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Uniform Solution for Uniform Polyhedra*

Zvi Har'El Department of Mathematics Technion − Israel Institute of Technology Haifa32000, Israel E-Mail: [email protected]

ABSTRACT

An arbitrary precision solution of uniform polyhedra and their duals is presented. The solution is uniform for all polyhedra givenbytheir kaleidoscopic construction, with no need to ‘examine’ each separately.

1. Introduction. Uniform polyhedra, whose faces are regular and vertices equivalent, have been studied since antiq- uity.Best known are the fivePlatonic solids and the 13 Archimedean solids. Wethen have the twoinfinite families of uniform prisms and . Allowing for star faces or vertices, we have the four Kepler- Poinsot regular star polyhedra and a rowof53nonconvex uniform polyhedra discovered in the 1880s and the 1930s. The complete set appeared in print for the first time in 1953, in a paper by Coxeter,Longuet- Higgins and Miller ([CLM], see also [S]). Magnus Wenninger’sdelightful book Polyhedron Models [W1], which appeared in 1971 but since has been reprinted manytimes, contains photos and building instructions for cardboard models of these 75 uniform polyhedra. Reading the book, makes the mathematically minded reader wonder: Howare the data for the models obtained? For example, what makes 1. 1600030093 the proper circumradius for a great ret- rosnub icosidodecahedron1 with length two? It is easy to check, that these data originate in [CLM, Table 7]. Some of the circumradii are exact, as theyare giveninterms of integers and radicals only,but few, asthe one mentioned above,are givenapprox- imately,toten decimal digits. This may be considered perhaps accurate enough, but if one wants to incor- porate polyhedra in a computer modeling software (cf. [Hy]), one would prefer to get the numbers in an arbitrary precision, or in the maximum precision the computer can handle. Furthermore, one is interested in exact, or maximum precision, values of other geometric data, such as the dihedral angles of the polyhe- dra, and for these the only available data are for the regular and the convex cases, and are accurate to 1′′ (cf. [CR, Table II] and [J, Table II]). This problem was treated by AndrewHume. His method is best described by a short quotation from his report [Hm]: In general, the data aresolutions of equations found by examining the polyhedra(for example [L, pp. 174-176]). The equations weresolved at least three times using symbolic algebrasys- tems... Uniform polyhedra for which symbolic algebra systems are useful are the so-called polyhedra, and the computations involvesolving cubic or quartic equations. Hume’ssolutions were used to create a database of polyhedra, which is publicly available at [email protected].

*Inmemoriam of my father,Gershon Har’El, who introduced me to spatial structures. Geometriae Dedicata 47: 57-110, 1993. 1 This is the [W2] version of the polyhedron name. -2-

Our approach is quite different. Rather then separately examining individual polyhedra, we suggest a uniform approach, which is easy to understand and easy to use, evenwith a hand-held calculator,and it eliminates the need for a database for uniform polyhedra and their duals, since the fast iterative algorithm may yield arbitrary precision data. Furthermore, it may be used in the same ease for convex aswell as for nonconvex polyhedra (which are not treated by [Hm]). A computer program, called kaleido (cf. [Ha]) and publicly available at [email protected],has been developed to compute the data of a uniform polyhedron (and its dual), giveneither the vertexconfiguration, i.e, the enumeration of the appearing as faces incident at a vertexinthe order in which theyare found (cf. [CR, §2.9.2]), or the so- called Wythoffsymbol which describes the kaleidoscopic construction of the polyhedron (cf. [CLM, §3]). Kaleido is also capable of computing the vertexand coordinates and displaying a rotating wire-frame image of each polyhedron. We would liketoexpress our gratitude to H. S. M. Coxeter,BrankoGrünbaum and AndrewHume, for the very useful and enlightening comments.

2. The Fundamental System. The uniform solution is based upon projecting the uniform polyhedron onto a concentric , decomposing each n-sided tile in the spherical tiling into 2n congruent right-angled spherical , set- ting the trigonometric equations and solving them iteratively.Weshall explain the solution algorithm in the next section. Here, we discuss the procedure of identifying the fundamental triangles and setting the funda- mental equations. Let us assume, for the moment, that we are investigating a convex uniform polyhedron, such that each of its vertices is incident to m faces, with the ith face being a regular ni-gon, customarily denoted by ... the Schläfli symbol {ni}. The cyclic list (n1. n2. .nm)isthe vertexconfiguration. Projecting the polyhedron on a concentric sphere, we get a spherical tiling. Choose a C,and = ... let Ai,for i 1, ,m,bethe incenter of the of the ith tile incident to C.Also, let Bi be the foot of the per- + + pendicular from Ai to the arc separating the ith and (i 1)st tiles (the (m 1)st being identified as the first). ... In fact, the spherical A1 A2 Am is the of the dual tiling (cf. [GS, §1.2]). Also, the pla- ... nar polygon B1 B2 Bm is the so called vertexfigure at C (cf. [C, §2.1]). This way,wedecompose the spherical tiling into right-angled spherical triangles, called the funda- mental triangles:Ateach vertex C we get 2m fundamental triangles, arranged in congruent pairs Ai Bi−1C = ... and Ai BiC,for i 1, ,m.The set of fundamental triangles will be called the fundamental triangulation. Note that the same triangulation also arise from a similar decomposition of the dual tiling (cf. [GS], section 2.7). π Denote the sides and the angles of A B C by a , b , c , α , , γ ,inthe obvious manner (the right i i i i i i 2 i angle is at Bi). From the definition of Ai,and the regularity of the face, we easily see = niα i π ,(1) for i = 1, ...,m.Furthermore, if we consider the angles of the fundamental triangles meeting at C,weget = Σ γ i π .(2)

Finally,weuse the fact that the edge CBi is shared by the twoneighboring fundamental triangles Ai BiC = and Ai+1 BiC,and get ai ai+1.Thus, all the ai’s are equal, and the common value, say a,satisfies cos α cos a = i (3) sin γ i for i = 1, ...,m,asmay be easily deduced from the spherical lawofcosines, or from Napier’ssecond rule of circular parts (cf. [M]). Thus, we get a system of 2m + 1simultaneous equations (where m + 1are linear and m are trigono- + metric) in the 2m 1unknowns α i, γ i and cos a.Wename it the fundamental system,and we shall describe its solution in the next section. Once the fundamental system has been solved, the rest of the sides of the fundamental triangles may be solved using similar formulas: -3-

= cos γ i cos ci sin α i

= 1 cos bi tan α i tan γ i The solution of the fundamental triangles reflects directly on the problem of finding all the metrical properties of the uniform polyhedron: Let R, ρ, ri, l, hi, θ i, φ i,berespectively the circumradius (the distance from the center to a vertex), the midradius (the distance from the center to an edge), the inradius (the dis- tance from the center to the face), the semiedge, the facial inradius, the facial angle and the dihedral angle ρ (the angle between the the i’thand the i + 1’st faces). Then it is not hard to verify that = cos a, R ri = l = hi = = − = − − cos ci, tan a, tan ci, θ i π 2α i, φ i π ci ci+1. ρ ρ ri The construction described above can be easily adjusted to accommodate nonconvex uniform polyhe- dra2.Such polyhedra may have star polygons as faces, or have the faces meeting at a vertexsurround it more than once. In the former case ni may be taken fractional (with the denominator being the of the star face, cf. [C]), and in the latter case the right hand side of (2) is a multiple of π (with the multiplier being the density of the star vertex). Finally,nonconvex polyhedra may have retrograde faces, as follows: Choose an arbitrary orientation on the circumcircle of the vertexfigure. The i’thface is said to be retro- grade if the shorter of the the twoarcs connecting Bi−1 to Bi is oppositely oriented. (Weassume that at least one face is not retrograde!) In this case we will represent a retrograde {n}byan{n′}, with n′ <2sat- 1 1 isfying + = 1. This will make α ′=π−α obtuse, and γ ′=−γ negative,asrequired. Wewill reiterate n n′ on these adjustments in the next section. It is worthwhile to note that [CLM], p. 420, utilize the circumcircle of the vertexfigure and elemen- tary trigonometry to derive a set of equations which is essentially identical to our fundamental sys- tem. Theyare also able to obtain, by elimination, a single quartic equation, and use it to to prove that their list of snub polyhedra is complete (cf. [CLM], table 4). However, from our point of view, such derivation obscures the tight relationship between uniform polyhedra and their fundamental triangulations, a relation- ship which will be instrumental in section 4. Furthermore, the elimination method is not as uniform as the iterative solution to be presented in section 3, since it has to be carried out afresh for each polyhedron type and requires quite a fewsymbolic manipulations, algebraic and trigonometric.

