Uniform Solution for Uniform Polyhedra* Zvi Har’El Department of Mathematics Technion − Israel Institute of Technology Haifa 32000, Israel E‐Mail: rl@math.technion.ac.il ABSTRACT An arbitrary precision solution of uniform polyhedra and their duals is presented. The solution is uniform for all polyhe‐ dra given by their kaleidoscopic construction, with no need to ‘examine’ each polyhedron separately. 1. Introduction. Uniform polyhedra, whose faces are regular and vertices equivalent, have been studied since antiquity. Best known are the five Platonic solids and the 13 Archimedean solids. We then have the two infinite families of uniform prisms and antiprisms. Allowing for star faces or vertices, we have the four Kepler‐Poinsot regular star polyhedra and a row of 53 nonconvex uniform poly‐ hedra 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’s delightful book Polyhedron Models [W1], which appeared in 1971 but since has been reprinted many times, contains photos and building instructions for cardboard models of these 75 uniform polyhedra. Reading the book, makes the mathematically minded reader wonder: How are the data for the models obtained? For example, what makes 1.1600030093 the proper circumradius for a great retrosnub icosidodecahedron1 with edge length two? It is easy to check, that these data originate in [CLM, Table 7]. Some of the circumradii are exact, as they are given in terms of integers and radi‐ cals only, but few, as the one mentioned above, are given approximately, to ten decimal digits. This may be considered perhaps accurate enough, but if one wants to incorporate polyhedra in a computer modeling software (cf. [Hy]), one would prefer to get the numbers in an arbitrary precision, or in the maxi‐ mum precision the computer can handle. Furthermore, one is interested in exact, or maximum precision, values of other geometric data, such as the dihe‐ dral angles of the polyhedra, 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]). ─────────── * In memoriam 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‐ This problem was treated by Andrew Hume. His method is best described by a short quotation from his report [Hm]: In general, the data are solutions of equations found by examining the polyhedra (for example [L, pp. 174‐176]). The equations were solved at least three times using symbolic algebra systems... Uniform polyhedra for which symbolic algebra systems are useful are the so‐called snub polyhedra, and the computations involve solving cubic or quar‐ tic equations. Hume’s solutions were used to create a database of polyhedra, which is publicly available at netlib@research.att.com. Our approach is quite different. Rather then separately examining indi‐ vidual polyhedra, we suggest a uniform approach, which is easy to understand and easy to use, even with 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 as well as for nonconvex polyhedra (which are not treated by [Hm]). A computer program, called kaleido (cf. [Ha]) and publicly available at ftp@ftp.math.technion.ac.il, has been developed to compute the data of a uniform polyhedron (and its dual), given either the vertex configu‐ ration, i.e, the enumeration of the polygons appearing as faces incident at a vertex in the order in which they are found (cf. [CR, §2.9.2]), or the so‐ called Wythoff symbol which describes the kaleidoscopic construction of the polyhedron (cf. [CLM, §3]). Kaleido is also capable of computing the vertex and face coordinates and displaying a rotating wire‐frame image of each poly‐ hedron. We would like to express our gratitude to H. S. M. Coxeter, Branko Grünbaum and Andrew Hume, for the very useful and enlightening comments. 2. The Fundamental System. The uniform solution is based upon projecting the uniform polyhedron onto a concentric sphere, decomposing each n‐sided tile in the spherical tiling into 2n congruent right‐angled spherical triangles, setting the trigonometric equations and solving them iteratively. We shall explain the solution algo‐ rithm in the next section. Here, we discuss the procedure of identifying the fundamental triangles and setting the fundamental 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) is the vertex configuration. Projecting the polyhedron on a concentric sphere, we get a spherical tiling. Choose a vertex C, and let Ai, for i=1,...,m, be the incenter of the of the ith tile incident to C. Also, let Bi be the foot of the perpendicular 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 polygon A1A2...Am is the pro‐ totile of the dual tiling (cf. [GS, §1.2]). Also, the planar polygon B1B2...Bm is the so called vertex figure at C (cf. [C, §2.1]). ‐3‐ This way, we decompose the spherical tiling into right‐angled spherical triangles, called the fundamental triangles: At each vertex C we get 2m funda‐ mental