertex. The rgument tully pplies to ny -type vertex. By Lemm, weonludetht ny vertex is either n -vertex or -vertex. Let {m,n } e the nglewise vertex omintion. Without loss of generlity, we my further ssume tht is different from,,.sinetileswithngles,,,, hve totl of ngle nd ngle,wegetm =, n =. Therefore m = ndn =. Depending on whether =, we get the third nd the fourth ses of the proposition. Cse. The ngles in the pentgon re,,,,, with,,,. Agin,,, re not yet ssumed to e distint. Lemm eomes =. Wefurtherdividetheseyonsideringwhethersome mong,,, re equl. Cse.. = = =. Lemm eomes =, ndweget =, ontrdition. Cse.. = =. Lemm eomes + =. Thereisno -vertex euse this implies + + =, nd we get = =. Similrrgumentshowstht, re not verties. By Lemm, the only vertex is,whihisimpossile. Cse.. = =. Lemm eomes + =. AnrgumentsimilrtoCse.showstht, re not verties. By Lemm, theonlyvertiesre nd,sothtthenglewisevertexomintion is {m,n }. Sinetileswithngles,,,, hve totl of ngle nd ngle, we get m + n = nd n =. The system hs no non-negtive solution, ontrdition.

16 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) 9 Cse.. =, nd,, re distint. Similr to Cse.,,,, re not verties. By the symmetry of nd in the urrent se nd the exlusion etween nd (see Lemm ), we my ssume tht the only possile -type vertex is.then y Lemm, the nglewise vertex omintion is { m,m,n,n,k,k }. Sine tiles with ngles,,,, hve totl of ngle, ngle, ngle nd ngle, we get m + m =, m + n + n =, m + n + k + k =, n + k + k =. Adding the third nd the fourth equtions together, we get m +(n +n )+(k +k ) =. Compred with the seond eqution, we get n = n = k = k = 0. Then the seond eqution eomes m =, ontrditing to the first eqution. Cse..,,, re distint. Similr to Cse., there is no -type vertex. By Lemm, theonlypossiletypesre,,.wefurtherdivideintotwosuses. Cse... ppers in two verties not involving nd with different ngle omintions. Without loss of generlity, we my ssume tht the two verties re i i nd j j, 0< i, j <. By Lemm, we hve i j. Therefore up to symmetry, we my ssume, re verties. Solving + = + = together with = from Lemm, weget =, + =, + =. Now the ngle must pper t some vertex. Sine there is no -type vertex, y Lemms nd, oneof,,,,,, must pper s the vertex involving. If is vertex,then + = nd = imply =, ontrdition.allotherpossiilitiesledto similr ontrditions. Cse... The opposite of Cse.. nd its symmetri permuttions. This mens tht, without loss of generlity, we my ssume tht the only possile verties not involving re nd.thenthenglewisevertexomintionis,, together with some -type verties. By Lemm, wefindonlytwopossiilities: { m,n,n,k,k } {, m,n,n,k,k }. In the first omintion, ounting the totl numer of eh ngle gives m + n + n = n + k = n + k = n + k = n + k =. We get k = k = m +, n = m, n = m. Inprtiulr,wehven, k, k > 0, so tht,, re verties, nd we get + =, + =, + =. The solution is =, =, =. On the other hnd, to mintin n 0, the possile vlues for m re 0,,. Then we get the orresponding nglewise vertex omintions, whih re the three omintions in the fifth se of the proposition. The seond omintion gives m + n + n = n + k = n + k = n + k = n + k =. The system hs no non-negtive solution.