3. The Iterative Solution. We now describe the iterative solution of the fundamental system, based on well-known Newton’s method for solving nonlinear equations. The general idea is to solve2nof the unknown in terms of the 2n + 1’st, referred later as the independent unknown, and to compute the latter by iterative approximations. One of the problems in such an approach is the choice of the independent unknown, and we will attend to it shortly.

First, we notice that the α i’s are readily available from (1): = π α i (4) ni for i = 1, ...,m.

Furthermore, once one of the γ i’s,say γ 1,iscomputed, all the rest are available from (3): cos α cos a = 1 (5) sin γ 1

cos α i γ = arc sin( )(6) i cos a

2 The corresponding spherical tiling consists of hollow tiles in the sense of [GMS] (cf. [GS], section 12.3), and the fundamental triangulation be viewed as a multiple tiling of the sphere by triangles, or a triangulation of aRiemann surface (cf. [GS], section 12.4)). -4-

for i = 2, ...,m. Note that the inverse sine function in (6) has seemingly twovalues in the range (0, π ). However, it may be easily observed that at most one of the γ i’s may be obtuse, and we may assume, without anyloss of = ... generality,that γ i,for i 2, ,m,are all acute. Thus, (2) may be rewritten as = − = δ π Σ γ i 0, (7) where δ is a well defined function of γ 1 only. We recall Newton’smethod for solving the equation f (x) = 0: Choose an initial approximation to the root and then define a sequence of closer approximations using the recursion f (x) xˆ = x − (8) f ′(x)

Assume initially that the fundamental triangles are approximately planar.This motivates the initial approximation π γ = − α i 2 i. for i = 1, ...,m. = This certainly satisfies (5) and (6) (with cos a 1), but as these values for the γ i’s are smaller than in reality (as the sum of the angles in a spherical exceeds π ), δ in (7) will not vanish, but rather be a positive number,which motivates naming it the excess function.The iterative procedure will convergeif we are able to makethe excess arbitrarily small. It may be shown, at least empirically,that in order to guar- antee convergence it is sometimes necessary to choose γ 1 as the biggest of the initial values, i.e., choose α 1 to be the smallest of the α i’s,or, inviewof(4), choose n1 to be the biggest of the ni’s.

This choice is the obvious the case for those polyhedra for which γ 1 turn to be obtuse, in viewofour observation about the double-valuedness of the inverse sine function. Also, it may be shown that in the general case, this choice also guarantees a faster convergence than then anyother choice. We’ll see whyit happens once we developed the recursion formula. To use (8), we have todifferentiate the excess function; we have from (7), dδ dγ − = 1 + Σ ′ i (9) dγ 1 dγ 1 (with the prime denoting summation overthe range 2, ...,m). However, inviewof(5) and (6),

= sin γ 1 cos α i sin γ i cos α 1 which in turn implies, using logarithmic differentiation, dγ tan γ i = i .(10) dγ 1 tan γ 1 Combining (7), (9) and (10) we get

− δ = tan γ 1 ′ δ δ Σ tan γ i (with summation again overthe full range 1, ...,m,and δ ′ is the derivative in(9)).

Finally,wearrive atthe required recursion for γ 1: = + tan γ 1 γˆ1 γ 1 δ Σ tan γ i In practice, we notice that the faces meeting at a vertexare not necessarily different, thus we may group similar faces together,and adjust the above formulas to reflect the fact that at each vertexthe polyhe- = dron can have mi faces each of which is a {ni}, with Σ mi m.Furthermore, we handle retrograde faces -5-

by orienting the vertexfigure so that the the first, thus biggest, face is not retrograde. Thus, ni is a positive rational, and mi is a positive integer. d,the vertexdensity,isanon-negative integer. To summarize, fundamental system = niα i π

= Σ miγ i π d

cos α i = cos a sin γ i = where n1 max ni,has the following iterative solution algorithm: Step 1. = π − = π Choose the initial approximation γ i α i,where α i . 2 ni Step 2. = − Compute the excess δ π d Σ miγ i.Ifitisnumerically small enough, finish. Otherwise, continue to step 3. Step 3. Compute the next approximation to the independent unknown γ 1 using the recursion = + tan γ 1 γˆ1 γ 1 δ Σ mi tan γ i Step 4. = ... Compute cos a and γ i, i 2, ,m,using cos α cos a = 1 sin γ 1

cos α i γ = arc sin( ) i cos a Step 5. Return to step 2.

We want to makeanobservation about the convergence of this algorithm. The increment in γ 1 in step 3may be very small evenwithout the excess δ being small. This happens if the denominator Σ mi tan γ i is large, unless the numerator tan γ 1 behavessimilarly.Insuch cases the algorithm converges slowly or even diverges, and our choice for the independent unknown prevent them from occurring. By the way,choosing cos a as the independent unknown creates similar problems and although it looks plausible and perhaps more elegant (compare with the well known Lagrangemultipliers ), it should be avoided. A mathemati- cally rigorous proof of convergence is out of the scope of this paper; Wewill have tobesatisfied by empiri- cal convergence for all the finitely manyuniform polyhedra (cf. [CLM], [S]). Note that in fewexceptional cases, the direct approach fails. In these cases, it happens that the vertex density d is zero, and the faces incident at a vertexoccur in several oppositely oriented pairs, and thus even the initial approximation satisfies the fundamental system, which is obviously impossible, since, as we π already remarked, α + γ > in a right-angled spherical triangle. It is interesting to note that these polyhe- i i 2 dra are exactly those which cannot be constructed directly using the , to be presented in the section 4. In section 5 we present an adjustment to the construction, which in turn leads to an adjust- ment to the fundamental system, which makes it possible to handle these exceptions.

4. The Kaleidoscope. We start by discussing the dihedral kaleidoscope.Itconsists of twohinged planar mirrors. The reflections in these mirrors generate a group of isometries, which is finite if the dihedral angle of the kalei- π doscope is a rational submultiple of π ,i.e., of the form ,where q is a rational greater than one. We now q -6-