triangles, arranged in congruent pairs AiBi−1C and AiBiC, 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). Denote the sides and the angles of AiBiC by ai, bi, ci, αi, π/2, γi, in the obvious manner (the right angle is at Bi). From the definition of Ai, and the regularity of the face, we easily see niαi=π, (1) for i=1,...,m. Furthermore, if we consider the angles of the fundamental tri‐ angles meeting at C, we get Σγi=π. (2) Finally, we use the fact that the edge CBi is shared by the two neighboring fundamental triangles AiBiC and Ai+1BiC, and get ai=ai+1. Thus, all the ai’s are equal, and the common value, say a, satisfies cosa=cosαi/sinγi (3) for i=1,...,m, as may be easily deduced from the spherical law of cosines, or from Napier’s second rule of circular parts (cf. [M]). Thus, we get a system of 2m+1 simultaneous equations (where m+1 are lin‐ ear and m are trigonometric) in the 2m+1 unknowns αi, γi and cosa. We name it the fundamental system, and we shall describe its solution in the next sec‐ tion. Once the fundamental system has been solved, the rest of the sides of the fundamental triangles may be solved using similar formulas: cosci=cosγi/sinαi cosbi=1/tanαitanγi The solution of the fundamental triangles reflects directly on the prob‐ lem of finding all the metrical properties of the uniform polyhedron: Let R,ρ,ri,l,hi,θi,ϕi, be respectively the circumradius (the distance from the center to a vertex), the midradius (the distance from the center to an edge), the inradius (the distance from the center to the face), the semiedge, the facial inradius, the facial angle and the dihedral angle (the angle between the the i’th and the i+1’st faces). Then it is not hard to verify that ρ/R=cosa, ri/ρ=cosci, l/ρ=tana, hi/ri=tanci, θi=π−2αi, ϕi=π−ci−ci+1. The construction described above can be easily adjusted to accommodate nonconvex uniform polyhedra2. Such polyhedra may have star polygons as faces, ─────────── 2 The corresponding spherical tiling consists of hollow tiles ‐4‐ or have the faces meeting at a vertex surround it more than once. In the for‐ mer case ni may be taken fractional (with the denominator being the density 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 vertex figure. The i’th face is said to be retrograde if the shorter of the the two arcs connecting Bi−1 to Bi is oppositely oriented. (We assume that at least one face is not retro‐ grade!) In this case we will represent a retrograde {n} by an {n′}, with n′<2 satisfying 1/n+1/n′=1. This will make α′=π−α obtuse, and γ′=−γ negative, as required. We will reiterate on these adjustments in the next section. It is worthwhile to note that [CLM], p. 420, utilize the circumcircle of the vertex figure and elementary plane trigonometry to derive a set of equa‐ tions which is essentially identical to our fundamental system. They are 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 relation‐ ship between uniform polyhedra and their fundamental triangulations, a rela‐ tionship 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 few symbolic 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 solve 2n of the unknown in terms of the 2n+1’st, referred later as the independent unknown, and to compute the latter by iterative approxima‐ tions. One of the problems in such an approach is the choice of the indepen‐ dent unknown, and we will attend to it shortly. First, we notice that the αi’s are readily available from (1): αi=π/ni (4) for i=1,...,m. Furthermore, once one of the γi’s, say γ1, is computed, all the rest are available from (3): cosa=cosα1/sinγ1 (5) γi=arcsin(cosαi/cosa) (6) for i=2,...,m. ─────────── in the sense of [GMS] (cf. [GS], section 12.3), and the fundamen‐ tal triangulation be viewed as a multiple tiling of the sphere by triangles, or a triangulation of a Riemann surface (cf. [GS], section 12.4)). ‐5‐ Note that the inverse sine function in (6) has seemingly two values in the range (0,π). However, it may be easily observed that at most one of the γi’s may be obtuse, and we may assume, without any loss of generality, that γi, for i=2,...,m, are all acute. Thus, (2) may be rewritten as δ=π−Σγi=0, (7) where δ is a well defined function of γ1 only. We recall Newton’s method for solving the equation f(x)=0: Choose an ini‐ tial approximation to the root and then define a sequence of closer approxima‐ tions using the recursion ^x=x−f(x)/f′(x) (8) Assume initially that the fundamental triangles are approximately planar. This motivates the initial approximation γi=π/2−αi. for i=1,...,m. This certainly satisfies (5) and (6) (with cosa=1), but as these values for the γi’s are smaller than in reality (as the sum of the angles in a spher‐ ical triangle exceeds π), δ in (7) will not vanish, but rather be a positive number, which motivates naming it the excess function. The iterative