Tilings of the sphere with right triangles, II: the asymptotically right families

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1 Tilings of the sphere with right tringles, II: the symptoticlly right fmilies Robert J. McG. Dwson Deprtment of Mthemtics nd Computing Science Sint Mry s University Hlif, Nov Scoti, Cnd Blir Doyle HB Studios Multimedi Ltd. Lunenburg, Nov Scoti, Cnd B0J C0 October 7, 005 Abstrct Sommerville [0] nd Dvies [] clssified the sphericl tringles tht cn tile the sphere in n edge-to-edge ( norml ) fshion. Reling this condition yields other tringles, which tile the sphere but hve some tiles intersecting in prtil edges. This pper determines which right sphericl tringles within certin fmilies cn tile the sphere. Keywords: sphericl right tringle, monohedrl tiling, non-norml, non-edge-to-edge, symptoticlly right AMS subject clssifiction: 5C0 Supported by grnt from NSERC Supported in prt by n NSERC USRA

2 Introduction A tiling is clled homohedrl if ll tiles re congruent, nd norml if two tiles tht intersect do so in single verte or n entire edge. In 93, D.M.Y. Sommerville [0] clssified the norml homohedrl tilings of the sphere with isosceles tringles, nd those with sclene tringles in which the ngles meeting t ny one verte re congruent. H.L. Dvies [] completed the clssifiction of norml homohedrl tilings by tringles in 967 (pprently without knowledge of Sommerville s work), llowing ny combintion of ngles t verte. (Coeter[] nd Dwson[5] both erred in filing to note tht Dvies does include tringles - notbly the hlf- nd qurter-lune fmilies - tht Somerville did not consider.) There re, of course, resons why the norml tilings re of specil interest; however, non-norml tilings do eist, including some [3,, 5] using tiles tht cnnot tile in norml fshion. In [3] complete clssifiction of isosceles sphericl tringles tht tile the sphere ws given. In [5] specil clss of right tringles ws considered, nd shown to contin only one tringle tht could tile the sphere. This pper nd its compnion ppers [6, 7, 8] continue the progrm of clssifying the tringles tht tile the sphere, by giving complete clssifiction of the right tringles with this property. Non-right tringles will be clssified in future work. Bsic results nd definitions In this section we gther together some elementry definitions nd bsic results used lter in the pper. We will represent the mesure of the lrger of the two non-right ngles of the tringle by β nd tht of the smller by γ. The lengths of the edges opposite these ngles will be B nd C respectively, with H s the length of the hypotenuse. (Note tht it my be tht β>90 nd B>H.) We will mke frequent use of the well known result 90 <β+ γ<70, β γ<90 () We will denote the number of tiles by N; this is of course equl to 70 /(β + γ 90 ). Let V = {(, b, c) Z 3 : α + bβ + cγ = 360,, b, c 0}. We cll the triples (, b, c) the verte vectors of the tringle nd the equtions verte equtions. The verte vectors represent the possible (unordered) wys to surround verte with the vilble ngles. We cll V itself the verte signture of the tringle. For right tringles V is lwys nonempty, contining t lest (, 0, 0). Any subset of V tht is linerly independent over Z nd genertes V is clled bsis for V. All bses for V hve the sme number of elements; if bses for V hve n + elements we will define the dimension of V, dim(v), to be n. An oblique tringle could in principle hve V = nd dim(v) = ; but such tringle could not even tile the neighborhood of verte. The dimension of V my be less thn the dimension

3 of the lttice {(, b, c) Z 3 : α+bβ +cγ = 360 } tht contins it, but it cnnot be greter. If tringle cn tile the sphere in non-norml fshion, it must hve one or more split vertices t which one or more edges ends t point in the reltive interior of nother edge. The ngles t such split verte must dd to 80, nd two copies of this set of ngles must give verte vector in which,b, nd c re ll even. We shll cll (, b, c) even nd (, b, c)/ split vector. We will cll (, b, c)/ nβ (resp. γ) split if b (resp c) is nonzero. If both re nonzero we will cll the split vector βγ split. It is esily seen tht if (3,,b) V, then lso (0,, b) V; nd if (,,b) V, then lso (0,, b) V. A verte vector with = 0 or will be clled reduced. If V contins vector (, b, c) such tht (, b, c)/ isβ split, γ split, or βγ split, then it must hve reduced vector corresponding to split of the sme type, which must necessrily hve =0. The following result ws proved in [5]: Proposition The only right tringle tht tiles the sphere, does not tile normlly, nd hs no split vector prt from (, 0, 0)/ is the (90, 08, 5 ) tringle. In fct, this tringle tiles in ectly three distinct wys. One is illustrted in Figure ; the others re obtined by rotting one of the equilterl tringles, composed of two tiles, tht cover the polr regions. This lets us prove: Figure : A tiling with the (90, 08, 5 ) tringle Proposition For ny right tringle tht tiles the sphere but does not tile normlly, dim(v) =. 3

4 Proof: A right tringle with dim(v) = 0 would hve (, 0, 0) s its only verte vector, which mens tht the neighborhood of β or γ corner could not be covered. Moreover, the lttice {(, b, c) Z 3 : α + bβ + cγ = 360 } is t most two-dimensionl; so dim(v). If dim(v) =, the other bsis vector V =(,b,c ) must hve b = c. If =,b = c or =0,b = c, we would hve β + γ 90 ; nd if = 0,b = c = the tringle is of the form (90,θ,80 θ) nd tiles normlly. Thus, under our hypotheses, there is no second split, nd the only such tringle tht tiles but not in norml fshion is (by the previous proposition) the (90, 08, 5 ) tringle. However, this hs V = {(, 0, 0), (,, ), (,, 3), (, 0, 5)}, nd dim(v) =. Corollry There re no continuous fmilies of right tringles tht tile the sphere but do not tile normlly. Proof: As dim(v) =, the system of equtions α +0β +0γ = 360 () α + b β + c γ = 360 (3) α + b β + c γ = 360 () hs unique solution (α,β,γ) whose ngles (in degrees) re rtionl. Note: Both the requirement tht the tringle be right, nd the requirement tht it llow no norml tiling of the sphere, re necessry. Consider the ( 360 n, 80 θ, θ) tringles where n is, in the first cse, odd, nd, in the second cse, equl to. In ech cse dim(v) = for lmost every θ nd the fmily is continuous. We my lso consider the tringles with α + β + γ =π; four of ny such tringle tile the sphere, lmost every such tringle hs dim V = 0, nd they form continuous two-prmeter fmily.. The irrtionlity hypothesis With few well-known eceptions such s the isosceles tringles, nd the hlfequilterl tringles with ngles (90, θ, θ/), it seems nturl to conjecture tht sphericl tringle with rtionl ngles will lwys hve irrtionl rtios of edge lengths. This irrtionlity hypothesis is probbly not provble without mjor dvnce in trnscendence theory. However, for our purposes it will lwys suffice to rule out identities of the form ph + qb + rc = p H + q B + r C where p,q,r,p,q,r re positive nd the sums re less thn 360. For ny specified tringle for which the hypothesis holds, this cn be done by testing rther smll number of possibilities, nd without ny gret precision in the rithmetic. This will generlly be done without comment.