consider the meaning of these particular angles from additional points of view. Intersect the kaleidoscope by a perpendicular plane, and choose a circle in this plane with center at the corner.The kaleidoscope encloses a circular arc, and we may ask for the cases in which repeated reflec- tions of this arc in the mirrors produce a finite covering of the circle. It is easy to see that a d-fold covering n obtained if q = ,where n and d are relatively prime positive integers, with n > d.Inthis case the the d π π d kaleidoscope generates D ,the of order 2n,and the = arc consists of d copies of the n q n π fundamental domain of the group, a arc. n Now, put a point object in a general position on the enclosed arc. Then, repeated reflections in the mirrors will yield 2n points. which may naturally be viewed as the vertices of a 2n-gon, where adjacent ... vertices belong to arcs which share an endpoint. Let the 2n-gon be C1C2 C2n,where C1 is the point object. = Note, that in twocases we get a smaller number of images: The first case is when C1 C2 or = C1 C2n,which means that C1 is a fixed point of one of the original reflections, or,geometrically,that the point object is located on one of the mirrors. Then, we get n pairs of coincident adjacent vertices, which means that n of the edges of the polygon are degenerate, and may be discarded. Thus, when the point object n is located on one of the mirrors, we get regular n-gon, with density d,customarily denoted by { }, i. e., d 1 1 {q}. Note, that if {q}and {q′}satisfy + = 1, theyare twooppositely oriented copies of the same q q′ polygon. + The other case is when Cn+1 = C1.This is only possible if the n 1st arc coincides with the first, but oppositely oriented. Thus, C1 is located on the bisecting plane of the kaleidoscope, and the reflection in this plane belongs to to the dihedral group generated by the kaleidoscope. In this case we have nodegenerate edges, but rather a regular n-gon which is transversed twice. As the density of the compound is d, d has to d be evenand each polygon has density .Thus, when the point object is located on the bisector,weget a n 2 double { }, i.e., a double {2q}. It is easy to check that if the point object is similarly located but the (d/2) denominator d is odd, it traces a single regular 2n-gon with density d,i.e., a single {2q}. In all the other cases the polygon is an irregular 2n-gon, howeverifweuse only rotations (or any ... ev ennumber of reflections) then the polygon C1C3 C2n−1 is still regular,a{q}. We now add a third mirror,and get the trihedral kaleidoscope,which consists of three concurrent planar mirrors. The problem of finding conditions on the dihedral angles for the kaleidoscope to generate a finite group of isometries was posed and solved by A. H. Schwarz in 1873. As we have already seen, it is necessary that these angles are rational submultiples of π ,but howeverthis is not sufficient. Intersecting the kaleidoscope by a sphere centered in its corner,Schwarz formulated the following problem: Enumerate π π π all convex spherical triangles PQR,with angles , and (p,q,r are all rational greater than one), with p q r the property that when reflected in its sides anynumber of times, PQR produces a finite covering of the sphere. Such atriangle is called a and is denoted by the symbol (pqr). It may be shown that unless twoofthe angles are π (in which case the kaleidoscope generates the dihedral symmetry group), 2 p, q,and r may have asnumerators the numbers 2, 3, 4 or 5, with the further stipulation that 4 and 5 cannot occur together.The denominator has to be smaller than the numerator,and this makes the number of nondi- hedral Schwarz triangles finite. Acomplete list of Schwarz triangles may be found in [C], table III, and in appendix I. Putting a point object C in the Schwarz triangle PQR,itisnow reflected repeatedly to produce a polyhedron, whose vertices are the images of the point object. As before, twovertices are considered adja- cent if theybelong to triangles which share an edge, and several vertices are incident to the same face if theybelong to triangles which share a vertex. Note that as in the previous discussion, the polyhedron may be degenerate: Two images may coincide without a similar coincidence of the Schwarz triangles. This means that the polyhedron may be split into several, not necessarily distinct, polyhedra. This phenomenon is investigated in [CLM], and we will not discuss it here anydetail (but see howeversection 5). -7-

The kaleidoscopic construction of polyhedra is named after Wythoff(1918) who was the first to use it (in four dimensions, cf. [CLM], p. 406). Assuming nondegeneracy, Wythoffconstruction is very useful in getting the topological invariants of the traced polyhedra, that is, the covering density D and the χ .Both values can be eas- ily computed for the for the multiple tiling of the sphere by Schwarz triangle: D may be found from the area of (pqr)and the order g of the symmetry group generated by the reflection in its sides: π π π 4π D = g( + + − π ), p q r or, g 1 1 1 D = ( + + − 1). 4 p q r Note that the value of g is 4n ,24, 48 or 120 if the group is n-fold dihedral, tetrahedral, octahedral or icosa- hedral, respectively.Aspointed out in [CLM] (p. 412, 418, 425), this is also the correct value for the den- sity of anypolyhedron derivable from (pqr), except for the cases where the pos- sesses faces which are hemispherical or concave (i.e., bigger than a hemisphere). In the former case, the planar faces are incident at the center,and the density is not well-defined. In the latter case, the spherical and planar faces occur on twodifferent sides of the center,and the density of the planar polyhedron may be computed by subtracting the density of the spherical polyhedron above from the number of concave faces (see the tables in appendix II; in our notation, a face is concave if γ is obtuse, see section 3). The Euler characteristic relates V,the number of vertices, E,the number of edges, and F,the number of faces of anymap drawn on a givenclosed surface by the famous Euler formula V − E + F = χ .Here, we are considering a D-sheeted , which is triangulated by the Schwarz triangles, so that = = 3g = + + π = g F g, E ,and V V p Vq Vr ,where Vq is the number of vertices with the angle ,i.e., Vq , 2 q 2nq where nq is the numerator of q.Thus, we get the formula g 1 1 1 χ = ( + + − 1) 2 n p nq nr This value is also inherited by anynon-degenerate polyhedron derivable from (pqr)(see the tables in ap- pendix II; see section 5 for the degenerate cases). It is very interesting to note that χ depends upon the numerators only.This means, for instance, that the convex Plato’sicosahedron and the star-cornered 5 Poinsot’sgreat , derivable from (2 35)and (2 3 ), have both Euler characteristic 2, and are 2 thus homeomorphic as closed surfaces, although their densities are much different (1 for the former,7for the latter). It is clear,that whereverweput a point C in the Schwarz triangle PQR,the traced polyhedron will be isogonal, i. e., all its vertices will be equivalent under the action of the isometry group generated by the three reflections. We will nowdiscuss the positions of C which guarantee that the traced faces will be regu- lar polygons, and study the corresponding spherical tiling and the fundamental triangulation and its triangu- ...... lar Ai BiC.Asbefore, A1 A2 will be the prototile of the dual tiling, and B1 B2 the vertexfig- ure.

4.1. C is at the vertex P of PQR.

Let A1 and A2 be the vertices Q and R respectively,and let B1 be the foot of the perpendicular from P on QR. n The resulting fundamental equations are, setting p = : d = = qα 1 rα 2 π ,

+ = nγ 1 nγ 2 π d...... By reflecting PQR repeatedly in its sides, we get the sequences of vertices A1 A2 A3 and B1 B2 B3 , -8-

which repeat themselves after 2n terms. The resulting polyhedron is denoted by the Wythoffsymbol p|qr.Its vertexconfiguration is (q. r. q. r. ....q.r)(2nterms), and its vertexdensity is d.The vertexfigure and the dual prototile are 2n- gons.

4.2. C is at the intersection of the side PQ with bisector of the opposite angle R.

Let A1, A2 and A3 be the vertices Q, R and P,and let B1 and B2 be the feet of the perpendiculars from C on QR and RP,respectively. The resulting fundamental equations are: = = = qα 1 2rα 2 pα 3 π ,

+ + = γ 1 2γ 2 γ 3 π .

By reflecting A2, B1 and B2 in QR,weget the images A4, B4 and B3,respectively. The resulting polyhedron is denoted by the Wythoffsymbol pq|r,and its vertexconfiguration is (p.2r.q.2r). The vertexfigure and the dual prototile are .

4.3. C is at the incenter of PQR.

Let A1, A2 and A3 be the vertices P, Q and R,and let B1, B2 and B3 be the feet of the perpendicu- lars from C on PQ, QR and RP,respectively. The resulting fundamental equations are: = = = 2pα 1 2qα 2 2rα 3 π ,

+ + = γ 1 γ 2 γ 3 π .

The resulting polyhedron is denoted by the Wythoffsymbol pqr|, and its vertexconfiguration is (2p.2q.2r). The vertexfigure and the dual prototile are triangles.