17 0 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Cse. The ngles in the pentgon re,,,,ζ,lldifferentfrom. Butthefivenglesrenot yet ssumed to e distint. Lemm eomes ζ =. Wefurtherdividetheseyonsideringwhether some ngles re equl. Cse.. = = = = ζ. Lemm implies =, ontrdition. Cse.. = = = ζ. The verties must e or, nd Lemm eomes + =. If is vertex, then + =, ndweget =,ontrdition.if is vertex, then we get the similr ontrdition. Cse.. = = = ζ. The verties must e or.wegetontrditionsimilrtocse.. Cse.. = = ζ,nd,, re distint. Lemm eomes + + =. Thisimpliestht is not vertex, so tht ll verties re -type. Sine there re three distint ngles, one ngle must pper in two verties with different ngle omintions. By Lemm nd the symmetry etween nd, we only need to onsider oth, pper, or oth, pper, or oth, pper. If, re verties, then + = nd + =. Cominedwith + + =, we get ll ngles equl, ontrdition. The other pirs of verties led to similr ontrditions. Cse.. =, = ζ,nd,, re distint. Lemm eomes + + =. Thisimpliestht,, re not verties. Sine there re three distint ngles nd ll verties re -type, one ngle must pper in two verties with different ngle omintions. By Lemm, thesymmetryetween nd,ndthefttht nd re not verties, we my ssume tht, re verties. This implies + = nd + =. Cominedwith + + =, wegetllnglesequl,ontrdition. Cse.. = ζ,nd,,, re distint. Lemm eomes =. Thisimpliestht is not vertex. If there is vertex of -type, then up to symmetry, we my ssume is vertex. Then we get + + = nd + = ( + + ) =. Thisimpliestht,, nnot form -type verties, nd, re not verties. So the only possile verties of -type re,,,.moreover,ylemm, thepperneof exludes nd.thenythesymmetry etween nd in our urrent sitution, we only need to onsider the nglewise vertex omintion {n,k, k }.Countingthetotlnumerofehnglegives n =, n + k = n + k = k + k =. The system hs no non-negtive solution. So ll verties re of -type, nd the nglewise vertex omintion is { m,m,m,n,n,n,k,k,k,k,k,k }. Counting the totl numer of the ngle gives (m + m + m ) + n + n + n =. If m = m = m = 0ndn > 0, then y Lemm, wegetn = n = 0ndn =. Therefore the totl numer of the ngle is n =, ontrdition. Similr rgument rules out n > 0nd n > 0. Therefore m,m,m nnot e ll 0, nd we my ssume m > 0withoutlossofgenerlity. By Lemm, this implies m = m = n = k = k = 0, nd we get m + n + n =. If n = n = 0, then we get m =. Thus the -verties lredy ontin ll ppernes of the ngle.therefore does not pper in ny other vertex, nd we re left with the nglewise vertex omintion {, k, k }.Byountingthetotlnumersofthengles nd, we get k + k = k + k =. This implies k = k = > 0. Therefore oth nd re verties, ontrditiontolemm. So we my ssume n > 0. By Lemm, wegetn = k = 0, so tht the nglewise vertex omintion eomes {m,n, k,k, k }.Byountingthetotlnumerofehngle, we get m + n =, m + k + k = n + k + k = k + k =.

18 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) By Lemm, wemusthveeitherk = 0ork = 0. If k = 0, the system gives k =.Ifk = 0, the system gives k =.Sinethenumersrenotintegers,wegetontrdition. Cse..,,,,ζ re distint. Cse... There re -type verties. We my ssume is vertex. This implies + + =, sotht,, nnot form - type verties. By Lemm, wehve + ζ = ( + + ) =, sotht,ζ nnot pper t the sme vertex. By Lemm, the only -type vertex is, ndtheonlywyfor,ζ to pper is to get omined with ngles from,,. If,ζ n e omined with the sme ngle from,,,thenuptosymmetry,wemyssume,, ζ re verties. We get + + = + = + ζ =. Togetherwith + ζ =, we get =, ontrdition. So,ζ must e omined with different ngles from,,.bylemm nd up to symmetry, there re three possile omintions. In the first omintion,,, ζ re verties. We get + + = + = + ζ =. Together with + ζ =, weget =, ontrdition. In the seond omintion,,, ζ re verties. Similr to the first omintion, we lso get the ontrdition =. In the third omintion,,, ζ re verties. We get + + = + = + ζ =. Together with + ζ =, wedonotgetontrditionsefore.however,if,, ζ re the only verties, then the totl numer of will e stritly igger thn the totl numer of. Sine oth totl numers re tully, we onlude tht these nnot e ll the verties. By Lemm, the only other possile verties re, ζ.if is vertex, then + =. Addingthistothe equtions ove, we get ll ngles equl, ontrdition. If ζ is vertex, then we get similr ontrdition. Cse... There is no -type vertex. We my ssume is vertex. Sine there is no -type vertex, y Lemm, the only other vertex involving is θ for unique θ, ndtheonlyothervertexinvolving is ρ for uniqueρ. Sineehofthreedistintngles,,ζ must pper t some vertex, there must e some piring mong,,ζ.withoutlossofgenerlity,therefore,wemyssumeoth, re verties. Moreover, up to symmetry, we my further ssume tht ζ is pired with or.giventhe existene of,theonlypossiilitiesreζ nd ζ.soeither,, ζ re verties, or,, ζ re verties. If,, ζ re verties, then + = + = + ζ =. Cominedwith ζ = from Lemm, wegetthefollowingpirwisereltions + =, =, =, + =, ζ =, + ζ =, + ζ =. The reltions exlude ll other verties, whih must e of -type. Therefore the nglewise vertex omintion is {m,n, kζ }.Thisimpliesthtthetotlnumerof is stritly igger thn the totl numer of.sineothnumersre,wegetontrdition. If,, ζ re verties, then + = + = +ζ =. Similrtothepreviousse,the equtions together with ζ = from Lemm exlude ll other verties. Therefore the nglewise vertex omintion is {m,n, k ζ }. Thenthetotlnumerof is more thn the totl numer of, ontrdition.. Clssifition of ngle ongruent tiling Proposition lssifies the numeris of spheril tilings y ngle ongruent pentgons. In this setion, we study how the numeril informtion n fit the omintoril struture in Fig. 9, similr to wht we hve done for edges in Setion. In Setion, weintroduedthenottionsp i, E ij, V ijk for the tiles, the edges nd the verties in Fig. 9. Inthissetion,wedenoteyA i, jk the ngle of the tile P i t the vertex V ijk.wewilleven write V ijk = if we know V ijk is n -vertex.

