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Envelope Polyhedra
J. Richard Gott, III Department of Astrophysical SciencesPrinceton UniversityPrinceton, NJ 08544, USA
ABSTRACTThis paper presents an additional class of regular polyhedra—envelope polyhedra—made of regularpolygons, where the arrangement of polygons (creating a single surface) around each vertex is identical;but dihedral angles between faces need not be identical, and some of the dihedral angles are 0 ◦ (i.e.,some polygons are placed back to back). For example, squares–6 around a point { } is produced bydeleting the triangles from the rhombicuboctahedron, creating a hollow polyhedron of genus 7 withtriangular holes connecting 18 interior and 18 exterior square faces. An empty cube missing its topand bottom faces becomes an envelope polyhedron, squares–4 around a point { } with a toroidaltopology. This definition leads to many interesting finite and infinite multiply connected regularpolygon networks, including one infinite network with squares–14 around a point { } and anotherwith triangles–18 around a point { } . These are introduced just over 50 years after my relatedpaper on infinite spongelike pseudopolyhedra in American Mathematical Monthly (Gott, 1967). Keywords: geometry, polyhedra INTRODUCTION—PSEUDOPOLYHEDRAMy work on envelope polyhedra grows directly out of my earlier work on pseudopolyhedra, which I will describe first.This was my high school science fair project which won 1st Place in mathematics at the (May, 1965) National ScienceFair International (now the Intel International Science and Engineering Fair). A picture of this project appears in mybook
The Cosmic Web (2016), along with a description. These were infinite spongelike polyhedra whose polygons wereall regular, whose vertices were congruent, and two polygons always shared only one edge. All had a sum of polygonangles around a vertex > ◦ , and corresponded to surfaces with negative curvature.Positively curved surfaces like the sphere can be approximated by regular polyhedra where the sum of angles ata vertex is < ◦ . The cube is a very rough approximation to a sphere. A cube is made up of squares meeting 3around a point. At the corner of a cube, three square faces meet at a vertex, and each square has a 90 ◦ angle atits corner, making the sum of the angles around a point at the corner 3 × ◦ , or 270 ◦ . This is 90 ◦ less than 360 ◦ .All 5 regular Platonic polyhedra: tetrahedron ( triangles–3 around a point ), octahedron ( triangles–4 around a point ),icosahedron ( triangles–5 around a point ), cube ( squares–3 around a point ), and dodecahedron ( pentagons–3 around apoint ) have a sum of angles around a vertex of < ◦ . A plane can be tiled by squares–4 around a point to makea checkerboard pattern, where the sum of angles around each point is therefore 4 × ◦ , or 360 ◦ degrees—this is asurface of zero curvature. Johannes Kepler recognized that the three long-known planar networks, triangles–6 arounda point , squares–4 around a point , and hexagons–3 around a point , were also regular polyhedra but with an infinitenumber of faces. (Kepler in addition allowed regular star polygons crossing through each other to count, creatingregular stellated polyhedra. With slightly more lenient rules you can find additional interesting structures.)In early 1965, found 7 regular spongelike polygon networks having a sum of angles around a point > ◦ , withan infinite number of faces and an infinite number of holes: triangles–8 around a point , triangles–10 around a point , squares–5 around a point , squares–6 around a point , pentagons–5 around a point , hexagons–4 around a point , and Corresponding author: J. Richard Gott, [email protected] a r X i v : . [ m a t h . M G ] A ug Gott
Figure 1.