4.4. C traces a snub polyhedron. In the previous positions, the symmetry group of the polyhedron is generated by reflections. In partic- ular,the polyhedron is reflexible. Wenow discuss a fourth position, which usually traces a chiral polyhe- dron, i.e., a polyhedron with a symmetry group which consists of rotations only3.This is done by consider- ing only images under an evennumber of reflections (as was done for the dihedral kaleidoscope). In this case, it is easier to study the dual tiling directly,and to infer the existence of C without an actual construc- tion. Let O be the Fermat point of the Schwarz triangle PQR,i.e, the point where the sides subtend equal 2π 2π angles (assuming it exists4). Reflect O in the sides and get an A A ...A with angles , , 1 2 6 3 p 2π 2π 2π 2π , , and respectively,where odd indices refer to the images of O,and evenindices refer to the 3 q 3 r vertices P, Q and R.Reflecting this hexagon anyevennumber of times, we get an isohedral, vertically reg- ular tiling of the sphere, i.e., the dual of a . The latter consists of a {p}, a {q}and a {r}with centers at P, Q and R,respectively,alternating with three, so-called snub,{3}’s. Let C be the common ver- texofthese six polygons (since the snub {3}’sare congruent, it is located at the circumcenter of A1 A3 A5), = ... and let Bi,for i 1, 2, ,6,bethe foot of the perpendicular from C to Ai Ai+1. The resulting fundamental equations are: = = = = 3α 1 pα 2 qα 4 rα 6 π ,

3 However, (5/2 3 3)and (3/2 3/2 5/2)being isosceles, theyyield reflexible snub polyhedra. 4 The existence of O is verified by a continuity argument, at least for the case where all the angles of PQR are smaller then 2π /3 (this is not the case for (3/2 3/2 5/2)and (3/2 5/3 2)). -9-

+ + + = 3γ 1 γ 2 γ 4 γ 6 π . Here, we counted one snub {3} thrice, because the snub {3}s are all congruent. The resulting snub polyhe- dron is denoted by the Wythoffsymbol |pqr,and its vertexconfiguration is (3. p.3.q.3.r). The vertex figure and the dual prototile are . π Note that in all the cases, the occurrence of α = implies that the corresponding tile is a {2}, i.e., a 2 , and may be discarded.

5. Exceptional Polyhedra. An polyhedron is orientable if its faces may be coherently oriented, that is, assigned orientations in such a way that the orientations induced on an edge common to twofaces are opposite. It is easy to verify, that the Wythoffconstruction always yields orientable polyhedra, assuming that vertices which belong to nonadjacent Schwarz triangles are considered distinct, eveniftheycoincide. The number of the polyhe- dron vertices may be readily found by dividing the order of the kaleidoscope symmetry group by the num- ber of copies of adjacent Schwarz triangles which share a vertex, i. e., by twofor pq|ror |pqr,and by n n for |qr. d We will nowdiscuss twoparticular cases where the Wythoffconstruction yields only pairs of coin- ciding vertices, that, when identified, are the vertices of a nonorientable, or one-sided,polyhedron. In these cases, the number of real vertices will be of course only a half of the expected number.The discussion is based on the phenomenon, described in section 4, that a point object on the bisector of a dihedral kaleido- π scope with angle , q rational with an evendenominator,yields a double {2q}. q Forconclusion, we shall have a brief, independent discussion about the solution for the only non- Wythoffian polyhedron in existence.

5.1. Polyhedra pqr|with an evendenominator. π Consider a Schwarz triangle PQR,with an angle at R, r rational with an evendenominator.Sup- r pose the point object C is located in the incenter of PQR.Inparticular,itlies on the bisector atR.Aswe have seen, the images of C trace a {2p}, a {2q}and a {2r}, with centers at P, Q and R,respectively.How- ev er, since r has an evendenominator,wefind that the face with center at R,which is traced by reflections at RP and RQ only,isadouble {2r}. When transversing it once, PQR is reflected in the bisector at R,and the faces with centers at P and Q are reflected into twosimilar faces with centers at the images of P and Q. Thus, in addition to the double {2r}, we have at C two{p}’sand two{q}’s, arranged in the cyclic order (2r. p. q.2r.q.p). By discarding the double {2r}’s, we get a single, one-sided polyhedron. The nonori- entability is due to orientation reversal: Trying to find a coherent orientation on the faces and considering the four faces incident at a vertex, we notice that twoofthem has to be retrograde, i.e., the vertexconfigura- tion is (p. q. p′. q′). However, when considering the faces adjacent to a discarded double {2r}, none of the four faces incident at a vertexmay be retrograde. It is interesting to note, that the enumeration of the faces at C does not define the polyhedron well, and the direct algorithmic approach in section 3 fails. However, bytaking the {2r}’sinto account, and solv- ing the equations set in section 4.3, the metrical properties of these exceptional polyhedra may be still com- puted. The Euler characteristic can be obtained from the basic value discussed in section 4: By considering the fundamental triangulation and removing the double {2r}’s, the balance of edges and faces doesn’t change, and the number of vertices is reduced by Vr (see section 4). Thus, we get for this case g 1 1 χ = ( + − 1). 2 n p nq The density is not-well define, as these polyhedra do not produce an evencovering of the sphere. -10-

5.2. Hemi polyhedra pp′|r. π π Consider a Schwarz triangle PQR,where the angles and are complementary,and assume that p q q = p′ <2

5 3 5 5.3. The last polyhedron |3 . 2 2 3 The only uniform polyhedron which has more than six faces at a vertex, and the only one which can- not be constructed by the Wythoffconstruction (or a minute modification of it), has the vertexconfiguration 5 5 3 (4. .4.3.4. .4. ). Again, the enumeration of faces incident at a vertexdoes not determine the polyhe- 2 3 2 dron, and therefore a direct use of the uniform solution fails. However, byconsidering the Schwarz triangle 5 5 (3 )and its reflection in the perpendicular bisector of its shortest side (which is an arc of symmetry of 2 3 the generated group), it may be shown (cf. [CLM], section 11), that the oppositely oriented pairs of trian- 5 3 5 5 5 gles and of |3 also belong to an enantiomorphous and vertexsharing pair of |3 ’s. 2 2 3 2 3 Actually,both pentagrams belong to both snub polyhedra, with roles interchanged, and each triangle is a non-snub triangle of one of the pair.Wecan conclude easily that the last uniform polyhedron is a second instance of an orientable hemi polyhedron. Furthermore, solving the snub polyhedra as in section 4.4, we can complete the solution by computing the triangulation of the extra using the fundamental equa- tion (3), section 2. The Euler characteristic has the bizarre value of χ =−56, computable from the actual counts of vertices, edges and faces. -11-

6. References. [BC] Ball, W. W. R & Coxeter,H.S.M., Mathematical Recreations and Essays,Thirteenth Edition, Dover, NewYork, 1987. [C] Coxeter,H.S.M., Regular ,Third Edition, Dover, New York, 1973. [CLM] Coxeter,H.S.M., Longuet-Higgins, M. S. & Miller,J.C.P., Uniform Polyhedra,Phil. Trans. Royal Soc. London, Ser.A,246 (1953), 401-409. [CR] Cundy,H.M.&Rollett, A. P., Mathematical Models,Third Edition, Tarquin, Stradbroke, 1981. [GMS] Grünbaum, B., Miller,J.C.P.&Shephard, G. C., Uniform Tilings with Hollow Tiles,inThe Geomet- ric Vein - The Coxeter Festschrift,C.Davis et al., eds., Springer-Verlag, NewYork, 1982, 17-64. [GS] Grübaum, B. & Shephard, G. C., Tilings and Patterns,Freeman, NewYork, 1987. [Ha] Har’El, Z., Kaleido - Kaleidoscopic Construction of Uniform Polyhedra,Department of Mathemat- ics, Technion, Haifa, 1992. [Hm] Hume, A., Exact Descriptions of Regular and Semi-regular Polyhedraand Their Duals,Computing Science Technical Report No. 130, AT&T Bell Laboratories, Murray Hill, 1986. [Hy] Huybers, P., Uniform Polyhedrafor Building Structures,Bull. Inter.Assos. Shells and Spatial Struc- tures, 21 (1980), 27-38. [J] Johnson, N. W., ConvexPolyhedrawith Regular Faces,Canad. J. Math., 18 (1966), 169-200. [L] Lines, L., Solid Geometry,Dover, New York, 1961. [M] Moritz, R., On Napier'sFundamental Theorem Relating to Right Spherical Triangles,Amer.Math. Monthly,22(1915), 220-222. [S] Skilling, J., The Complete Set of Uniform Polyhedra,Phil. Trans. Royal Soc. London, Ser.A,278 (1975), 111-135. [W1] Wenninger,M.J., Polyhedron Models,Cambridge University Press, Cambridge, 1971. [W2] Wenninger,M.J., Dual Models,Cambridge University Press, Cambridge, 1984. -12-