proce‐ dure will converge if we are able to make the excess arbitrarily small. It may be shown, at least empirically, that in order to guarantee convergence it is sometimes necessary to choose γ1 as the biggest of the initial values, i.e., choose α1 to be the smallest of the αi’s, or, in view of (4), choose n1 to be the biggest of the ni’s. This choice is the obvious the case for those polyhedra for which γ1 turn to be obtuse, in view of our 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 any other choice. We’ll see why it happens once we developed the recursion formula. To use (8), we have to differentiate the excess function; we have from (7), −dδ/dγ1=1+Σ′dγi/dγ1 (9) (with the prime denoting summation over the range 2,...,m). However, in view of (5) and (6), sinγi=sinγ1cosαi/cosα1 which in turn implies, using logarithmic differentiation, ‐6‐ dγi/dγ1=tanγi/tanγ1. (10) Combining (7), (9) and (10) we get −δ/δ′=δtanγ1/Σtanγi (with summation again over the full range 1,...,m, and δ′ is the derivative in (9)). Finally, we arrive at the required recursion for γ1: ^γ1=γ1+δtanγ1/Σtanγi In practice, we notice that the faces meeting at a vertex are not neces‐ sarily different, thus we may group similar faces together, and adjust the above formulas to reflect the fact that at each vertex the polyhedron can have mi faces each of which is a {ni}, with Σmi=m. Furthermore, we handle retro‐ grade faces by orienting the vertex figure so that the the first, thus big‐ gest, face is not retrograde. Thus, ni is a positive rational, and mi is a positive integer. d, the vertex density, is a non‐negative integer. To summarize, fundamental system niαi=π Σmiγi=πd cosαi/sinγi=cosa where n1=maxni, has the following iterative solution algorithm: Step 1. Choose the initial approximation γi=π/2−αi, where αi=π/ni. Step 2. Compute the excess δ=πd−Σmiγi. If it is numerically small enough, fin‐ ish. Otherwise, continue to step 3. Step 3. Compute the next approximation to the independent unknown γ1 using the recursion ^γ1=γ1+δtanγ1/Σmitanγi Step 4. Compute cosa and γi, i=2,...,m, using cosa=cosα1/sinγ1 ‐7‐ γi=arcsin(cosαi/cosa) Step 5. Return to step 2. We want to make an observation about the convergence of this algorithm. The increment in γ1 in step 3 may be very small even without the excess δ being small. This happens if the denominator Σmitanγi is large, unless the numerator tanγ1 behaves similarly. In such cases the algorithm converges slowly or even diverges, and our choice for the independent unknown prevent them from occurring. By the way, choosing cosa as the independent unknown creates similar problems and although it looks plausible and perhaps more ele‐ gant (compare with the well known Lagrange multipliers ), it should be avoided. A mathematically rigorous proof of convergence is out of the scope of this paper; We will have to be satisfied by empirical convergence for all the finitely many uniform polyhedra (cf. [CLM], [S]). Note that in few exceptional cases, the direct approach fails. In these cases, it happens that the vertex density d is zero, and the faces incident at a vertex occur in several oppositely oriented pairs, and thus even the initial approximation satisfies the fundamental system, which is obviously impossible, since, as we already remarked, αi+γi>π/2 in a right‐angled spherical triangle. It is interesting to note that these polyhedra are exactly those which cannot be constructed directly using the Wythoff construction, to be presented in the section 4. In section 5 we present an adjustment to the construction, which in turn leads to an adjustment to the fundamental system, which makes it pos‐ sible to handle these exceptions. 4. The Kaleidoscope. We start by discussing the dihedral kaleidoscope. It consists of two hinged planar mirrors. The reflections in these mirrors generate a group of isometries, which is finite if the dihedral angle of the kaleidoscope is a rational submultiple of π, i.e., of the form π/q, where q is a rational greater than one. We now 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 reflections of this arc in the mirrors produce a finite covering of the circle. It is easy to see that a d‐fold covering obtained if q=n/d, where n and d are relatively prime positive integers, with n>d. In this case the the kaleidoscope generates Dn, the dihe‐ dral group of order 2n, and the π/q=πd/n arc consists of d copies of the fun‐ damental domain of the group, a π/n arc. 