Regular Pseudopolyhedrons from Gott (1967). hexagons–6 around a point . I called these pseudopolyhedrons (Fig. 1), after the pseudosphere which is a surface ofconstant negative curvature encountered in the non-Euclidean geometry of Nikolai Lobachevsky and Janos Bolyai.After I got to Harvard, math instructor Tom Banchoff, later famous for his professional friendship with SalvadorDali, encouraged me to submit a paper on pseudopolyhedra to the American Mathematical Monthly, which I did.The referees report was quite positive, but noted that three of my polygon networks had been discovered before. Thereference was to a paper, which I had never heard of, by H. S. M. Coxeter (1937). It described how the first of thesefigures— squares–6 around a point —was found in 1926 by John Petrie, who also discovered hexagons–4 around a point (the one I found first when I was 18). Petrie gets credit for discovering this entire class of figures. Coxeter himselfdiscovered hexagons–6 around a point . Petrie and Coxeter did this work in 1926 when both were 19. In addition todemanding that the configuration of polygons around each vertex be identical (as I did), their criteria for regularityalso demanded that the dihedral angles between all adjacent pairs of faces also be equal and the vertex have rotationalsymmetry. With those conditions they were able to prove that the three examples they found were the only regularfigures of this type. They called them regular skew polyhedra. I was happy to add the Coxeter/Petrie reference. Mypaper was still publishable, the referee said, because I had discovered four new pseudopolyhedrons. I still required theconfiguration of polygons around each vertex to be identical, but allowed the dihedral angles between adjacent facesto vary. Some had some dihedral angles of 180 ◦ , for example. I rediscovered all three structures discovered by Petrieand Coxeter as well as finding four new ones allowed by my more lenient rules. My paper appeared in print in 1967. nvelope Polyhedra Penguin Dictionary of Curious and Interesting Geometry (Wells, 1991). In my paper Iused an anglicized plural form, pseudopolyhedrons, thinking this would be modern, but the Latin plural has remainedhealthy over the past half century, so I will surrender and here refer to them in the plural as pseudopolyhedra. Theseare today sometimes also called infinite polyhedra, spongelike polyhedra, or infinite skew polyhedra. These have aninfinite number of faces and therefore belong to the set of apeirohedra (along with the regular planar networks andcylindrical networks).When Siobhan Roberts wrote her definitive biography of Coxeter,
King of Infinite Space , in 2006, I was happyto contribute my story of the astronomical applications these figures later had in understanding the distribution ofgalaxies in space. In the early 1980s there were two schools of thought about how galaxies were clustered in space.The American school, headed by Jim Peebles, maintained that there was a hierarchical pattern of clusters of galaxiesfloating in a low density void, like isolated meatballs in a low density soup. The Soviet school, headed by YakovZeldovich, maintained that galaxies formed on a giant honeycomb punctuated by isolated voids. I realized neithermodel was consistent with the new theory of inflation, which showed that fluctuations in density in the early universewere produced by random quantum fluctuations. In this case, the regions of above-average and below-average densityshould have the same topology. This could occur with a spongelike topology, which divided space into two equivalentparts. I knew this because it occurred in some of my pseudopolyhedra ( triangles–10 around a point , pentagons-5around a point , squares–6 around a point , hexagons–4 around a point , hexagons–6 around a point ). We showed (Gott,Melott, and Dickinson 1986) that the spongelike initial conditions required by inflation would grow under the influenceof gravity into a spongelike structure of galaxy clusters connected by filaments of galaxies, with low density voidsconnected by tunnels, a structure now known as the cosmic web and verified by many surveys. I tell the story of thisdiscovery in my book The Cosmic Web (Gott, 2016). I gave an invited lecture on this at the Royal Institution whichcan be seen on YouTube.Additional regular pseudopolyhedra have been discovered by crystallographer A. F. Wells: including triangles–7around a point ; triangles–9 around a point ; and triangles–12 around a point . These are illustrated in Wells’s 1969paper and his 1977 book, Three Dimensional Nets and Polyhedra . Wells, like me, did not demand equal dihedralangles between adjacent faces. All are spongelike with an infinite number of faces and an infinite number of holes.Melinda Green rediscovered my pentagons–5 around a point , and has illustrated many pseudopolyhedra (see references).Avraham Wachmann, Michael Burt, and Menachem Kleinman (abbreviated WBK) have discovered many semi-regularspongelike polyhedra, composed of polygons of more than one kind, for example, two squares and two hexagonsaround each point. (But they failed to find pentagons–5 around a point .) These are to the Petrie/Coxeter/Gott/Wellspseudopolyhedra as the Archimedean polyhedra are to the 5 classic Platonic polyhedra and are illustrated in their1974 book