Appendix I: Schwarz Triangles. To facilitate the enumeration of Schwarz triangles, one notices that for the covering to be simple, = = ... (pqr)must be a Möbius triangle,that is, one of the triangles D1/n (2 2 n)(with n 2, 3, ), = = = T1 (2 3 3), O1 (2 3 4) and I1 (2 3 5), which are the fundamental domains of the full dihedral, tetrahe- dral, octahedral and groups, respectively.ASchwarz triangle which produces a d- fold covering of the sphere can be decomposed to d copies of one of the Möbius triangles. These proper- ties may be used to synthesize Schwarz triangles, as we nowexplain. Let = S1 (pxr1;yv w1) and = ′ S2 (x qr2;uyw2) be twoSchwarz triangles (where we included in the symbol also the sides opposite the respective angles), π π which have a common edge y and complementary angles and .Then, S and S can be pasted along x x′ 1 2 the common edge, and produce a newtriangle + = S1 S2 (pqr;uv w), where we have set 1 1 1 = + , r r1 r2 = + w w1 w2. This is a Schwarz triangle if r >1.Asusual, 2S is a shorthand for S + S.

In fewcases, S1 and S2 can be pasted together in more then one way,toproduce several different Schwarz triangles. In particular,ifS1and S2 are twocopies of the same right-angled triangle (pq2;uv w), with p, q >2 and p ≠ q,theycan be pasted along each of the legs, to produce twoisosce- p q les triangles, with the same area but different perimeter,namely (qq ; ww2u)and (pp ; ww2v). 2 2 Starting from the Möbius triangles we get the denumerable family of dihedral Schwarz triangles n D = (2 2 )(with d = 1, ...,n−1), and the tetrahedral, octahedral and icosahedral triangles listed in d/n d tables 1 through 3. The triangles are arranged by increasing area: The area of Id ,for instance, is d times the area of I1.Inthe cases where there are several triangles with the same area, theyare ordered by increasing perimeter and distinguished from one another by a letter,e.g., I2a and I2b.The sides are givenin 5 terms of the legs s, u and v of T1 and I1 .Because of space considerations, we list just one of the several ways to split each Schwarz triangle into smaller ones. It is interesting that every triangle (except the small- est) can always be split into twotriangular pieces, with only one exception: The the trirectangular triangle D1/2,ifconsidered as an icosaheral Schwarz triangle, can only be split into four pieces or more. The listed 5 5 5 splitting is the symmetric one: An , ( ), surrounded by three copies of its half, the 2 2 2 5 right-angled triangle (5 2). Triangles which belong to a proper symmetry subgroup are flagged. 2

1 √ 5 √ 5 5 It may be shown that cos 2s =− ,cos 2u = and cos 2v = . 3 3 5 -13-

Table 1: Tetrahedral Schwarz Triangles symbol sides splitting − T1 (3 3 2) ssπ2s 3 T(3 3 ) π − 2s π − 2s 2s 2T 22 1 3 T (3 2 ) s 2s π − sT+T 3 2 12 3 3 T(2 ) π − 2s π − s π − sT+T 5 2 2 23 33 3 T( )2s 2s 2s 2T 622 2 3

Table 2: Octahedral Schwarz Triangles symbol sides splitting O (4 3 2) π − s π s 1 2 4 − O2a (3 3 2) ssπ2s2O1 T1 3 O(4 4 ) ssπ2O 2b2 2 1 O (4 2 2) π π π O + O D 3 4 2 2 1 2a 1/4 3 O (3 3 ) π − 2s π − 2s 2s 2O T 4a 2 2a 2 4 π O (4 3 ) s π − sO+O 4b 3 2 2a2b 3 π 3π O (4 2 ) − s π − s O + O 5 2 2 4 1 4b O (2 2 2) π π π 2O D 6a 2 2 2 3 1/2 3 O (3 2 ) s 2s π − sO+O T 6b 2 2a4a 3 4 π π O (3 2 ) π − s + sO+O 7 3 4 2 34b 4 3 O (2 2 ) π π π O + O D 9 3 2 2 4 4b 5 3/4 3 3 O (2 ) π − 2s π − s π − s 2O T 10 2 2 5 5 3 4 3 O (2 ) s π π + sO+O 11 2 3 4 2 6b5 3 3 3 O ( )2s 2s 2s 2O T 12 2 2 2 6b 6 3 4 4 π O ( ) π − s π − s 2O 14 2 3 3 2 7 -14-

Table 3: Icosahedral Schwarz Triangles symbol sides splitting I (5 3 2) uv π− u − v 1 2 5 I (3 3 ) π − u − v π − u − v 2u 2I 2a 2 2 2 1 3 I (5 5 ) π − u − v π − u − v 2v 2I 2b 2 2 2 1 5 I (5 2) v π − v 2vI+I 3 2 2 12b 5 I(5 3 ) π − u − v 2v π + u − vI+I 43 2 2 2a2b 55 5 I ( )2v 2v 2v 2I 6a 22 2 3 3 I (5 3 )2u π+ u − v π − u + vI+I 6b 2 2 2 2a4 5 I (5 5 )2v 2v π−2v 2I 6c 4 3 5 I (3 2) π − v π − u π + u − vI+I 7 2 2 2 2 34 5 3 π π I (5 ) − u − v − u + v π − 2vI+I 8 2 2 2 2 2a6b 5 π I(5 2 ) v π − 2v + vI+I 9 3 2 36c 5 5 I(3 )2v π+ u − v π − u + vI+I 10a 2 3 2 2 46a 5 π π I (5 3 ) − u − v π − 2v + u + vI+I 10b 4 2 2 46c 3 π I (5 2 ) u + u + v π − vI+I 11 2 2 110b 5 I (3 2 ) π − v π − u + v π + uI+I 13 3 2 2 2 6b7 5 5 3 π π I( ) + u − v + u − v π − 2v 2I 14a 2 2 2 2 2 7 5 π π I (3 3 ) + u − v + u − v π − 2u 2I 14b 4 2 2 7 I (2 2 2) π π π I + 3I D 15 2 2 2 6a 3 1/2 5 5 I (3 )2v π− u + v π + u + vI+I 16 2 4 2 2 6a10a 5 3 I ( 2 ) π − u π − u + v π + vI+I 17 2 2 2 2 2 89 5 5 5 I( )2v π−2v π−2v 2I 18a 2 3 3 9 5 3 π π I (3 ) − u − v π − 2u + u + vI+I 18b 3 2 2 2 810b 5 π I (3 2 ) v + u + v π − uI+I 19 4 2 910b 5 5 π I ( 2 ) − v π − 2v π − vI+ I 21 2 4 2 10b 11 5 3 3 I ( )2u π+ u + v π + u + v 2I 22 2 2 2 2 2 11 5 3 I (2 ) π + u − v π + u π + vI+ I 23 3 2 2 2 2 10a 13 5 5 3 π π I ( ) − u + v − u + v π − 2v 2I 26 3 3 2 2 2 13 5 5 π I (2 )2v + v π − vI+ I 27 3 4 2 11 16 5 5 π I (2 ) − u − v π − v π − uI+ I 29 2 4 2 11 18b 5 3 5 π π I ( ) + u − v π − 2v + u + vI+ I 32 3 2 4 2 2 14a 18a 3 3 5 π π I ( ) − u + v − u + v π − 2u 2I 34 2 2 4 2 2 17 3 5 5 I ( )2v π+ u + v π + u + v 2I 38 2 4 4 2 2 19 5 5 5 I ( ) π − 2v π − 2v π − 2v 2I 42 4 4 4 21 -15-