5 Note: The possibility tht some liner combintion pa + qb + rc of edge lengths will hve rtionl mesure in degrees is not ruled out, nd in fct this is sometimes the cse. For instnce, the (90, 60, 0 ) tringle hs H +B +C = 80. Note: It will be seen below tht, while norml tilings tend to hve mirror symmetries, the symmetry groups of non-norml tilings re usully chirl. The irrtionlity hypothesis offers n eplntion for this. Frequently there will only be one wy (up to reversl) to fit tringles together long one side of n etended edge of given length without obtining n immeditely impossible configurtion. If the configurtion on one side of n etended edge is the reflection in the edge of tht on the other, the tiling will be loclly norml. A non-norml tiling must hve n edtended edge where this does not hppen; the configurtion on one side must either be completely different from tht on the other or must be its imge under 80 rottion bout the center of the edge. Note: It my be observed tht ll known tilings of the sphere with congruent tringles hve n even number of elements. This is esily seen for norml tilings, s 3N =E (where E is the number of edges.) The irrtionlity hypothesis, if true, would eplin this observtion in generl. A miml rc of gret circle tht is contined in the union of the edges will be clled n etended edge. Ech side of n etended edge is covered by sequence of tringle edges; the sum of the edges on one side is equl to tht on the other. In the bsence of ny rtionl dependencies between the sides, it follows tht one of these sequences must be rerrngement of the other, so tht 3N is gin even. In light of this, one might wonder whether in fct every tringle tht tiles the sphere dmits tiling tht is invrint under point inversion nd thus corresponds to tiling of the projective plne; however, while some tiles do dmit such tiling, others do not. For instnce, it is shown below tht the (90, 75, 60 ) tringle dmits, up to reflection, unique tiling; nd the symmetry group of tht tiling is Klein -group consisting of the identity nd three 80 rottions.. Clssifiction of β sources It follows from Proposition tht the verte signture of every tringle tht tiles but does not do so in norml fshion must contin t lest one vector with b>,cnd t lest one with c>,b. We will cll such vectors β sources nd γ sources respectively; nd we my lwys choose them to be reduced. Henceforth, then, we will ssume V to hve bsis consisting of three vectors V 0 =(, 0, 0), V =(, b, c), nd V =(,b,c ), with, <, b>c, nd b <c. (For some tringles, more thn one bsis stisfies these conditions; this need not concern us.) The restrictions tht β>γnd b>cleve us only finitely mny possibilities for V. In prticulr, if b+c >7, then 360 = bβ +cγ > β +γ nd β +γ <90, 5

6 which is impossible. Similrly, if = we must hve b + c 5. We cn lso rule out the vectors (0,, 0), (0,, 0), nd (,, 0), ll of which force β 80. We re left with possibilities for V. We my divide them into three groups, depending on whether lim c β is cute, right, or obtuse. The symptoticlly cute V re (0, 7, 0), (0, 6, ), (0, 6, 0), (0, 5, ), (0, 5, ), (0, 5, 0), (, 5, 0), (,, ), nd (,, 0). As for lrge enough c these yield Eucliden or hyperbolic tringles, there re only finitely mny vectors V tht cn be used in combintion with ech of these. The symptoticlly right V re (0,, 3), (0,, ), (0,, ), (0,, 0), (, 3, ), (, 3, ), nd (, 3, 0). Ech of these vectors forms prt of bsis for V for infinitely mny sphericl tringles. The symptoticlly obtuse V re (0, 3, ), (0, 3, ),(0, 3, 0),(0,, ),(,, ), nd (,, 0). For lrge enough c these yield triples of ngles tht do not stisfy the second ngle inequlity; so gin there re only finitely mny possible V to consider. In the reminder of this pper, we will clssify the tringles tht tile the sphere nd hve verte signtures with symptoticlly right V. One prticulrly lengthy subcse is delt with in compnion pper [6]. A pper now in preprtion will clssify the right tringles tht tile the sphere nd hve verte signtures with symptoticlly obtuse or cute V, completing the clssifiction of right tringles tht tile the sphere. 3 The min result The min result of this pper is the following theorem, the proof of which will be deferred until the net section. Theorem The right sphericl tringles which hve verte signtures with symptoticlly right V re i). (90, 90, 80 n ), ii). (90, 60, 5 ), iii). (90, n iv). (90, n v). (90, 75, 60 ), vi). (90, 60, 0 ), vii). (90, 75, 5 ), nd viii). (90, 78 3, 33 3 )., 360 n ) for even n,, 360 n ) for odd n>, 6

7 The first three of these tile normlly, though they lso dmit non-norml tilings. The remining five hve only non-norml tilings. We now emine the tiles listed bove in more detil. i-iii) The three norml cses Both Sommerville nd Dvies included the (90, 90, 360 n ) nd (90, 60, 5 ) tringles in their lists; but Somerville did not include the (90, n, 360 n ) tringles, which re not isosceles nd do not dmit tiling with ll the ngles equl t ech verte. Figure : Emples of norml tilings Sommerville nd Dvies give two norml tilings with the first fmily of tringles when n is even, nd Dvies gives second norml tiling with the (90, 60, 5 ) tringle. In ech cse these re obtined by twisting the tiling shown long gret circle composed of congruent edges,until vertices mtch up gin. (For cler ccount of these the reder is referred to Ueno nd Agok [].) There re lso lrge number of non-norml tilings with these tringles, which we shll not ttempt to enumerte here; some of the possibilities re described in [3]. iv) The (90, n, 360 ) qurterlunes (n odd) n When n is odd, there is no norml tiling with the (90, n, 360 ) tringle. However, there re tilings, in which the sphere is divided into n lunes with polr ngle 360 n, ech of which is subdivided into four (90, n, 360 n ) tringles. This my be thought of s further subdivision of the tiling with n ( n, 360 n, 360 n ) tringles, given in in [3]. There re two wys to divide lune into four tringles, mirror imges of ech other, nd this choice my be mde independently for ech lune. When two djcent dissections re mirror imges, then the edges mtch up correctly 7