Infinite Polyhedra . WBK also allow different dihedral angles between faces as I and Wells did, and inaddition allowed networks to contain pairs of mirror vertices, ones where the arrangement of polygons was congruentonly under mirror reflection. Envelope polyhedra containing such mirror vertices will be discussed in the second halfof the Appendix. FINDING ENVELOPE POLYHEDRA—DIHEDRAI thought of envelope polyhedra in 1991, while visiting Aspen, Colorado to attend a seminar on cosmology, to talkabout my two-moving cosmic string solution in general relativity which allowed time travel to the past. I was thinkingabout the classic regular polyhedra as approximations to a sphere. For this reason, there are polyhedral maps of theEarth. Perhaps the most famous and successful is the Gnomonic Cahill Butterfly map. It maps the Earth onto aregular octahedron. Then one unfolds the 8 triangular faces in a butterfly pattern sitting on a plane. It shows relativelylow distortion but has a number of “interruptions.” Another successful polyhedron map was invented by the famousarchitect Buckminster Fuller, the inventor of the geodesic dome. Fuller mapped the Earth onto a 20-sided icosahedron.He unfolded the 20 triangular faces to sit on the plane. I remembered that there was also a conformal projectioninvented by Emile Guyou in 1887, which maps the two hemispheres of the Earth onto two squares sitting side by side.If one folds the two squares together as one folds a billfold closed, and seals them one would create an envelope withthe Western Hemisphere mapped on the front and the Eastern Hemisphere on the back. This envelope has two squarefaces taped together along their edges back to back. I realized this is also a polyhedron. It has 2 square faces (the frontand back of the envelope), 4 edges (which form the edges of the envelope) and 4 vertices (which form the four cornersof the envelope. This obeys Euler’s rules for convex polyhedra, namely that the number of faces minus the number of
Gott edges plus the number of vertices equals 2: F − E + V = 2. This envelope has F = 2, E = 4, V = 4, so 2 − ◦ . Only 2 faces cometogether at a vertex: the front of the envelope and the back. This is a polyhedron we would designate as squares–2around a point . This has a Schl¨afli symbol { } and in the WBK nomenclature would be designated 4 . Imagine anant tethered to a vertex (one of the four corners) with a tiny string. It stays at a constant distance from the vertex asit circles it. An ant crawling around this vertex would traverse an angle of 90 ◦ on the front square, then go over anedge and start crawling on the back square through another 90 ◦ . So the total angle at the vertex is 90 ◦ + 90 ◦ = 180 ◦ .This is 180 ◦ less than we would get circling a point on a plane which is 360 ◦ , giving an angle deficit for this vertex of180 ◦ . There are 4 such vertices or corners, and so the total angle deficit is 4 × ◦ = 720 ◦ , just as Descartes wouldhave figured. This is of course a polyhedron with zero volume, which is why the ancients did not count it. In general,one has envelope polyhedra which are N -gons–2 around a point , for all N ≥
3. Each has 2 Faces: triangles–2 arounda point , squares–2 around a point , pentagons–2 around a point , hexagons–2 around a point , and so forth.Years later, I found that this was just a rediscovery on my part of dihedra which have already been accepted aspolyhedra for some time. (Coxeter, 1937) mentioned dihedra, for example (but did not include them in his lists ofregular polyhedra). They all have the topology of a sphere. (One can therefore make a conformal map of the Earthon two hexagons: one covering the northern hemisphere, one covering the southern hemisphere. This answers in theaffirmative the gamers perennial question: can the sphere be tessellated with identical geodesic hexagons? Yes, withtwo. Each has six geodesic sides, six 60 ◦ geodesic arcs, along the equator.) There are an infinite number of envelopepolyhedra with 2 faces, N edges, and N vertices. These all satisfy the F − E + V = 2 rule for convex polyhedra, inthe most transparent way possible. They also satisfy the Coxeter-Petrie condition that all dihedral angles be equal (inthis case 0 ◦ ) and that the vertex figure should have rotational symmetry (in this case n = 2). These are not new. Butthey are just a subset of the larger class of envelope polyhedra that are the subject of this paper. From my work onpseudopolyhedra I already knew in 1991 that if one allowed dihedral angles of 0 ◦ , there would be many new envelopepolyhedra of zero and negative curvature, both finite and infinite with 360 ◦ around a vertex and > ◦ around avertex. And this would make for many additional interesting structures. I started adding these to my list of envelopepolyhedra. If dihedra can have dihedral angles between their two faces of 0 ◦ , then this should be allowed for dihedralangles in general. A WEALTH OF ENVELOPE POLYHEDRAI was used to polyhedra approximating surfaces that divided space into two regions: the inside and the outside in thecase of finite polyhedra, into two regions in the case of the plane tessellations: above the plane and below the plane,and into two spongelike interlocking regions in the case of the pseudopolyhedra. Envelope polyhedra, do not dividespace into two parts. They have some dihedral angles (angles between two faces) that are 0 ◦ . But they do representa surface an ant could crawl over. I don’t allow the surface to cross itself (I am not considering starred polyhedra).A regular envelope polyhedron has faces that are regular polygons, and the arrangement of polygons touching eachvertex must be identical. Some of the angles between adjacent faces in envelope polyhedra will be 0 ◦ —those faces willbe back to back as in an envelope. This leads to a wealth of forms. FINITE ENVELOPE POLYHEDRA WITH HOLESSee Fig. 2, squares–6 around a point . This and following photographs are stereo pairs with the left-eye view on theleft and the right-eye view on the right. Place your nose on the page and the left eye view will be in front of your lefteye and the right eye view will be in front of your right eye. You will see a blurry 3D view, slowly back away from thepage and the fused central image will come into clear focus, with side images to the left and right. These may also beviewed with a standard stereograph viewer. The models in Figs. 2–5 are made from Polydron plastic polygons (frompolydron.com), ignore their slightly serrated edges which allow them to be hinged together to make polyhedra.