Appendix II: Uniform Polyhedra In tables 4 to 8, we list the seventy-fivenondihedral uniform polyhedra, as well as the fivepentagonal prisms and antiprisms, grouped by generating Schwarz triangles. For each polyhedron we indicate a figure number in this appendix, its Wythoffand vertexconfiguration symbols, and the values of its Euler charac- teristic χ and density D (if well-defined). Nonorientable polyhedra are marked by starring their character- istic. For reference, we indicate for each polyhedron a figure number in [CLM], pp. 439-448, and a model number in [W1]. The polyhedron names listed in tables 5 to 8 are taken from [W2], where theyare attrib- uted to Norman W.Johnson, but note theysometimes vary among authors. Our figures represent wire- frame images of the listed uniform polyhedra and their duals (the latter are indicated by a starred figure number6). Although the figures are twodimensional, depth is faithfully represented by the variable thick- ness of the wires. The drawings were made using the UNIX7 utility pic,from directivesgenerated by kaleido (cf. [Ha]).

Table 4: Dihedral Uniform Polyhedra name fig symbol χ D dual pentagonal 125|2 (4. 4. 5) 21 pentagonal dipyramid D 1/5 pentagonal 2|225 (3. 3. 3. 5) 21 pentagonal deltohedron

5 5 pentagrammic prism 32|2 (4. 4. )22 2 2 pentagrammic dipyramid D2/5 5 5 pentagrammic antiprism 4|22 (3. 3. 3. )22 2 2 pentagrammic deltohedron 5 5 pentagrammic crossed antiprism D 5|22 (3. 3. 3. )23 3/5 3 3 pentagrammic concave deltohedron

Table 5: Tetrahedral Uniform Polyhedra name fig symbol χ D ref dual 63|2 3 (3. 3. 3) 2115,1 tetrahedron T 1 723|3 (6. 6. 3) 2116,6 triakistetrahedron

3 3 octahemioctahedron T 8 3|3 (6. .6.3)037,68 2 2 2 octahemioctacron 3 3 tetrahemihexahedron T 9 3|2 (4. .4.3)1*36,67 3 2 2 tetrahemihexacron

6 The ten unbounded polyhedra in figures 8*,9*,20*,54*,56*,66*,68*,75*,76*,80* are different from their [W2] counterparts. Theyuse the morecorrectly shown faces of [W2], p. 103. 7 UNIX is a trademark of AT&T Bell Laboratories. -16-

Table 6: Octahedral Uniform Polyhedra name fig symbol χ D ref dual 10 4|2 3(3. 3. 3. 3) 2117,2 cube 11 3|2 4(4. 4. 4) 2118,3 octahedron 12 2|3 4(3. 4. 3. 4) 2119,11 rhombic 13 2 4|3 (6. 6. 4) 2120,7 tetrakishexahedron O 1 14 2 3|4 (8. 8. 3) 2121,8 triakisoctahedron 15 3 4|2 (4. 3. 4. 4) 2122,13 deltoidal 16 2 34|(4. 6. 8) 2123,15 disdyakisdodecahedron 17 |2 34 (3. 3. 3. 3. 4) 2124,17 pentagonal icositetrahedron

3 3 small cubicuboctahedron O 18 4|4 (8. .8.4)−4238,69 2b 2 2 small hexacronic icositetrahedron 4 8 8 great cubicuboctahedron 19 3 4| ( .3. .4)−4450,77 3 3 3 great hexacronic icositetrahedron 4 4 cubohemioctahedron O4 20 4|3 (6. .6.4)−2* 51,78 3 3 hexahemioctacron 4 8 cubitruncated cuboctahedron 21 34|(.6.8)−4452,79 3 3 tetradyakishexahedron 3 3 great rhombicuboctahedron 22 4|2 (4. .4.4)2559,85 2 2 great deltoidal icositetrahedron O5 3 8 4 small rhombihexahedron 23 24|(8. 4. . )−6* 60,86 2 7 3 small rhombihexacron 4 8 8 stellated truncated 24 2 3| ( . .3)2766,92 3 3 3 great triakisoctahedron O7 4 8 great truncated cuboctahedron 25 23|(.4.6)2167,93 3 3 great disdyakisdodecahedron 4 3 8 4 8 great rhombihexahedron O 26 2| (4. . . )−6* 82,103 11 3 2 3 3 5 great rhombihexacron -17-

Table 7: Icosahedral Uniform Polyhedra name fig symbol χ D ref dual icosahedron 27 5|2 3(3. 3. 3. 3. 3) 2125,4 dodecahedron dodecahedron 28 3|2 5(5. 5. 5) 2126,5 icosahedron 29 2|3 5(3. 5. 3. 5) 2128,12 30 2 5|3 (6. 6. 5) 2127,9 pentakisdodecahedron I 1 31 2 3|5 (10. 10. 3) 2129,10 triakisicosahedron 32 3 5|2 (4. 3. 4. 5) 2130,14 deltoidal truncated icosidodechedon 33 2 35|(4. 6. 10) 2131,16 disdyakistriacontahedron 34 |2 35 (3. 3. 3. 3. 5) 2132,18 pentagonal hexecontahedron

5 55 5 small ditrigonal icosidodecahedron 35 3| 3(.3. .3. .3)−8239,70 2 22 2 small triambic icosahedron 5 5 small icosicosidodecahedron I2a 36 3|3 (6. .6.3)−8240,71 2 2 small icosacronic hexecontahedron 5 5 small snub icosicosidodecahedron 37 | 33 (3. .3.3.3.3)−8241,110 2 2 small hexagonal hexecontahedron 3 3 small dodecicosidodecahedron I 38 5|5 (10. .10. 5) −16 2 42,72 2b 2 2 small dodecacronic hexecontahedron 5 5 5 5 5 5 small stellated dodecahedron 39 5|2 ( . . . . )−6343,20 2 2 2 2 2 2 5 great dodecahedron 40 |2 5 (5. 5. 5. 5. 5)/2 −6344,21 2 small stellated dodecahedron 5 55 dodecadodecahedron 41 2| 5(.5. .5)−6345,73 2 22 medial rhombic triacontahedron 5 5 truncated great dodecahedron I3 42 2 |5 (10. 10. )−6347,75 2 2 small stellapentakisdodecahedron 5 5 rhombidodecadodecahedron 43 5|2 (4. .4.5)−6348,76 2 2 medial deltoidal hexecontahedron 5 10 4 small rhombidodecahedron 44 2 5| (10. 4. . )−18* 46,74 2 9 3 small rhombidodecacron 5 5 snub dodecadodecahedron 45 |2 5(3. 3. .3.5)−6349,111 2 2 medial pentagonal hexecontahedron -18-