8 Figure 3: An odd qurterlune tiling on the common boundry; but with n odd, this cnnot be done in every cse. (However, it is interesting to note tht double cover of the sphere with n lunes cn be tiled normlly.) As shown in [3], there re ppproimtely n /n essentilly different tilings of this type. The symmetry group depends on the choice of tiling; most tilings re completely symmetric. We hve V = {(, 0, 0), (,, ), (,, n+ ), (0,, ), (0, 0,n)} in ll cses (see section 5). It my be shown tht no tiling with this tile cn contin n entire gret circle within the union of the edges; s the tile itself is symmetric, no tiling cn hve mirror symmetry. The lrgest possible symmetry group is thus the proper dihedrl group of order n We do not t present know whether there re other tilings with these tringles, s there re when n is even. Despite the eistence of two verte vectors not used in ny of the known tilings, we conjecture tht there re not. v) The (90, 75, 60 ) tringle This tringle subdivides the (50, 60, 60 ) tringle. It ws shown in [3] tht eight copies of the ltter tringle tile the sphere; thus, siteen (90, 75, 60 ) tringles tile. This tiling is unique up to mirror symmetry (Proposition 5). Its symmetry group is the Klein -group, represented by three 80 rottions nd the identity. (As this does not include the point inversion, we conclude tht the (90, 75, 60 ) tringle fils to tile the projective plne.) An interesting feture of this tiling (nd the one it subdivides) is the long etended edge, of length 6.3+, visible in the figure. 8

9 Figure : The tiling with the (90, 75, 60 ) tringle vi) The (90, 60, 0 ) tringle This tringle tiles the sphere (N = 7) in mny wys. Two copies mke one (80, 60, 60 ) tringle, which ws shown in [] to tile the sphere in three distinct wys. Moreover, four copies yield the (0, 60, 0 ) tringle, nd si copies yield the (0, 60, 0 ) tringle. Both of these tile s semilunes, giving tilings of the 0 nd 60 lunes respectively (the ltter lredy non-norml). Figure 5: Some tilings with the (90, 60, 0 ) tringle Five copies yield the (90, 00, 0 ) tringle, nd seven yield the (90, 0, 0 ) tringle. While neither of these tiles, either combines with the (0, 60, 0 ), yielding the (90, 0, 60 ) nd (90, 0, 80 ) tringle respectively; nd com- 9

10 bining ll three gives 90 lune (Figure 6), which does tile. It is interesting to note tht this (unique; we leve this s n eercise to the reder!) tiling of the 90 lune hs no internl symmetries; usully when lune cn be tiled it my be done in centrlly symmetric fshion.. Figure 6: The unique tiling of the 90 lune with the (90, 60, 0 ) tringle Furthermore, si tiles cn lso be ssembled into n (80, 80, 80 ) tringle, which, while it does not tile on its own, yields tilings in combintion with three 00 lunes, ech ssembled out of one 0 nd one 60 lune. It seems probble tht the most symmetric tiling is the one with nine 0 lunes, with symmetry group of order 8 nd orbits; vrious other symmetries re possible, including completely symmetric tilings. Some tilings (such s the one on the left in Figure 5) hve centrl symmetry, so this tringle tiles the projective plne s well s the sphere. A complete enumertion of the tilings with this tile remins n interesting open problem. vii: The (90, 75, 5 ) tringle Eight copies of this tringle tile 0 lune, in rottionlly symmetric fshion (Figure 7). There re ectly two distinct wys to fit three such lunes together, forming non-norml tilings with N =. Either the three lunes hve the sme hndedness, in which cse edges do not mtch on ny of the three meridin boundries nd the symmetry group of the tiling is of order 6; or one lune hs different hndedness thn the other, edges mtch on two of the three meridins, nd the symmetry group hs order. It is conjectured tht there re no other tilings. A double cover of the sphere eists with 8 tiles in si lunes, lternting hndedness; this double cover is norml. viii: The (90, 78 3, 33 3 ) tringle This tringle is conjectured to tile uniquely (N=3) up to reflection (Figure 8). The symmetry group of the only known tiling is the Klein -group, represented 0

11 Figure 7: A tiling with the (90, 75, 5 ) tringle by three 80 rottions nd the identity. The tiles re prtitioned into eight orbits under this symmetry group; this ppers to be the lrgest possible number of orbits for mimlly symmetric tiling. Figure 8: A tiling with the (90, 78 3, 33 3 ) tringle Proof of Theorem The proof of Theorem breks up nturlly into sequence of propositions, deling seprtely with ech possible V. The nontrivil symptoticlly right V re (0,, 3), (0,, ), (0,, ), (, 3, ), nd (, 3, ); there re lso the trivil cses (, 0, 0) nd (0, 3, ) for which the tringle is isosceles with two right ngles. It is shown in [3] tht these tringles tile the sphere precisely when the third

12 ngle divides 360 ; nd in these cses there is lwys norml tiling [, 0]. For ech remining V, we will begin by determining n ehustive set of V, nd, for ech of these, find the rest of V. In some cses the lck of split vector othe rthn (, 0, 0)/ will then eliminte the tringle from considertion; in other cses we will need to emine the geometry eplicitly.. The (0,, 3) fmily Proposition 3 If right tringle tiles the sphere nd hs V =(0,, 3), then without loss of generlity V =(0, 0,c ) or (,,c ). Proof: Consider ny reduced γ source V =( V,b V,c V ); by definition, V =0 or. If V =nd<b V, we hve c V 3. Then W =V (0,, 3) (, 0, 0) hs W = 0 nd c W >b W > 0 nd is gin reduced γ source in V. If V = nd b V = 0, then W =V (, 0, 0) hs W = b W = 0 nd is lso reduced γ source in V. Thus, without loss of generlity, V =0or V = b V =. Now suppose V = 0 nd b V > 0. As V is γ source, we must hve b V =,, or 3. If b V =, then W =V (0,, 3) is reduced γ source in V nd hs W = b W = 0. If b V = 3, then W =V 3(0,, 3) is reduced γ source in V with W = b W = 0. Finlly, if b V =, we solve the system of equtions α β = (5) 0 c V γ 360 to obtin α = 90 ( 360cV 080 β = c V 3 ( ) 080 γ = c V 3 ) so tht 70 N = α + β + γ 80 = 3c V 8 3 but this is only n integer when 3 c V. As we hve ssumed tht the tringle tiles, this must be the cse; nd W = 3 V 3 (0,, 3) is reduced γ source in V with W = b W =0. Proposition If right tringle hs V =(0,, 3) nd V =(0, 0,c ), then c 8 nd V consists of the vectors in the pproprite set below tht hve ll