Squares–6 around a point , as illustrated in Fig. 2 starts with a rhombicuboctahedron with 18 square faces and 8triangular faces. Remove the triangular faces to create triangular holes into the interior. This leaves an envelopepolyhedron that is hollow with 18 exterior square faces and 18 interior square faces. The polygonal faces thus partiallyenvelop a hollow void. Thus, the name envelope polyhedron seems appropriate here too. Each vertex connects 3exterior and 3 interior squares. An ant crawling around a vertex would visit all 6 of these squares. [The pseudorhom-bicuboctahedron, a Johnson polyhedron, where the top in Fig. 2 is twisted relative to the bottom by 45 ◦ providesan alternate way to start the construction and gives an alternate form (2)]. All vertices are congruent with a sum of nvelope Polyhedra Figure 2.
The finite envelope polyhedron, squares–6 around a point . angles of 6 × ◦ > ◦ around each, so this approximates a negatively curved surface. It has 36 faces (18 exteriorplus 18 interior faces), 6 × × × × F − E + V = 2 × (1 − genus) = 36 −
72 + 24 = −
12. This gives it a genus of 7 and a topology equal to that of a spherewith 7 handles. It has negative curvature and so could be considered a finite envelope pseudopolyhedron. Note thatthe faces around each vertex must create a single surface around each vertex that an ant tethered to the vertex couldvisit. Thus two cubes touching at a point would not be considered squares–6 around a point because an ant tetheredto the vertex would circle the vertex visiting 3 squares on one cube and complete her circuit and return to where shestarted visiting only those 3 squares and never visit the other cube. Likewise, a cube with a square fin attached at anedge would not count as part of a squares–5 around a point structure because although one ant tethered at the vertexwould visit 5 outside square faces, another ant tethered on the inside of the cube would visit only 3 squares, againcreating multiple surfaces with different angle deficits at a single vertex. We are not allowing such multiple surfaces.The envelope polyhedra is one continuous two-dimensional surface.Here is another interesting example: octagons–4 around a point . Get an empty cubical cardboard box. Cut off eachof its corners with a saw. This will make a truncated cube. Each of the 6 square faces of the cube will have its cornerscut off, becoming an octagon. Cutting the corners of the cubical box will create 8 triangular holes where the 8 cornersof the box used to be. This has 12 octagonal faces (6 inside, 6 outside), 48 edges (12 outside edges, 12 inside edges,and 24 edges on the 8 triangular holes connecting the inside and outside), and 24 vertices (3 on each of the triangularholes). F − E + V = 12 −
48 + 24 = −
12. Thats 2 × (1 − genus) as predicted. This has a genus of 7. It has the samenumber of triangular openings as the squares–6 around a point envelope polyhedron in Fig. 2 so it also has a genus of7. Some dihedral angles between adjacent octagons are 90 ◦ (when both are either outside or both inside) and someare 0 ◦ (when one is inside and one is outside).An ant tethered to one of the vertices will circle it by crawling over two outside-facing octagons and two inside-facingoctagons as it goes around the outside and then through the triangular hole to visit the inside, making 4 octagonsaround each vertex. The interior angle in an octagon is 135 ◦ . So the total angle the ant traverses circling the vertex is4 × ◦ = 540 ◦ > ◦ . This is a negatively curved surface with the curvature all concentrated at the vertices. It evenlooks like a saddle-shaped surface. One can imagine an ant sitting on one of the vertices and draping its little hind legsover each side, one inside and one outside, like he was riding a horse. This negatively curved finite polyhedron, like Gott
Figure 3.
Envelope polyhedron, hexagons–4 around a point . Figure 4.