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. ‐8‐ Note, that in two cases 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 dis‐ carded. Thus, when the point object is located on one of the mirrors, we get regular n‐gon, with density d, customarily denoted by {n/d}, i. e., {q}. Note, that if {q} and {q′} satisfy 1/q+1/q′=1, they are two oppositely ori‐ ented copies of the same 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 no degenerate edges, but rather a regular n‐gon which is transversed twice. As the density of the compound is d, d has to be even and each polygon has density d/2. Thus, when the point object is located on the bisector, we get a double {n/(d/2)}, i.e., a double {2q}. It is easy to check that if the point object is similarly located but the 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, however if we use only rotations (or any even number 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 con‐ sists 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 however this is not sufficient. Intersecting the kaleidoscope by a sphere centered in its corner, Schwarz formulated the following problem: Enumerate all convex spherical triangles PQR, with angles π/p, π/q and π/r (p,q,r are all rational greater than one), with the property that when reflected in its sides any number of times, PQR produces a finite covering of the sphere. Such a triangle is called a Schwarz triangle and is denoted by the symbol (pqr). It may be shown that unless two of the angles are π/2 (in which case the kaleidoscope generates the dihedral symmetry group), p, q, and r may have as numerators 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 nondihedral Schwarz triangles finite. A complete 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, it is now reflected repeatedly to produce a polyhedron, whose vertices are the images of the point object. As before, two vertices are considered adjacent if they belong to tri‐ angles which share an edge, and several vertices are incident to the same face if they belong to triangles which share a vertex. Note that as in the previ‐ ous 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 any detail (but see however section 5). ‐9‐ 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, Wythoff construction is very useful in getting the topological invariants of the traced polyhedra, that is, the covering den‐ sity D and the Euler characteristic χ. Both values can be easily 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: 4πD=g(π/p+π/q+π/r−π), or, D=g/4(1/p+1/q+1/r−1). Note that the value of g is 4n , 24, 48 or 120 if the group is n‐fold dihe‐ dral, tetrahedral, octahedral or icosahedral, respectively. As pointed out in [CLM] (p. 412, 418, 425), this is also the correct value for the density of any polyhedron derivable from (pqr), except for the cases where the spherical polyhedron possesses 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 spheri‐ cal and planar faces occur on two different sides of the center, and the den‐ sity 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 any map drawn on a given closed sur‐ face by the famous Euler formula V−E+F=χ. Here, we are considering a D‐ sheeted Riemann surface, which is triangulated by the Schwarz triangles, so that F=g, E=3g/2, and V=Vp+Vq+Vr, where Vq is the number of vertices with the angle π/q, i.e., Vq=g/2nq, where nq is the numerator of q. Thus, we get the formula χ=g/2(1/np+1/nq+1/nr−1) This value is also inherited by any non‐degenerate polyhedron derivable from (pqr) (see the tables in appendix 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’s icosahedron and the star‐cornered Poinsot’s great icosahedron, derivable from (235) and (235/2), have both Euler characteristic 2, and are thus homeomorphic as closed surfaces, although their densities are much different (1 for the former, 7 for the latter). It is clear, that wherever we put a point C in the Schwarz triangle PQR, the traced polyhedron will be isogonal, i. e., all its vertices will be equiv‐ alent under the action of the isometry group generated by the three reflec‐ tions. We will now discuss the positions of C which guarantee that the traced faces will be regular polygons, and study the corresponding spherical tiling and the fundamental triangulation and its triangular prototiles AiBiC. As before, A1A2... will be the prototile of the dual tiling, and B1B2... the ‐10‐ vertex figure. 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. The resulting fundamental equations are, setting p=n/d: qα1=rα2=π, nγ1+nγ2=πd. By reflecting PQR repeatedly in its sides, we get the sequences of vertices A1A2A3... and B1B2B3..., which repeat themselves after 2n terms. The resulting polyhedron is denoted by the Wythoff symbol p|qr. Its ver‐ tex configuration is (q.r.q.r.....q.r) (2n terms), and its vertex density is d. The vertex figure 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: qα1=2rα2=pα3=π, γ1+2γ2+γ3=π. By reflecting A2, B1 and B2 in QR, we get the images A4, B4 and B3, respec‐ tively. The resulting polyhedron is denoted by the Wythoff symbol pq|r, and its vertex configuration is (p.2r.q.2r). The vertex figure and the dual prototile are quadrilaterals. 