19 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Fig.. Some tilings for the ngle omintion. Inside speified tile, we n lso nme vertex y its ngle nd n edge y the two ngles t the ends. For exmple, in Fig., A, is -ngle, A, nd A, re -ngles. Moreover, E nd E re -edges of P. In this setion, we let = π nd ssume,,,... re lwys distint. There is not muh to lssify for the ngle omintion.bsilly,wessign to ll ngles in Fig. 9. Thisnerelizedytheregulrdodehedrontiling. There re mny ngle ongruent tilings for the ngle omintion.fortuntely,thelssifition is not needed in the proof of the min theorem. We simply present some exmples in Fig.. Note tht for ny thik line in ny tiling, we n exhnge the ngles nd long the line nd still get n ngle ongruent tiling. Proposition. The spheril tilings y ngle ongruent pentgons with ngles,,,,,where = π nd,, re distint, re given y Fig. up to symmetry. Note tht in the tiling on the right, we n exhnge the four ngles nd round ny thik line nd still get tiling. As n exmple of the exhnge, on the right of Fig., wenexhnge nd round the edge E to get A, = A, = nd A, = A, =. ButweshouldkeepA, = A, =. The result is lso n ngle ongruent tiling.

20 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Fig. 9. Four possile rrngements for the ngle omintion. Fig. 0. Tilings for the ngle omintion. Proof. We re in the third se of Proposition. Thenglewisevertexomintionis{, }, whih implies the following AVC (for nglewise vertex omintion) ondition. AVC(for ): Any vertex is either or. Up to symmetry, there re four possile wys of rrnging the ngles in the pentgon, desried in Fig. 9. WestrtyssumingthenglesofthetileP in Fig. 9 re rrnged s in Fig. 9. Arrngement. The se is the left of Fig. 0. ByA, = nd the AVC ondition, we get V =,sothta, = A, =. ByA, = A, = nd the AVC ondition, we get V = V =, so tht A, = A, = A, = A, =. ByA, =, A, =, wegetllthenglesofp.then y A, = nd the AVC ondition, we get V =,sothtthererethree in P,ontrdition. Arrngement. The se is the right of Fig. 0. ByA, = nd AVC, we get A, = A, =. By A, = nd AVC, we get V =,sothta, = or.sinethetwo in P re not djent, we get A, =.TogetherwithA, =, wegetllthenglesofp.bythesmereson, we get ll the ngles of P.ThenyAVC,wegetV = V = V =,sothtthererethree in P,ontrdition. Arrngement. The se is the left of Fig.. ByAVC,wegetllthenglestV, V, V. By A, = A, =, A, = A, =, wegetllthenglesofp, P.ByAVC,wefurthergetll the ngles t ll the verties of P, P.Thenglestthesevertiesthenfurtherdeterminellthe ngles of P, P, P, P, P.ByAVC,wegingetllthenglestllthevertiesofthesetiles. Then we further get ll the ngles of P, P 9, P 0.FinllywegetllthenglesofP y AVC. Arrngement. The se is the right of Fig.. ByAVC,wegetllthenglestV, V, V. We lso find V =, so tht A, = or. Sine A, = nd the two in P re not djent, we get A, = nd then ll the ngles of P. By AVC, we further get ll the ngles t V, V 9, V 0.Wenotethtwemyexhngethefourngles, round E without ffeting the susequent disussion. Now we experiment with the ngle A,. By AVC, it must e either or.ifa, =,then A, = nd we get ll the ngles of P.ByAVC,wegetV =.SineA, = nd the two in P re not djent, we get A, = nd then ll the ngles of P. By AVC, we further get ll the ngles t V, V, V.IfA, =, thena, = nd we my rry out the similr rgument (first on P nd then on P ). The result is the sme piture with the four ngles, round E exhnged.