Finite envelope polyhedra: Front, squares–2 around a point ; Back row, left to right, squares–4 around a point , triangles–6 around a point , squares–4 around a point , triangles-8 around a point . squares–6 around a point in Fig. 2, is multiply connected. But all of my original multiply connected pseudopolyhedrawere infinite. Hexagons–4 around a point . Take an octahedron shaped box and saw off its corners. You will be left with a truncatedoctahedron shape made of hexagons with 6 square holes. See Fig. 3. An ant circling a vertex will traverse two hexagonson the exterior, before visiting another two hexagons on the interior to make hexagons–4 around a point .Take a triangular prism and remove the top and bottom triangles, to create the envelope polyhedron squares–4 arounda point . Take a regular octahedron and remove two opposite faces, to create the envelope polyhedron triangles–6 arounda point . Take a cube and remove two opposite faces to create the envelope polyhedron squares–4 around a point . Thesehave the topology of a doughnut, and 360 ◦ and zero curvature at the vertices. Take an icosahedron box and removefour triangles to create the envelope polyhedron triangles–8 around a point . These are illustrated in Fig. 4. INFINITE ENVELOPE POLYHEDRA
Squares–10 around a point . The hollow rhombicuboctahedron envelope polyhedron shown in Fig. 2 fits perfectly inan imaginary cube. Stack cubes like this to fill space with the rhombicuboctahedra glued back to back at a singlesquare (red in Fig. 2) in a repeating pattern. Each vertex connects two of these rhombicuboctahedra, and as an antcircles the vertex, it travels on an exterior square face of the first, then three interior faces of the first, then anotherexterior face of the first, an exterior face of the second, three interior faces of the second, and another exterior face ofthe second—giving 10 squares around the vertex. nvelope Polyhedra Figure 5.
Squares–14 around a point . Octagons–8 around a point . Take an infinite number of the truncated cubical cardboard boxes ( octagons–4 arounda point ) discussed above and fill all of space with them by stacking them like cubical boxes in a warehouse. We will begluing the boxes together. Two adjacent boxes will be glued together on their outside octagons. These will disappearfrom the surface. All that will be left are the inside octagons of each cube. Four boxes (numbered 1, 2, 3, 4) will fittogether at their edges. A vertex will connect 8 octagons. The vertex is at the end of an edge that connects two insideoctagons of cube 1. Cubes 2, 3, and 4 also come together at this vertex. An ant tethered to the vertex will circle thevertex by first traversing the two inside octagons of cube 1, then it will cross an edge of a triangular hole, to entercube 2 and traverse two of its inside octagons, then enter cube 3 traversing two of its interior octagons, before enteringcube 4 and traversing two of its interior octagons as she returns to the place she started. This makes a complicatedsaddle shaped surface that goes up and down, up and down, up and down, and up and down. It is a saddle with afour-fold symmetry, like a +. It is a saddle a horse could sit on, hanging its legs down, one each into boxes 1, 2, 3,4. It is cradled by the four triangular holes. The angle around each vertex is 8 × ◦ = 1080 ◦ . This is an infinitepolyhedron: it has an infinite number of faces, an infinite number of cubical cells with triangular holes. A fly couldfly through the whole structure, visiting any cell he wanted.(Note: Pellicer and Schulte (2009) and Schulte and Weiss (2016) considered skeletal polyhedra. These had skeletalfaces consisting only of edges, based on the idea of Gr¨unbaum (1994) that one could consider polyhedra consistingof polygons with edges only without membranes spanning them to form a surface. These form additional polyhedra,having skew polygons, and even infinite helical polygons. In these polyhedra skeletal polygons come together twoat an edge. But Pellicer and Shulte (2009) also consider regular polygonal complexes where more than two skeletalfaces meet at an edge. If we removed all the faces from octagons–8 around a point , we would have 4 skeletal octagonsmeeting at an edge. But to be clear, we are not considering such skeletal structures here—envelope polyhedra allhave surfaces. In octagons–8 around a point we have 8 octagons and 8 edges meeting at a point. In the surface, twooctagons always share an edge. Edges can be placed back to back, of course, in the structure, just as polygons canbe placed back to back. These constitute, in the surface, separate polygons and separate edges which an ant wouldencounter circling the vertex.) Triangles–18 around a point . (1) Start with the pseudopolyhedron triangles–10 around a point shown in Fig. 1,then remove triangles in the planes of triangles that are connected to other triangles in the same plane (with dihedralangles of 180 ◦ ) along each of their three edges. This creates triangular holes in the planes of triangles. Each vertexused to have 10 triangles around it on one convoluted surface; take one triangle away and that leaves 9 triangles, butthe hole allows the ant to traverse to the opposite side of the plane and visit the 9 opposite sides of these triangles tomake a total of 18 around a point. The angle around each vertex is 18 × ◦ = 1080 ◦ . There are left- and right-handedversions of this. Squares–12 around a point . Make a single layer of triangular prisms that cover a plane. Delete the triangular topsand bottoms of the prisms, leaving a triangular grid of cells. Six triangular prisms meet at a vertex, and each has twointerior squares that touch the vertex, giving 12 squares around a point.