13 components positive: {(, 0, 0), (0,, 3), (0, 0,c ), (, 0, 3c c c c ), (, 0, ), (3, 0, ), (,, 3 )} if c 0 mod {(, 0, 0), (0,, 3), (0, 0,c ), (0,, c + ), (,, c +3 9 c 9 c ), (, 3, ), (0, 6, )} if c mod {(, 0, 0), (0,, 3), (0, 0,c ), (, 0, c ), (,, c +6 )} if c mod {(, 0, 0), (0,, 3), (0, 0,c ), (0,, 3c +3 ), (0,, c +3 ), (0, 3, c +9 5 c ), (0, 5, )} if c 3 mod. (To be eplicit, (,, 3 c ) is present for c =8, ; (, 3, 9 c 9 c ) nd (0, 6, ) for c = 9; nd 0, 5, 5 c ) for c =, 5.) Proof: (i) Solving, s bove, for β nd γ, we hve 360c 080 > 080 nd c 8. (ii) The eqution of the plne Π V contining V is c =c c (c 3)b. (6) We need to find the positive integer points on this plne. Substituting the lower bounds c 8, c 0 into this, we obtin 8 +5b < 3 (7) On the other hnd, we note tht, regrdless of the vlue of c, Reducing(6) modulo, we obtin + b c>0. (8) ( + b)c b (mod ). (9) The finl step depends on the congruence clss of c (mod ). c 0: In this cse, (9) reduces to b 0, nd we hve (, b) {(0, 0), (, 0), (, 0), (3, 0), (, 0), (0, ), (, )} The first si of these pirs stisfy (8) nd thus give rise to solutions (s listed bove) for ll c ; the lst gives c 0 only for c =8,. c : Now, (9) reduces to b, nd we hve (, b) {(0, 0), (0, ), (0, ), (, ), (, 0), (0, 6), (, 3)} Agin, the first five of these give rise to solutions for ll c ; the lst gives c 0 for c = 9 only. 3

14 c : gives us b, nd the only solutions re All of these stisfy (8). c 3: mkes 0, nd we hve (, b) {(0, 0), (0, ), (0, ), (, ), (, 0), (, 0)} (, b) {(0, 0), (0, ), (0, ), (0, 3), (0, ), (, 0), (0, 5)} All but the lst of these stisfy (8); for (0, 5) we get c 0 only for c =, 5. Computing the vrious vlues of c completes the proof. It ws shown in [5] tht the only right tringle to tile the sphere with no split verte other thn (, 0, 0)/ is the (90, 08, 5 ) tringle (which hs V = (,, )). Any other tringle is thus shown not to tile s soon s it is shown tht it hs no second split. In prticulr, the tringles considered bove with c 3 (mod ) never tile. We lso hve the following: Proposition 5 No right tringle tht hs V =(0,, 3) nd V =(,,c ) tiles the sphere. Proof: The eqution of of Π V is 8c =6c (c 3) (c 9)b. (0) Computing modulo 8, we obtin +3b (mod 8) when c is odd, nd 3 + b (mod 8) when c is even. Multiplying either of these congruences by 3 gives the other, so they hve the sme solutions. The requirement tht β>γgives c 5, nd substituting this nd c 0 into (0) gives us the inequlity +8b 56. But the only pirs (, b) tht stisfy this inequlity nd the congruences re (0, ),(, ), nd (, 0), so V never hs ny elements other thn the given bsis. Moreover, mong these, only (, 0, 0) corresponds to split, so by [5] none of these tringles tile the sphere. Proposition 6 The only right tringle tht hs V sphere is the (90, 60, 0 ) tringle. = (0,, 3) nd tiles the Proof: On the strength of the previous two propositions, we my ssume tht V =(0, 0,c ) with c 8 nd c 3 (mod ). When c = 9 we hve the (90, 60, 0 ) tringle. If c (mod ) nd c > 9, the only verte vector corresponding to split is (0,, c +3 ), in which γ ngles outnumber β ngles by t lest 3; nd the only β source is (0,, 3). Let the verte O be one such β source. At lest one of the three tringles contributing γ verte to O must hve its medium edge O

15 b b b c d c b O O O O b c d O e Figure 9: Configurtions ner (0,, 3) verte pired with hypotenuse or short edge, not nother medium edge. If the other edge Ob is hypotenuse (Figure 9,b), b is necessrily split verte. If Ob is short,(figure 9c,d), there must gin be n ssocited split verte, on the etended edge bc. In every cse, the split verte hs surplus of t lest three γ ngles. Emining the four configurtions, we see tht it is not possible for the identified split verte to be relted in ny of these four wys to two (0,, 3) vertices O, O unless certin reltions hold mong the edge lengths which re esily ruled out by numericl computtion - for instnce, in Figure 0, only medium edge could fill the gp bb without β split; but it is esily verified tht 3B H. O b b O Figure 0: Two (0,, 3) vertices ttempting to shre split verte We conclude tht every (0,, 3) verte is ssocited in - fshion with (0,, n)/ split verte; but this requires the number of γ ngles in the whole tiling to be greter thn the number of β ngles, which is impossible. Thus, when c (mod ) nd c > 9, the tringle does not tile. If c is even, there re no β splits, nd unless c = 8 or, the only β source is (0,, 3). As bove, every such verte must hve n unpired medium edge. If this edge is covered by hypotenuse Ob, this must be oriented s in Figure 9, s the β split in Figure 9b is impossible; nd there is γ split t b. If it is covered by short edge, there is right-ngle gp t b. In the bsence of β splits, this cnnot be filled by nother right ngle (Figure 9c,d - numericl clcultion rules out B C for ny tringle in this fmily) nd must therefore be right-γ split (Figure 9e). It is esily verified tht no split verte cn be relted s in Figure 9 or 9e to two (0,, 3) vertices; so ech (0,, 3) is ssocited with split verte tht it does not shre with ny other (0,, 3), nd between 5