21 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Fig.. Tilings for the ngle omintion, ontinued. 9 0 Fig.. Tilings for the ngle omintion. We my suessively rry out the sme experiment round eh of E, E, E 90, E.Wefind tht we n mke similr independent hoies of,,ndehhoiedeterminesllthenglesof the pir of tiles on the two sides of the edge. Proposition 9. The spheril tiling y ngle ongruent pentgons with ngles,,,,,where = π nd,,, re distint, is given y the right of Fig. up to symmetry. Note tht in the tiling on the right, we n exhnge the four ngles nd round ny thik line nd still get tiling. Proof. We re in the fourth se of Proposition. Thenglewisevertexomintionis{, }, whih implies the following AVC ondition. AVC(for ): Any vertex is either or. Up to symmetry, there re two possile wys of rrnging the ngles in the pentgon. Arrngement. In this rrngement, the two re djent. See the tile P on the left of Fig.. By AVC, we get V = V = nd V =. ThusA, = or,ndisdjentto in P.

22 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Fig.. Eight possile rrngements for the ngle omintion. Therefore A, =. TogetherwithA, =, wegetllthenglesofp.thena, = implies V =,ndthererethree in P,ontrdition. Arrngement. In this rrngement, the two re not djent. See the tile P on the right of Fig.. ByAVC,wegetV = V = nd V =. IfA, =, A, =, thenwegetll the ngles of P, P nd ll the ngles t V, V, V.IfA, =, A, =,thenwegetthe similr result, exept tht the four ngles, round E re exhnged. Suh exhnge does not ffet the susequent disussion. By A, =, weknowa,, A, re,.wemymkeeitherhoiefora,, A, nd then get ll the ngles of P, P.SimilrdisussionsnthenesuessivelyrriedoutroundE, E 90, E nd determine ll the ngles. Proposition 0. If the ngles in spheril tiling y ngle ongruent pentgons re,,,,, with = π nd ll ngles distint, then up to symmetry, the tiling must hve nglewise vertex omintion {,,, } nd is given y one of the tilings on the left of Figs.,,,. Proof. We re in the lst se of Proposition. The following AVC ondition pplies to ll three possile nglewise vertex omintions in the se. AVC(for ): Any vertex is one of,,,,. Aonsequenewewilloftenuseistht,,, re foridden t ny vertex. All three nglewise vertex omintions re symmetri with respet to the simultneous exhnge of with nd with. Modulosuhsymmetry,therereeightpossilerrngementsofthengles in the pentgon, desried in Fig.. WedisussehseyssumingthenglesofP in Fig. 9 re rrnged s in Fig.. Sine the vertex ppers in ll three nglewise vertex omintions, we will lso ssume V =. ThisledstothepossiilitiesA, =, A, = nd A, =, A, =. Sine ll ngles re distint, there is no miguity for us to ll the orienttion of ngles given y Fig. ounterlokwise, ndlltheotherorienttionlokwise. Arrngement, Cse. A, =, A, =. The se is the left of Fig.. SineA, is djent to A, = in P,wegetA, = or. By AVC, we get A, = nd ll the ngles of P. By A, =, A, = nd AVC, we get A, =. Sine A, is djent to in P, we get A, = or. ByAVC,wegetA, = nd ll the ngles of P. By A, =, A, = nd AVC, we get A, =. Sine A, is djent to in P, we get A, = or. ByAVC,wegetA, = nd ll the ngles of P. By A, =, A, = nd AVC, we get A, =. SineA, is djent to in P,weget A, = or. ByAVC,wegetA, = nd ll the ngles of P. By AVC, we further get A, =,whihisdjenttoa, = in P.Thisisontrdition. We remrk tht the onlusion of the se is the following: If the ngles in tile (of first rrngement) re ounterlokwise oriented, nd the vertex t is, then,, must e lokwise