Squares–14 around a point . Take the previous structure and add east-west fins of back-to-back squares above thissingle layer of cells in a dashed pattern. See Fig. 5. This way each vertex has the 12 squares around a point that
Gott it had before plus the two squares in the fin. These fins are then connected to another plane of squares–12 arounda point triangular cell layer above it. Layers of fins and cells alternate vertically forever. Each of the 14 squares hasa vertex angle of 90 ◦ , so the total angle around a vertex is (14 × ◦ ) = 1260 ◦ , a trick for an Olympic snowboarder.That is 3 / rotations. That gives it an angle excess of 900 ◦ (above 360 ◦ ), the most found for any polyhedron so far.There are many other envelope polyhedra. The Appendix gives a list of those I have found so far and a summaryof how these additional ones are constructed. In general they are found by: (a) deleting faces from regular polyhedrato create polygonal holes (as illustrated by some of the examples in Fig. 4); (b) deleting polygons from semi-regularpolyhedra containing faces of different types (like the triangular prism in Fig. 4) and the Archimedean polyhedra tocreate polygonal holes; (c) deleting polygons from regular and semi-regular planar networks and cylindrical networks;(d) deleting polygons from regular pseudopolyhedra like those shown in Fig. 1, and from semi-regular pseudopolyhedralike those shown in WBK, and (e) Creating planar networks of open-ended prisms with fins (like Fig. 5) and withoutfins, based on regular and semi-regular (Archimedean) planar networks. SUMMARYPolyhedra fall into three groups depending on whether the sum of face angles around a vertex are < ◦ , = 360 ◦ ,or > ◦ (i.e., whether the curvature at each vertex is positive, zero, or negative. Dihedral angles do not all haveto be identical. But the arrangement of polygons around each vertex must be identical. Envelope Polyhedra, whichwe introduce here, just have some dihedral angles which are 0 ◦ . Using the nomenclature of WBK, where N -gons– M around a point with Schl¨afli symbol { N, M } are designated N M , we find the following structures so far: < ◦ , 3 , 3 , 4 ,5 (The classical Platonic Polyhedra) N where N ≥ ◦ , 4 , 6 (Infinite Planar and/or Cylindrical Tessellations)3 , 4 (Finite Envelope Polyhedra with Toroidal Geometries)3 , 4 (Infinite Envelope Polyhedra with Filmstrip Geometries) > ◦ , 3 , 3 , 3 , 3 , 4 , 4 , 5 , 6 , 6 (Infinite Pseudopolyhedra/Skew Polyhedra)3 , 3 , 4 , 6 , 8 , 10 (Finite Envelope Pseudopolyhedra)3 , 3 , 3 , 3 , 4 , 4 , 4 , 4 , 6 , 8 , 8 (Infinite Envelope Pseudopolyhedra)These are summarized in Table 1.A number of these have several geometrical forms as described in the Appendix. Open symbols refer to envelopepolyhedra which have some dihedral angles equal to 0 ◦ : the symbol (cid:5) indicates a finite number of sides and refersto polyhedra with a finite number of faces, ◦ is a circle and refers to envelope polyhedra with an infinite number offaces. Closed symbols refer to Polyhedra: + (which are finite), and • which are either infinite Planar or CylindricalNetworks, or infinite Pseudopolyhedra which have no dihedral angles equal to 0 ◦ . ENVELOPE POLYHEDRA WITH MIRROR VERTICESBelow, and in the Appendix, we also list additional infinite structures (3 , 4 , 4 , 4 , 6 ) with mirror vertices.All the previous polyhedra we have discussed have vertices that are identical. But one may obtain a considerablenumber of additional structures by allowing vertices and their mirror images to be considered as congruent. Oneexample starts with the pseudopolyhedron hexagons–4 around a point shown in Fig. 1. Now remove the same set oftwo opposite faces on each truncated octahedron box. You get a repeating pattern of hexagons–6 around a point : 6 ,but with pairs of mirror vertices whose vertex figures are not identical (except under mirror reflection) which is shownin Fig. 6. Because it includes the mirror image vertices, this structure 6 is not included in Table 1 which includesonly structures with vertices that are identical. nvelope Polyhedra N (cid:5) N N N N N N N N N (cid:5) ◦ (cid:5) (cid:5) (cid:5) (cid:5) (cid:5) (cid:5)◦ (cid:5) (cid:5) • (cid:5)• • ◦ (cid:5) + 5 • (cid:5) + 4 (cid:5) • ◦ • (cid:5) • ◦ ◦ (cid:5) + 3 + 3 + 3 (cid:5) • ◦ • (cid:5)• • •◦ ◦ ◦ •◦ ◦ ◦ + = Finite Polyhedron (cid:5) = Finite Envelope Polyhedron • = Infinite Planar & Cylindrical Networks, or Pseudopolyhedra ◦ = Infinite Envelope Polyhedron Table 1.