16 them the number of γ ngles is gin greter thn the number of β ngles. Thus none of these tringles (including those with c =8, ) tile using (0,, 3) s the sole β source. The only remining possibilities for tilings involve tringles with c =8or c =, using (,, ) nd (,, 0) respectively s β sources. We shll show tht these vertices, too, re necessrily ssocited with γ splits. When c = 8, we obtin the (90, 56, 5 ) tringle. Between them, the ngles meeting t (,, ) verte O hve five hypotenuses, t lest one of which must be unpired. The β ngle of the unpired hypotenuse must be t O, or its other end would require β split. If we ssume tht the unpired hypotenuse meets the short edge of the neighboring tringle, tringles, nd 3 of Figure re forced in turn by voiding β splits. If it meets long edge, one of Figure b,c is forced. b b 3 b O O b c O Figure : Configurtions ner (,, ) In ech of these three cses, the indicted two split vertices must eist. We must now consider whether t lest one of these split vertices my form prt of nother configurtion of the sme type, nother type from Figure, Figure 9, or Figure 9e. Most of the pirings re impossible unless the edge lengths stisfy simple equtions tht re esily ruled out, s in Figure 0. The only three cses in which the designted split vertices cn be shred re shown in Figure (b with nother of the sme type), Figure b (c with nother of the sme type) nd Figure c (b with 9). In ech cse, t lest one right-ngled gp bounded by medium edges or hypotenuses eists, nd this cnnot be filled by right ngle without requiring β split. In figure,b, then, two (,, ) vertices shre split vertices conting totl of eight γ ngles. In Figure c, one (,, ) verte nd one (0,, 3) verte shre split vertices contining si γ ngles. In every other cse, single (,, ) verte hs sole custody of two split vertices with t lest four γ ngles. In every cse, these configurtions require more γ ngles thn β ngles; so the (90, 56, 5 ) tringle does not tile. We now consider the (90, 67, 30 ) tringle, which hs c =. As shown bove, it cnnot tile without using vertices (,, 0) s β sources. Any verte of 6

17 O O O O O O b c Figure : Two β sources O, O, shring split vertices this type hs single right ngle, with n unpired medium edge. This cnnot meet the djcent tringle (, in Figure ) on short edge, s voiding β splits gives us tringles,3,, nd 5, nd then β split t cnnot be voided. If it meets hypotenuse, the β ngle of the djcent tringle must be t O, giving us 3b with two split vertices; nd these re the only two possibilities. O O O O b O c O d Figure 3: Configurtions ner (,, 0) vertices Agin, there re two specil cses in which these γ splits my be shred with nother β source. In Figure 3c, two (,, 0) vertices O, O shre both their split vertices. The tringles 3,3, re then forced; the right-ngled gps cnnot be filled with right ngles without β split; nd so O nd O shre twelve γ ngles. A similr rgument shows tht the (,, 0) nd (0,, 3) in Figure 3d shre nine γ ngles. In every cse, γ ngles outnumber β ngles, so tht the tringle cnnot tile.. The (0,, ) fmily Proposition 7 If right tringle tiles the sphere nd hs V =(0,, ) then without loss of generlity V =(0, 0,c ). Conversely, every tringle with (0,, ), (0, 0,c ) 7

18 V tiles the sphere. Proof: s for Proposition 3.3 The (0,, ) fmily Proposition 8 If right tringle hs V =(0,, ) nd tiles the sphere then without loss of generlity V =(0, 0,c ) or (,,c ). Proof: s for Proposition 3 Proposition 9 If right tringle hs V =(0,, ) nd V =(0, 0,c ) nd tiles the sphere, then 3 c, c 6 nd V = {(, 0, 0), (0,, ), (0, 0,c ), (, 0, 3c {(, 0, 0), (0,, 3), (0, 0,c ), (0,, 3c + ), (0,, c + ), (0, 3, c +3 {(, 0, 0), (0,, 3), (0, 0,c ), (,, c + {(, 0, 0), (0,, 3), (0, 0,c ), (0,, c + ), (,, c + Proof: s for Proposition c c ), (, 0, ), (3, 0, ), } if c 0 mod )} if c mod c ), (, 0, ), } if c mod ), } if c 3 mod. Proposition 0 No right tringle with V =(0,, ) nd V =(,,c ) tiles the sphere. Proof: s for Proposition 5; there is never ny second split. Proposition The only right tringle tht hs V =(0,, ) nd tiles the sphere is the (90, 75, 60 ) tringle. Proof: From Proposition 9 we see tht there is no second split unless c is divisible by 6, in which cse we hve (0, 0,c )/. We lso hve (, 0, c )/ if c ; there re no other splits. In the cse c = 6 we obtin the (90, 75, 60 ) tringle; henceforth, then, we suppose c. The possible splits re then (0, 0, m)/ with m 6 nd (, 0, n)/ with n 3. We see lso tht (0,, ) is the only β source, so such verte must pper in ny tiling with this tringle. We emine the neighborhood of ny such verte O (see Figure ). Let tringle contribute the γ ngle. Consider the tringle, which covers the long leg of ner O. If the short leg of meets (Figure nd b) nd the gp is filled by right ngle, then we need β split, which is impossible. (It is esily checked tht C B for ny tringle in this fmily.) If the gp is filled by γ ngles (Figure c), or if the long leg of is covered by the hypotenuse of (Figure d), there is γ split t. This split cnnot be relted in the sme wy to ny other (0,, ) verte. 8

19 O O O O b c d Figure : The split verte ssocited with O Unless the split verte X is of the form (, 0, 6)/ there re more thn three γ ngles t X, nd it follows tht O nd X between them hve surplus of γ ngles; thus the entire tiling hs surplus of γ ngles, which is impossible. If X is (, 0, 6)/, there must be right ngle t X. If X is s shown in Figure c, tringles 3 nd must be s shown in Figure 5 to void β split; but then whichever wy we plce the third tringle between them, β split is required. If X hs the configurtion of Figure d, nd the right ngle is between two γ ngles (Figure 5b),β split is required (t y); if not (Figure 5c), we must either hve β split t z or z, or hve H + C =B, which is esily shown not to hold for ny tringle in this fmily. We conclude tht no other tringle in this fmily tiles the sphere. y y O O O b c Figure 5: cnnot hve right ngle. The (, 3, ) fmily Proposition If right tringle hs V =(, 3, ) nd tiles the sphere, then without loss of generlity V =(0, 0,c ), (0,,c ), (0,,c ),or(, 0,c ). Proof: s for Proposition 3 9