23 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) 0 Fig.. Tilings for the first rrngement. oriented t the vertex. By symmetry, we lso know tht if the ngles in tile re lokwise oriented, then,, must e ounterlokwise oriented t the vertex. Arrngement, Cse. A, =, A, =. The se is the middle nd the right of Fig.. SineA, is djent to in P,wegetA, = or. If A, =, thenwegetllthenglesofp nd A, = in prtiulr. By AVC, we lso get A, =. BeingdjenttoA, = in P, one of A,, A, is.thisontrditstoavctv or V. Therefore we must hve A, =. ThenwegetllthenglesofP nd A, =.Nowonsider the two possile orienttions of P. In the middle of Fig., P is ounterlokwise oriented. By AVC, we get A, =, ndthengle A, djent to in P must e. ThisdeterminesllthenglesofP.NowP is ounterlokwise oriented, nd,, is lso ounterlokwise oriented t V.Thisontrditstotheonlusionof Cse. On the right of Fig., P is lokwise oriented. By AVC, we get A, =. ThenthengleA,0 djent to in P is or. If A,0 =, then P is lokwise oriented, nd,, re lso lokwise oriented t V 0.ThisontrditstotheonlusionofCse.Thereforewemusthve A,0 =. ThenwegetllthenglesofP nd y AVC, we further get A, =. Thelokwise orienttion of,, t V nd the onlusion of Cse imply tht P must e ounterlokwise oriented. This determines ll the ngles of P.ThenyAVC,wegetA, =,ndthereretwo in P,ontrdition. Arrngement, Cse. A, =, A, =. Consider the two possile orienttions of P.TheounterlokwiseseistheleftofFig.. By AVC, we my suessively determine ll the ngles of P, P, P, P, P, P 0, P, P, P, P 9. There is no ontrdition, nd we get n ngle ongruent tiling with nglewise vertex omintion {,,, }. On the right of Fig., P is lokwise oriented. By AVC, we get A, =, thengetllthengles of P,ndfurthergetA, =. NowoneofA,, A,9 djent to in P is.thisontrdits to AVC t V or V 9. Arrngement, Cse. A, =, A, =. By AVC, we get ll the ngles of P nd A, =. IfP is ounterlokwise oriented, s on the left of Fig., thenyavc,wegeta, =. Now one of A,, A,0 djent to in P is. This ontrdits to AVC t V or V 0.IfP is lokwise oriented, s on the right of Fig., then y AVC, we get ll the ngles of P, P nd further get A, =. TogetherwithA, =,weget A, =. ThisontrditstoAVCtV. Arrngement, Cse. A, =, A, =. If P is lokwise oriented, s on the upper right of Fig., thenyavc,wemysuessively determine ll the ngles of P, P, P nd get A, =. Thus we find nd djent in P, ontrdition.similrugmentshowsthtp nnot e ounterlokwise oriented. Thus we get ll the ngles of P, P s desried on the left of Fig.. ThenyAVC,wemysuessivelydetermine

24 H. Go et l. / Journl of Comintoril Theory, Series A 0 (0) Fig.. Tilings for the seond rrngement. 0 Fig.. Tilings for the seond rrngement, ontinued. ll the ngles of P, P, P, P 0, P, P, P, P 9, P.Thereisnoontrdition,ndwegetnngle ongruent tiling with nglewise vertex omintion {,,, }. Arrngement, Cse. A, =, A, =. The se is the lower right of Fig.. ByAVC,wegetllthenglesofP, P nd further get A, =. ThenthengleA, djent to in P is either or. EitherwyontrditstoAVC t V. Arrngement, Cse. A, =, A, =. By AVC, we get ll the ngles of P, P nd A, =. IfP is ounterlokwise oriented, s on the left of Fig., thenyavc,wegetllthenglesofp nd A, =. Then nd re djent in P,ontrdition.IfP is lokwise oriented, s on the right of Fig., thenyavc,wegetll the ngles of P 0 nd A,0 =. ThenoneofA,, A,0 djent to A,0 in P must e. This ontrdits to AVC t V or A 0. Arrngement, Cse. A, =, A, =. If P is ounterlokwise oriented, s on the left of Fig. 9, thenyavc,wesuessivelygetll the ngles of P, P, P nd A, =, A, =. Wefind nd djent in P,ontrdition. If P is lokwise oriented, s on the right of Fig. 9, thenyavc,wesuessivelygetllthengles of P, P, P nd A, = A, =. Wefindtwo in P,ontrdition. Arrngement, Cse. A, =, A, =. We tret the seond se first euse the onlusion will e useful for the first se. By AVC, we get ll the ngles of P, P.IfP is lokwise oriented, s on the left of Fig. 0, thenyavc,we suessively get ll the ngles of P, P, P 9, P 0.ThenwegetontrditiontoAVCtV 0.