Nomenclature follows that of WBK: For example squares–6 around a point appears as 4 . This has a Schl¨afli symbol { } . The symbols (cid:5) and ◦ in that box indicate that both finite and infinite envelope polyhedra of this type exist; • indicatesthat an infinite pseudopolyhedron of this type exists. N represents all integers N ≥ Also starting with squares–6 around a point in Fig. 1, on can delete horizontal squares that are tops and bottomsof cubes open in the east-west direction to get squares–10 around a point , which also includes mirror vertices (withmirror symmetric vertex figures).Many additional such structures with mirror vertex figures can be obtained by deleting polygons from structuresin WBK. I should mention that WBK consider mirror image vertices as congruent and many of their semi-regularpseudopolyhedra have vertices that are congruent only under mirror reflection. In plane geometry we are used toconsidering a triangle and its mirror image as congruent, but in three dimensions they are actually identical sincethey can be superimposed by rotation in the third dimension. Chemists know that in 3D, glucose comes in distinctmirror image right-handed ( D -glucose = dextrose) and left-handed ( L -glucose) forms. Dextrose can be digested bythe body to produce energy while L -glucose cannot. In 3D solid Euclidean geometry mirror image vertices cannot besuperimposed. So some might regard those with mirror image vertices as less regular. Therefore I have separated theenvelope polyhedra into two classes, those whose vertices have identical vertex figures, listed in the previous sections Figure 6.
Hexagons–6 around a point with mirror image vertices. Gott and the first part of the Appendix, and those containing mirror image vertex figures listed in this section and in aseparate section of the Appendix.No doubt there are more envelope polyhedra yet to be discovered. I leave that as a challenge for interested readers—to add to the list. Surprisingly, pseudopolyhedra eventually had an application in astrophysics. Will any of theseenvelope polyhedra eventually find applications, in chemistry, biology, art, or architecture perhaps? Time will tell.I thank Robert J. Vanderbei for 3D photography of the models, and Wesley N. Colley for help in preparation of themanuscript. APPENDIX A. ADDITIONAL ENVELOPE POLYHEDRA • Triangles–6 around a point. (1) Start with an N -gon antiprism ( N ≥
3) and remove the top and bottom N -gons.This leaves a band of triangles connecting them, like the corrugated circumference of a drum without the top andbottom. Each vertex links 3 outside triangles with 3 interior triangles. These have 2 N faces with the topology ofa doughnut. The N = 5 case is an icosahedron with its top and bottom caps of 5 triangles each deleted, leavinga ring of 10 exterior triangles and 10 interior triangles. (2) A straight filmstrip of triangles that is infinitely long.They alternate between triangles with points up and points down. Each vertex links 3 triangles on one side with3 triangles on the other side. These vertices have zero curvature and 360 ◦ around a vertex. (3) Such an infinitefilmstrip may be twisted to form a helix, there are left- and right-handed versions of these. • Triangles–8 around a point. (1) Start with a hollow snub cube, one of the Archimedean polyhedra where 4triangles and one square meet at each vertex, and delete the square faces. This leaves 32 exterior triangularfaces, 32 interior triangular faces, and 6 square holes connecting the interior and exterior. There are left andright handed versions of this. (2) Start with a hollow snub dodecahedron, another Archimedean polyhedron,and delete the pentagonal faces. This leaves 80 exterior triangular faces, 80 interior triangular faces, and 12pentagonal holes connecting the interior and exterior. This has the topology of a sphere with 11 handles. Thereare both left and right handed versions of this. • Triangles–12 around a point. (1) Start with squares–8 around a point , a single plane layer of cubical boxes withopen ends. This is a checkerboard of cubical boxes with open tops and bottoms. The vertical sides of the boxesare oriented either east-west or north-south. All the boxes have vertical edges. Distort this by leaning all thevertical edges in the northwest direction until they are at 45 ◦ to the vertical, like the diagonal of a verticaloctahedron. The boxes now have rhombus sides that can be tiled by two equilateral triangles. Each cube is nowleaning, with 4 rhombus interior sides made up of two equilateral triangles each. The rhombus sides of eachopen-ended box have angles of 120 ◦ (where two triangles meet) and 60 ◦ where there is one triangle. An antcircling a vertex on top of the layer would cover 2 interior triangles in the