20 Proposition 3 If right tringle hs V =(, 3, ) nd V =(0, 0,c ) nd tiles the sphere, then c, c 8 nd V consists of ll vectors in {(, 0, 0), (, 3, ), (0, 0,c ), (0, 3, c +8 tht hve nonnegtive integer components. Proof: s for Proposition ), (0, 6, c 8 ), (, 0, 3c c ), (, 0, ), (, 3, c 8 ),,(3, 0, c )} Proposition The only right tringle tht hs V =(, 3, ) nd V =(0, 0,c ) nd tiles the sphere is the (90, 60, 5 ) tringle, which tiles edge-to-edge. Proof: As observed bove, c must be even nd t lest 8. When c =8we obtin the known (90, 60, 5 ) tile, so we consider the cse when c 0. The minimum number of γ ngles t split other thn (, 0, 0)/ is 3, chieved by the (, 0, 6)/ split when c =. The only β source is the rther wek (, 3, ); so such verte (cll it O) must pper in ny tiling. Between them, the ngles t O hve short edges, 3 medium edges, nd 5 hypotenuses. There is thus t lest one unpired hypotenuse. This cnnot be covered ectly by other edges; C <H<B+ C, nd in the bsence of β split we cnnot hve more thn two short edges on n etended edge. The other end of this hypotenuse is therefore t split, necessrily involving t lest three γ ngles. The split verte is contined in n etended edge which termintes t O. At most two (, 3, ) vertices cn be relted to one split in such wy; but between them these three vertices hve seven γ ngles nd only si β ngles. Thus no such tiling is possible. Note: In fct, it is probbly true tht the split verte could not be relted even to second (, 3, ) verte, but it is esier to concede the point. Proposition 5 No right tringle with V =(, 3, ) nd V =(0,,c ) tiles the sphere. Proof: Clcultion shows tht N =8c 6 3, which is never n integer. Proposition 6 No right tringle with V =(, 3, ) nd V =(0,,c ) tiles the sphere. (The proof of this proposition is lengthy, nd is crried out in the compnion pper [6].) Proposition 7 The only right tringle with V =(, 3, ) nd V =(, 0,c ) tht tiles the sphere is the (90, 60, 5 ), lredy listed in Proposition 0. 0

21 Proof: If c 0 (mod 3) we hve (0, 0, c 3 ) V); by Proposition this gives the (90, 60, 5 ) tile for c = 6 nd tringles tht do not tile otherwise. If c (mod 3), we hve (0,, c + 3 Vnd by Proposition none of these tringles tile. Finlly, if c (mod 3), nd the only split is (, 0, 0)/..5 The (, 3, ) fmily Proposition 8 If right tringle hs V =(, 3, ) nd tiles the sphere, then without loss of generlity V =(0, 0,c ), (0,,c ), (, 0,c ) or (,,c ). Proof: s for Proposition 3 Proposition 9 If right tringle hs V =(, 3, ) nd V =(0, 0,c ) nd tiles the sphere, then 3 c, c 6 nd V consists of ll vectors in {(, 0, 0), (, 3, ), (0, 0,c ), (0, 3, c + tht hve integer components. Proof: s for Proposition ), (, 0, 3c c c ), (, 0, ), (3, 0, )} Proposition 0 The only right tringle tht hs V =(, 3, ), V =(0, 0,c ), nd tiles the sphere is the (90, 75, 5 ) tringle. Proof: We note tht there is no second split when c is odd; so by Proposition we my ssume c to be even. There is never β split. As result, there cn be no (, 0, 3c c )or(3, 0, ) vertices. The right ngles t such verte would hve between them n odd number of short edges terminting in β ngles, t lest one of which would be unmtched. By the sme token, (, 0, c ) verte cn only eist if c nd the two short edges of the right tringles re pired, s in Figure 6. q q p b c Figure 6: Split verte ner (,3,) verte

22 We will now show tht (with this interprettion) split verte involving γ ngles lwys eists. (Note tht Proposition sttes tht tile other thn (90, 08, 5) must hve second split vector, but not tht it is necessrily used in the tiling.) Suppose, for contrdiction, tht there is tiling tht does not use ny γ split. As (0, 3, c + ) is never β source, there must eist t lest one (, 3, ) verte. In the bsence of non-right-ngle splits, the only possible configurtion t such verte is s shown in Figure 6b. But then the edge pq must be covered by nother medium edge. The right ngle cnnot go t B, s no verte vector hs two right ngles nd β; we thus hve the configurtion of Figure 6c, in which q is the required (, 0, c ) verte. However, we will see tht unless n = 8 no split verte is possible. It is cler tht when we put fn of γ ngles t verte, ll β ngles must either be t the end of the fn or pired, s otherwise there will be n overhng nd gp tht cnnot be filled without β split. (Figure 7). For (0, 0,c )/ split, c 0 this results in two pirs of djcent edges such s pq, qr. In the bsence of β split, the only wy to cover either of these etended edges is with nother pir of tringles s shown; but this leves n impossible four β ngles t B. Similr problems occur for (, 0, c )/ splits with c 0 (Figure 7b). p q p q r b r Figure 7: Fns of tringles led to illegl configurtions There remin three cses when c = 6,, nd 6. When c = 6, the only split vector is (0, 0, 6)/. Avoiding overhngs t β ngles forces the configurtion of Figure 8. If the etended edge pq etended beyond q, we would hve n overhng or fourth β ngle t r; neither is permitted. It follows tht pq is complete etended edge nd must be covered by nother hypotenuse nd short edge on the other side; this imples second copy of the sme configurtion. If the second copy were mirror imge, we would hve four β ngles t p; the only lterntive is the configurtion of Figure 8b. The edge qr must be mtched, but we cnnot hve two right ngles nd β together, so the tringle (nd the corresponding ) must be s shown. This leves γ gp t r which must be filled by tringle s shown, putting β ngle t s. But then the edge ps cnnot be covered without creting n illegl combintion of ngles t one end or the other. When c = we hve lredy ruled out (0, 0, )/ but we must show tht the (, 0, 6)/ split lso leds to illegl configurtions. By rguments similr to