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 perpendiculars from C on PQ, QR and RP, respectively. The resulting fundamental equations are: 2pα1=2qα2=2rα3=π, ‐11‐ γ1+γ2+γ3=π. The resulting polyhedron is denoted by the Wythoff symbol pqr|, and its vertex configuration is (2p.2q.2r). The vertex figure and the dual prototile are triangles. 4.4. C traces a snub polyhedron. In the previous positions, the symmetry group of the polyhedron is gener‐ ated by reflections. In particular, the polyhedron is reflexible. We now dis‐ cuss a fourth position, which usually traces a chiral polyhedron, i.e., a polyhedron with a symmetry group which consists of rotations only3. This is done by considering only images under an even number 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 con‐ struction. Let O be the Fermat point of the Schwarz triangle PQR, i.e, the point where the sides subtend equal angles (assuming it exists4). Reflect O in the sides and get an hexagon A1A2...A6 with angles 2π/3, 2π/p, 2π/3, 2π/q, 2π/3 and 2π/r respectively, where odd indices refer to the images of O, and even indices refer to the vertices P, Q and R. Reflecting this hexagon any even number of times, we get an isohedral, vertically regular tiling of the sphere, i.e., the dual of a uniform tiling. 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 vertex of these six polygons (since the snub {3}’s are congruent, it is located at the circumcenter of A1A3A5), and let Bi, for i=1,2,...,6, be the foot of the perpendicular from C to AiAi+1. The resulting fundamental equations are: 3α1=pα2=qα4=rα6=π, 3γ1+γ2+γ4+γ6=π. Here, we counted one snub {3} thrice, because the snub {3}s are all congruent. The resulting snub polyhedron is denoted by the Wythoff symbol |pqr, and its vertex configuration is (3.p.3.q.3.r). The vertex figure and the dual pro‐ totile are hexagons. Note that in all the cases, the occurrence of α=π/2 implies that the cor‐ responding tile is a {2}, i.e., a digon, and may be discarded. ─────────── 3 However, (5/233) and (3/23/25/2) being isosceles, they yield 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 (3/23/25/2) and (3/25/32)). ‐12‐ 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 two faces are opposite. It is easy to verify, that the Wythoff construction always yields orientable polyhedra, assuming that vertices which belong to nonadjacent Schwarz triangles are considered distinct, even if they coincide. The number of the polyhedron vertices may be readily found by dividing the order of the kaleidoscope symmetry group by the number of copies of adjacent Schwarz triangles which share a vertex, i. e., by two for pq|r or |pqr, and by n for n/d|qr. We will now discuss two particular cases where the Wythoff construction yields only pairs of coinciding vertices, that, when identified, are the ver‐ tices of a nonorientable, or one‐sided, polyhedron. In these cases, the num‐ ber 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 kaleidoscope with angle π/q, q rational with an even denominator, yields a double {2q}. For conclusion, we shall have a brief, independent discussion about the solution for the only non‐Wythoffian polyhedron in existence. 5.1. Polyhedra pqr| with an even denominator. Consider a Schwarz triangle PQR, with an angle π/r at R, r rational with an even denominator. Suppose the point object C is located in the incenter of PQR. In particular, it lies on the bisector atR. As we have seen, the images of C trace a {2p}, a {2q} and a {2r}, with centers at P, Q and R, respec‐ tively. However, since r has an even denominator, we find that the face with center at R, which is traced by reflections at RP and RQ only, is a double {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 two similar faces with centers at the images of P and Q. Thus, in addition to the double {2r}, we have at C two {p}’s and 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 nonorientability is due to orientation reversal: Trying to find a coherent orientation on the faces and considering the four faces inci‐ dent at a vertex, we notice that two of them has to be retrograde, i.e., the vertex configuration is (p.q.p′.q′). However, when considering the faces adjacent to a discarded double {2r}, none of the four faces incident at a ver‐ tex may 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, by taking the {2r}’s into account, and solving the equations set in section 4.3, the metrical properties of these exceptional polyhedra may be still computed. 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 ‐13‐ χ=g/2(1/np+1/nq−1). The density is not‐well define, as these polyhedra do not produce an even cov‐ ering of the sphere. 5.2. Hemi polyhedra pp′|r. Consider a Schwarz triangle PQR, where the angles π/p and π/q are comple‐ mentary, and assume that q=p′<2