Table 7: Icosahedral Uniform Polyhedra name fig symbol χ D ref dual

5 55 5 ditrigonal dodecadodecahedron 46 3| 5(.5. .5. .5)−16 4 53,80 3 33 3 medial triambic icosahedron 5 10 10 great ditrigonal dodecicosidodecahedron 47 3 5| ( .3. .5)−16 4 54,81 3 3 3 great ditrigonal dodecacronic hexecontahedron 5 5 small ditrigonal dodecicosidodecahedron 48 3|5 (10. .10. 3) −16 4 55,82 3 3 small ditrigonal dodecacronic hexecontahedron I4 5 5 icosidodecadodecahedron 49 5|3 (6. .6.5)−16 4 56,83 3 3 medial icosacronic hexecontahedron 5 10 icositruncated dodecadodecahedron 50 35|(.6.10) −16 457,84 3 3 tridyakisicosahedron 5 5 snub icosidodecadodecahedron 51 | 35 (3. .3.3.3.5)−16 4 58,112 3 3 medial hexagonal hexecontahedron 3 great ditrigonal icosidodecahedron 52 |3 5 (3. 5. 3. 5. 3. 5)/2 −8661,87 2 great triambic icosahedron 3 3 great icosicosidodecahedron 53 5|3 (6. .6.5)−8662,88 2 2 great icosacronic hexecontahedron I6b 3 3 small icosihemidodecahedron 54 3|5 (10. .10. 3) −4* 63,89 2 2 small icosihemidodecacron 3 10 6 small dodecicosahedron 55 35|(10. 6. . )−28* 64,90 2 9 5 small dodecicosacron 5 5 small dodecahemidodecahedron I 56 5|5 (10. .10. 5) −12* 65,91 6c 4 4 small dodecahemidodecacron 5 5 5 5 great stellated dodecahedron 57 3|2 ( . . )2768,22 2 2 2 2 great icosahedron 5 great icosahedron 58 |2 3 (3. 3. 3. 3. 3)/2 2 769,41 2 great stellated dodecahedron 5 55 great icosidodecahedron 59 2| 3(.3. .3)2770,94 2 22 great rhombic triacontahedron I7 5 5 great truncated icosahedron 60 2 |3 (6. 6. )2771,95 2 2 great stellapentakisdodecahedron 5 6 4 rhombicosahedron 61 2 3| (6. 4. . )−10* 72,96 2 5 3 rhombicosacron 5 5 great snub icosidodecahedron 62 |2 3(3. 3. .3.3)2773,113 2 2 great pentagonal hexecontahedron 5 10 10 small stellated truncated dodecahedron 63 2 5| ( . .5)−6974,97 3 3 3 great pentakisdodekahedron 5 10 truncated dodecadodecahedron I9 64 25|(.4.10) −6 375,98 3 3 medial disdyakistriacontahedron 5 5 inverted snub dodecadodecahedron 65 | 25 (3. .3.3.5)−6976,114 3 3 medial inverted pentagonal hexecontahedron 5 5 10 5 10 great dodecicosidodecahedron 66 3| ( . . .3)−16 10 77,99 2 3 3 2 3 great dodecacronic hexecontahedron 5 5 5 5 small dodecahemicosahedron 67 |3 (6. .6. )−8* 78,100 3 2 3 2 small dodecahemicosacron I10a 5 5 10 6 10 great dodecicosahedron 68 3| (6. . . )−28* 79,101 2 3 3 5 7 great dodecicosacron 5 5 5 5 great snub dodecicosidodecahedron 69 | 3(3. .3. .3.3)−16 10 80,115 3 2 3 2 great hexagonal hexecontahedron -19-

Table 7: Icosahedral Uniform Polyhedra name fig symbol χ D ref dual

5 5 great dodecahemicosahedron I 70 5|3 (6. .6.5)−8* 81,102 10b 4 4 great dodecahemicosacron 5 10 10 great stellated truncated dodecahedron 71 2 3| ( . .3)213 83,104 3 3 3 great triakisicosahedron 5 5 great rhombicosidodecahedron 72 3|2 (4. .4.3)213 84,105 3 3 great deltoidal hexecontahedron I13 5 10 great truncated icosidodecahedron 73 23|(.4.6)213 87,108 3 3 great disdyakistriacontahedron 5 5 great inverted snub icosidodecahedron 74 | 23 (3. .3.3.3)213 88,116 3 3 great inverted pentagonal hexecontahedron 5 5 5 10 5 10 5 great dodecahemidodecahedron I 75 | ( . . . )−12* 86,107 18a 3 2 3 3 3 3 2 great dodecahemidodecacron 3 5 10 3 10 great icosihemidodecahedron I 76 3| ( . . .3)−4* 85,106 18b 2 3 3 2 3 great icosihemidodecacron 3 3 5 3 3 small retrosnub icosicosidodecahedron I 77 | (3. .3. .3.5)−83891,118 22 2 2 2 2 2 small hexagrammic hexecontahedron 3 5 10 4 10 great rhombidodecahedron 78 2| (4. . . )−18* 89,109 2 3 3 3 7 great rhombidodecacron I23 3 5 3 5 great retrosnub icosidodecahedron 79 | 2(3. .3. .3)237 90,117 2 3 2 3 great pentagrammic hexecontahedron

Table 8: Non-Wythoffian Uniform Polyhedra name fig symbol χ D ref dual

3 5 5 5 5 3 great dirhombicosidodecahedron 80 | 3 (4. .4.3.4. .4. )−56 92,119 2 3 2 3 2 2 great dirhombicosidodecacron -20-