23 r s r p q p q b Figure 8: The (0, 0, 6)/ split those used in the lst cse, we obtin the configurtion of Figure 9. If we put right ngle nd γ ngle into the gp t p, the short edge t the right ngle will be unpired nd will require n illegl β split; it follows tht p must be (0, 3, ) verte. p p b c q Figure 9: The (, 0, 6)/ split There re only two wys to plce the γ ngles without β split; in Figure 9b, n overhng is creted tht mkes it impossible to cover the remining edge of tringle, while in Figure 9c, we eventully get four β ngles t q (s in Figure 7). Finlly, when c = 6, we first note tht we cnnot hve (0, 3, 5) verte. The set of edges of the ngles meeting t such verte would contin eight hypotenuses, five of them from γ ngles nd hence terminting in β ngle. All of these must be pired with other hypotenuses to prevent β split, nd t lest two of them must be pired with ech other, s tringles, re in Figure 0. We now emine the medium edges of these ngles. Either one of these edges is mtched (s t left), in which cse there is n etended edge pq which must be mtched ectly by two more short edges; or it is not, in which cse there is n overhng (s t r). In ny cse, we end up with two more β ngles t p, which is impossible. Now we show tht we cnnot hve (, 0, 8)/ split. Suppose we did; by rguments similr to those used bove, its neighborhood would hve the config- 3

24 urtion of Figure 0b. The etended edge st must be covered s shown (tringles 3, in Figure 0c). Tringle 5 is then forced, s the only wy to cover tu without creting β split. s s v 3 6 p q r t u t 5 u b c Figure 0: The (, 0, 8)/ split Then t must be (, 3, ) verte. The remining γ ngle is provided by tringle 6. The hypotenuse of tht tringle must be pired with tht of tringle to void β split; but then verte v hs t lest two γ ngles nd β ngle. However, the only such verte vector is (0, 3, 5), nd we hve seen tht this cnnot occur in tiling. y y b Figure : The (, 3, ) verte in the bsence of the (, 0, 8)/ split But if we hve no split in the tiling ecept for (, 0, 0)/, it is esily seen tht every (, 3, ) verte must hve the configurtion of Figure. The etended edge y must be covered ectly, nd this cn only be done s shown in Figure b. But then verte hs two γ ngles nd β ngle, nd we hve seen tht this is impossible. We conclude tht this tringle fils to tile the sphere. Proposition If right tringle hs V =(, 3, ) nd V =(0,,c ) nd tiles the sphere, then c nd V consists of ll vectors in {(, 0, 0), (, 3, ), (0,,c ), (0, 5, c tht hve positive integer components. Proof: s for Proposition ), (, 0, 3c ), (,, c )}

25 Proposition The only right tringles tht hve V =(, 3, ), V =(0,,c ), nd tile the sphere re the (90, 7, 5 ) tringle nd the (90, 78 3, 33 3 ) tringle. Proof: From the previous proposition, c must be even nd greter thn or equl to. When c = we get the (90, 7, 5 ) tringle, which is qurterlune (lthough, typiclly, it hs θ>80 θ, so its polr ngle is β nd not γ.) For c = 6 we get the (90, 78 3, 33 3 ) tringle. This tiles the sphere with N = 3. We will now show tht when c 8 the (0,,c )/ split is not relizble. Firstly, if there is such split, then without loss of generlity there eists one with the short edge of the β ngle on the etended edge contining the split, s in Figure ; for if it is locted s in Figure b,c, then there is, s shown, nother split of the required form nerby, t the point mrked. (Note tht for c 8 both the medium edge nd hypotenuse re more thn twice s long s the short edge, nd tht in every cse the only split involving right ngles is (, 0, 0)/.) b c Figure : A defult form for the (0,,c )/ split We will now show by induction tht ll the edges in the fn of γ ngles re mtched. Consider the γ ngle djcent to the β ngle. If this is positioned s Tringle in Figure 3,b, right-ngled split is creted. If this were filled s in Figure 3, then either wy of covering the hypotenuse of tringle would require split with two β ngles. The only lterntive, in Figure 3b, forces the hypotenuse of tringle to be mtched by tringle 3, s shown. There cnnot be fourth β ngle t p; we thus hve either γ ngle (not shown) or right ngle net to tringle 3 t p. Any choice of ngle nd orienttion forces tringle 5 s shown, nd the overhng t q, which mkes it impossible to cover the hypotenuse of tringle 5. (As c > 6, this hypotenuse is not the other side of the split.) We thus hve Tringle positioned s shown in Figure 3c. If the medium leg of tringle were not mtched, we would hve n overhng t r, forcing the γ ngle of tringle. Tringle 5, with its uncoverble hypotenuse, gin follows. Thus, the medium edges must mtch (Figure 3d). Agin, the hypotenuses of the third nd fourth tringles in the split must mtch. Suppose not; if the short edge of tringle is not mtched, its hypotenuse cnnot be covered (figure ). If it is mtched (tringle 5 in figure b), tringles 6 nd 7, the γ ngle 8, nd tringle 9 re forced in tht order. However, 5

26 3 3 p 5 q r b c d Figure 3: A defult form for the (0,,c )/ split: the second nd third tringles this cretes n impossible combintion of ngles t v. We conclude tht the first four ngles of the fn must be s in figure c u 5 7 v b c Figure : The fourth tringle We cn now proceed inductively to show tht the other edges in the split re lso mtched. Suppose, for contrdiction, tht the first unmtched edge to be between tringles numbered n + nd n +, n, s in Figure 5. The overhng nd split t nd the gmm ngle lbelled re forced, resulting in n impossible configurtion t y. If, on the other hnd, the first unmtched edge is between tringles n nd n +, tringle of Figure 5 is forced; we then get the overhng nd split t z, tringle, the γ ngle 3, nd tringle. However, the hypotenuse of tringle cnnot be covered. It follows, then, tht ll the edges between the ngles t (0,,c )/ split with the β ngle s shown re mtched. The net step is to show tht in fct no split configurtion of this type (nd hence no (0,,c )/ split whtsoever) eists in tiling of the sphere. If c /is even, we hve configurtion something like Figure 6. There cnnot be n overhng t p, becuse the new split would require tringle (s shown), which would prevent the originl verte from being split s hypothesized. It follows tht the etended edge pq is covered by the short edges of two more tringles, necessrily positioned s shown in Figure 6b. 6