25|2 (4.4.5) 25|2 (4.4.5) Fig. 1 Fig. 1*

|2 2 5 (3.3.3.5) |2 2 5 (3.3.3.5) Fig. 2 Fig. 2*

25/2|2 (4.4.5/2) 25/2|2 (4.4.5/2) Fig. 3 Fig. 3* -21-

|2 2 5/2 (3.3.3.5/2) |2 2 5/2 (3.3.3.5/2) Fig. 4 Fig. 4*

|2 2 5/3 (3.3.3.5/3) |2 2 5/3 (3.3.3.5/3) Fig. 5 Fig. 5*

3|2 3 (3.3.3) 3|2 3 (3.3.3) Fig. 6 Fig. 6* -22-

23|3 (6.6.3) 23|3 (6.6.3) Fig. 7 Fig. 7*

3/2 3|3 (6.3/2.6.3) 3/2 3|3 (6.3/2.6.3) Fig. 8 Fig. 8*

3/2 3|2 (4.3/2.4.3) 3/2 3|2 (4.3/2.4.3) Fig. 9 Fig. 9* -23-

4|2 3 (3.3.3.3) 4|2 3 (3.3.3.3) Fig. 10 Fig. 10*

3|2 4 (4.4.4) 3|2 4 (4.4.4) Fig. 11 Fig. 11*

2|3 4 (3.4.3.4) 2|3 4 (3.4.3.4) Fig. 12 Fig. 12* -24-

24|3 (6.6.4) 24|3 (6.6.4) Fig. 13 Fig. 13*

23|4 (8.8.3) 23|4 (8.8.3) Fig. 14 Fig. 14*

34|2 (4.3.4.4) 34|2 (4.3.4.4) Fig. 15 Fig. 15* -25-

234| (4.6.8) 234| (4.6.8) Fig. 16 Fig. 16*

|2 3 4 (3.3.3.3.4) |2 3 4 (3.3.3.3.4) Fig. 17 Fig. 17*

3/2 4|4 (8.3/2.8.4) 3/2 4|4 (8.3/2.8.4) Fig. 18 Fig. 18* -26-

34|4/3 (8/3.3.8/3.4) 34|4/3 (8/3.3.8/3.4) Fig. 19 Fig. 19*

4/3 4|3 (6.4/3.6.4) 4/3 4|3 (6.4/3.6.4) Fig. 20 Fig. 20*

4/3 3 4| (8/3.6.8) 4/3 3 4| (8/3.6.8) Fig. 21 Fig. 21* -27-

3/2 4|2 (4.3/2.4.4) 3/2 4|2 (4.3/2.4.4) Fig. 22 Fig. 22*

3/2 2 4| (8.4.8/7.4/3) 3/2 2 4| (8.4.8/7.4/3) Fig. 23 Fig. 23*

23|4/3 (8/3.8/3.3) 23|4/3 (8/3.8/3.3) Fig. 24 Fig. 24* -28-

4/3 2 3| (8/3.4.6) 4/3 2 3| (8/3.4.6) Fig. 25 Fig. 25*

4/3 3/2 2| (4.8/3.4/3.8/5) 4/3 3/2 2| (4.8/3.4/3.8/5) Fig. 26 Fig. 26*

5|2 3 (3.3.3.3.3) 5|2 3 (3.3.3.3.3) Fig. 27 Fig. 27* -29-

3|2 5 (5.5.5) 3|2 5 (5.5.5) Fig. 28 Fig. 28*

2|3 5 (3.5.3.5) 2|3 5 (3.5.3.5) Fig. 29 Fig. 29*

25|3 (6.6.5) 25|3 (6.6.5) Fig. 30 Fig. 30* -30-

23|5 (10.10.3) 23|5 (10.10.3) Fig. 31 Fig. 31*

35|2 (4.3.4.5) 35|2 (4.3.4.5) Fig. 32 Fig. 32*

235| (4.6.10) 235| (4.6.10) Fig. 33 Fig. 33* -31-

|2 3 5 (3.3.3.3.5) |2 3 5 (3.3.3.3.5) Fig. 34 Fig. 34*

3|5/2 3 (5/2.3.5/2.3.5/2.3) 3|5/2 3 (5/2.3.5/2.3.5/2.3) Fig. 35 Fig. 35*

5/2 3|3 (6.5/2.6.3) 5/2 3|3 (6.5/2.6.3) Fig. 36 Fig. 36* -32-

|5/2 3 3 (3.5/2.3.3.3.3) |5/2 3 3 (3.5/2.3.3.3.3) Fig. 37 Fig. 37*

3/2 5|5 (10.3/2.10.5) 3/2 5|5 (10.3/2.10.5) Fig. 38 Fig. 38*

5|2 5/2 (5/2.5/2.5/2.5/2.5/2) 5|2 5/2 (5/2.5/2.5/2.5/2.5/2) Fig. 39 Fig. 39* -33-

5/2|2 5 (5.5.5.5.5)/2 5/2|2 5 (5.5.5.5.5)/2 Fig. 40 Fig. 40*

2|5/2 5 (5/2.5.5/2.5) 2|5/2 5 (5/2.5.5/2.5) Fig. 41 Fig. 41*

25/2|5 (10.10.5/2) 25/2|5 (10.10.5/2) Fig. 42 Fig. 42* -34-

5/2 5|2 (4.5/2.4.5) 5/2 5|2 (4.5/2.4.5) Fig. 43 Fig. 43*

25/2 5| (10.4.10/9.4/3) 25/2 5| (10.4.10/9.4/3) Fig. 44 Fig. 44*

|2 5/2 5 (3.3.5/2.3.5) |2 5/2 5 (3.3.5/2.3.5) Fig. 45 Fig. 45* -35-

3|5/3 5 (5/3.5.5/3.5.5/3.5) 3|5/3 5 (5/3.5.5/3.5.5/3.5) Fig. 46 Fig. 46*

35|5/3 (10/3.3.10/3.5) 35|5/3 (10/3.3.10/3.5) Fig. 47 Fig. 47*

5/3 3|5 (10.5/3.10.3) 5/3 3|5 (10.5/3.10.3) Fig. 48 Fig. 48* -36-

5/3 5|3 (6.5/3.6.5) 5/3 5|3 (6.5/3.6.5) Fig. 49 Fig. 49*

5/3 3 5| (10/3.6.10) 5/3 3 5| (10/3.6.10) Fig. 50 Fig. 50*

|5/3 3 5 (3.5/3.3.3.3.5) |5/3 3 5 (3.5/3.3.3.3.5) Fig. 51 Fig. 51* -37-

3/2|3 5 (3.5.3.5.3.5)/2 3/2|3 5 (3.5.3.5.3.5)/2 Fig. 52 Fig. 52*

3/2 5|3 (6.3/2.6.5) 3/2 5|3 (6.3/2.6.5) Fig. 53 Fig. 53*

3/2 3|5 (10.3/2.10.3) 3/2 3|5 (10.3/2.10.3) Fig. 54 Fig. 54* -38-

3/2 3 5| (10.6.10/9.6/5) 3/2 3 5| (10.6.10/9.6/5) Fig. 55 Fig. 55*

5/4 5|5 (10.5/4.10.5) 5/4 5|5 (10.5/4.10.5) Fig. 56 Fig. 56*

3|2 5/2 (5/2.5/2.5/2) 3|2 5/2 (5/2.5/2.5/2) Fig. 57 Fig. 57* -39-

5/2|2 3 (3.3.3.3.3)/2 5/2|2 3 (3.3.3.3.3)/2 Fig. 58 Fig. 58*

2|5/2 3 (5/2.3.5/2.3) 2|5/2 3 (5/2.3.5/2.3) Fig. 59 Fig. 59*

25/2|3 (6.6.5/2) 25/2|3 (6.6.5/2) Fig. 60 Fig. 60* -40-

25/2 3| (6.4.6/5.4/3) 25/2 3| (6.4.6/5.4/3) Fig. 61 Fig. 61*

|2 5/2 3 (3.3.5/2.3.3) |2 5/2 3 (3.3.5/2.3.3) Fig. 62 Fig. 62*

25|5/3 (10/3.10/3.5) 25|5/3 (10/3.10/3.5) Fig. 63 Fig. 63* -41-

5/3 2 5| (10/3.4.10) 5/3 2 5| (10/3.4.10) Fig. 64 Fig. 64*

|5/3 2 5 (3.5/3.3.3.5) |5/3 2 5 (3.5/3.3.3.5) Fig. 65 Fig. 65*

5/2 3|5/3 (10/3.5/2.10/3.3) 5/2 3|5/3 (10/3.5/2.10/3.3) Fig. 66 Fig. 66* -42-

5/3 5/2|3 (6.5/3.6.5/2) 5/3 5/2|3 (6.5/3.6.5/2) Fig. 67 Fig. 67*

5/3 5/2 3| (6.10/3.6/5.10/7) 5/3 5/2 3| (6.10/3.6/5.10/7) Fig. 68 Fig. 68*

|5/3 5/2 3 (3.5/3.3.5/2.3.3) |5/3 5/2 3 (3.5/3.3.5/2.3.3) Fig. 69 Fig. 69* -43-

5/4 5|3 (6.5/4.6.5) 5/4 5|3 (6.5/4.6.5) Fig. 70 Fig. 70*

23|5/3 (10/3.10/3.3) 23|5/3 (10/3.10/3.3) Fig. 71 Fig. 71*

5/3 3|2 (4.5/3.4.3) 5/3 3|2 (4.5/3.4.3) Fig. 72 Fig. 72* -44-

5/3 2 3| (10/3.4.6) 5/3 2 3| (10/3.4.6) Fig. 73 Fig. 73*

|5/3 2 3 (3.5/3.3.3.3) |5/3 2 3 (3.5/3.3.3.3) Fig. 74 Fig. 74*

5/3 5/2|5/3 (10/3.5/3.10/3.5/2) 5/3 5/2|5/3 (10/3.5/3.10/3.5/2) Fig. 75 Fig. 75* -45-

3/2 3|5/3 (10/3.3/2.10/3.3) 3/2 3|5/3 (10/3.3/2.10/3.3) Fig. 76 Fig. 76*

|3/2 3/2 5/2 (3.3/2.3.3/2.3.5/2) |3/2 3/2 5/2 (3.3/2.3.3/2.3.5/2) Fig. 77 Fig. 77*

3/2 5/3 2| (4.10/3.4/3.10/7) 3/2 5/3 2| (4.10/3.4/3.10/7) Fig. 78 Fig. 78* -46-

|3/2 5/3 2 (3.3/2.3.5/3.3) |3/2 5/3 2 (3.3/2.3.5/3.3) Fig. 79 Fig. 79*

|3/2 5/3 3 5/2 (4.5/3.4.3.4.5/2.4.3/2) |3/2 5/3 3 5/2 (4.5/3.4.3.4.5/2.4.3/2) Fig. 80 Fig. 80*