27 y n+ n+ n n+ 3 b z Figure 5: The inductive step q r q s r q p p b c Figure 6: Noneistence of the (0,,c )/ split when c If c the net two short edes must be covered in the sme wy nd we immeditely hve n impossible four β ngles t q. Ifc = 8 we cn void this only by hving n overhng nd split t r. Completing this split gives us Figure 6c; but there is no wy to cover the etended edge qs without n illegl split or fourth β t q. We conclude tht there is no (0,,c )/ split when c /is even. When c / is odd, we hve configurtion like tht of Figure 6. Agin, if c we immeditely get verte with four β ngles nd we re done. When c = 0, we cn hve (, 3, ) verte t y nd n overhng nd split t z (Figure 6b); but ny wy of putting right ngle or γ ngle long yw cretes n impossible configurtion. A right ngle with the short edge on yw gives the configurtion of Figure 6c; the split t p must hve tringles nd 3 s shown, nd ny ttempt to fill the right ngle gp t q creted n illegl overhng t either r or w. On the other hnd, if tringle is plced with its right ngle t y nd its long edge long yw, there must be n overhng t s nd its hypotenuse cn only be covered s shown. After tringle is plced, the remining ngles t the split t t re ll γ ngles, forcing ngle 3 s shown; tringle is then forced, nd its hypotenuse cnnot be covered. The two cses with γ ngles on yw re ruled out by similr rguments. We conclude tht for c > 6 no (0,,c ) split cn be relized. From this it is strightforwrd to rule out ll γ sources, nd thus to show tht tiling is impossible. Proposition 3 The only right tringles tht hve V =(, 3, ), V =(, 0,c ), 7

28 z y w z y w b c 3 r 3 s w q p y w t y d Figure 7: Noneistence of the (0,,c )/ split, c nd tile the sphere re those listed in Propositions 0 nd. Proof: If c 0 (mod 3), then (0, 0, c 3 ) V. This gives the (90, 75, 5 ) tile for c = 6 nd Proposition 0 shows tht the tringle does not tile otherwise. If c (mod 3), then (0,, c + 3 V; we get the (90, 7, 5 ) tile for c =5, the (90, 78 3, 33 3 ) for c = 8, nd (by Proposition ) tringle tht does not tile for ll other c. Finlly, if c (mod 3), following the methods of Proposition we find tht s well, yielding no split ecept (, 0, 0). Proposition No right tringle with V =(, 3, ) nd V =(,,c ) tiles the sphere. Proof: Agin, every element of V hs (mod 3), so tht the only split is (, 0, 0). 5 Other results Proposition 5 The tiling shown in Figure is (up to reflection) the only tiling of the sphere with the (90, 75, 60 ) tringle. Proof: Clcultion shows tht V T = {(, 0, 0), (, 0, 3), (,, ), (0,, ), (0, 0, 6)}; s observed bove, there is no β split, so there cnnot be n overhng on either side of β ngle. We look t possible covers for the short leg pq of tringle. If pq is covered by longer edge, there is n overhng s shown in Figure 8. The gp t p must be filled by right ngle, in one of two positions. As 8

29 q p q p Figure 8: An impossible configurtion H/C = nd B/C =.3..., either of these must result in β split t, which is not possible. If pq is covered by nother short leg with the opposite orienttion (Figure 9), the edge pr cnnot etend pst r. Suppose, for the ske of contrdiction, tht it did; tringle would be forced. The remining ngles t q would be γ s, forcing n overhng t s, so tht tringle must be s shown. There cnnot be n overhng t t, nd the etended edge qt must be covered by two more long legs, s no other combintion of edges equls B. However, q lredy hs two β ngles nd right ngle, nd cnnot ccept nother of either type. q s p r t Figure 9: Another impossible configurtion Thus, if pq is covered s shown, pr must be covered by nother edge of the sme length; s there cnnot be nother right ngle t p, we hve the configurtion of Figure 30. The ngle upv must be filled with no overhng t u or v; this forces (essentilly) the configurtion of Figure 30b, nd, s in Figure 8, filling the 90 gp t will force β split. v q v q p r p r u b u Figure 30: A third impossible configurtion We conclude tht pq is pired with nother short leg, oriented in the sme wy. It follows tht the tringles in the tiling re prtitioned into mirror-imge pirs, forming (50, 60, 60 ) tringles; s shown in [3], these tile the sphere 9

30 uniquely up to reflection. 6 Conclusion This pper lists ll the symptoticlly right right tringles tht tile the sphere, including one infinite fmily, nd four spordic tringles, tht tile only in nonnorml fshion. It is prt of sequence of ppers (long with [7] nd [8]) tht will give complete clssifiction of the right tringles tht tile the sphere. References [] Coeter, H. S. M., review of [], Mth. Rev. 36 # 335 [] Dvies, H.L., Pckings of sphericl tringles nd tetrhedr. Proc. Colloquium on Conveity (Copenhgen, 965) 5 [3] Dwson, R. J. McG., Tilings of the Sphere with Isosceles Tringles Disc. nd Comp. Geom. 30 (003)67-87 [] Dwson, R. J. McG., A tringle tht tiles the sphere in ectly three wys Disc. nd Comp. Geom. 30 (003)59-66 [5] Dwson, R. J. McG., Single-split tilings of the sphere with right tringles, in Discrete Geometry: In Honor of W. Kuperberg s 60th Birthdy (A. Bezdek, ed.; Feb. 003, Dekker Lecture Notes in Pure nd Applied Mthemtics), 07- [6] Dwson, R. J. McG., nd Doyle, B., Tilings of the sphere with right tringles III: the (, 3, ), (0,,n) fmily, preprint [7] Dwson, R. J. McG., nd Doyle, B., Tilings of the sphere with right tringles IV: the symptoticlly obtuse cse, preprint [8] Dwson, R. J. McG., nd Doyle, B., Tilings of the sphere with right tringles V: the symptoticlly cute cse, in preprtion [9] Grünbum, B., nd Shepherd, G. C., Tilings nd Ptterns (New York, 987) [0] Sommerville, D.M.Y., Division of spce by congruent tringles nd tetrhedr, Proc. Roy. Soc. Edinburgh 3 (93), 85 6 [] Ueno, Y.; Agok, Y., Clssifiction of the tilings of the -dimensionl sphere by congruent tringles, Technicl Report 85, Division of Mthemticl nd Informtion Sciences, Hiroshim University, 00 30