southeast box, 3 interior trianglesin the northeast box, 4 interior triangles in the northwest box, and 3 interior triangles in the southwest boxbefore returning to where it started, making 12 triangles around a point. Flip the layer of boxes over to see thatthe vertices on the bottom of the layer are identical to those on the top. The total angle around the vertex is720 ◦ , just as in the squares–8 around a point structure we started with. Both are topologically equivalent. Bothstructures are flexible. • Triangles–14 around a point . (1) Start with the Gott (1967) pseudopolyhedron triangles–10 around a point , andpull out for consideration two adjacent planes of triangles and the tunnels that connect them. Each vertex inthe plane of triangles will have a triangular hole in it where the column would have been to connect it to thenext plane of triangles in the triangles–10 around a point configuration. The two planes of triangles, with theirtriangular holes are connected to each other by tunnels that are antiprisms (octahedrons with two missing [topand bottom] opposite faces). For example an ant circling each vertex on the top plane would traverse 1 triangleon the bottom of that plane of triangles, 3 triangles that are part of the exterior of a tunnel, 3 more triangles onthe bottom of the top plane, then through the open triangular hole onto the top of the top plane, where it wouldtraverse 3 triangles, on the top of the plane, 3 triangles in the interior of the tunnel, 1 more triangle on the topof the plane, and it is back where it started. That gives 14 triangles around a point. There are left and right nvelope Polyhedra made of snub cubes connected at theirmissing square faces in a cubic array. Left and right handed snub cubes alternate like black and white cubes ina 3D chessboard. Left handed ones join to right handed ones at the vertices of the missing squares. Now deletefrom the array of triangles those triangles on left-handed snub cubes that touch three different missing squares.That leaves left-handed snub cubes with 8 triangular holes. Each vertex lies on the corner of a square tunnelconnecting a left and right handed snub cube. It is missing one triangle out of 8 giving its ant a triangular holeto allow it to visit both sides of the 7 remaining triangles to give a surface with 14 triangles around a point.There are left and right-handed versions of this structure (depending on whether one takes the triangles outof the left-handed, or right-handed snub cubes. But within each case the vertices are all identical—either allleft-handed or all right-handed.) • Squares–4 around a point . (1) Start with an N -gon prism (for N >
2) with an N -gon top, N square sides, andan N -gon bottom, then remove the N -gon top and bottom. This leaves an envelope polyhedron with 2N faces, Noutside square faces and N inside square faces. There are an infinite number of these which all have the topologyof a doughnut-like the sides of a drum with the top and bottom missing. Each vertex joins two outside squaresand two inside squares. Thus, the polyhedron has an angle deficit of 0 ◦ . For N = 4, this is a cube with thetop and bottom missing. (2) There is also an arrangement where one has two strips of squares back to backextending in a straight line like a filmstrip. This has the topology of an infinite cylinder. It has zero curvature.There are also arrangements where the squares line up back to back in a zig-zag patterns, like a filmstrip withfolds in it back and forth. • Squares–6 around a point. (1) Start with just the top plane of Gott’s (1967) pseudopolyhedron squares–5 arounda point (4 ). This is a single checkerboard plane punctuated by holes so that three squares on the top surfaceof the plane surround each vertex, while three more squares surround it on the bottom surface of the plane.As the ant circles the vertex it visits 6 squares giving 4 —or the holes can be in a staggered arrangement asin the front plane of squares in 4 in WBK at the bottom of page 2, or a plane of squares in 4 in WBK attop right of page 16. (2) A single honeycomb layer of hexagonal prisms tiling a planar layer, with the tops andbottoms removed. An ant circles a vertex visiting the two interior squares of each of three hexagonal cells making squares—6 around a point . This is like Fig. 5 without the fins and with hexagonal cells instead. (3) Squaresand octagons (4 × ) tile the plane, to use the nomenclature of WBK. Convert them into a layer of octagonaland square prisms (cubes) without their tops and bottoms. Each vertex is surrounded by two interior squares ofthe cube and two interior squares of each of two octagonal prisms: 6 squares around a point. (4) Triangles anddodecagons (3 × ) tile the plane. Similarly, make them into triangular and dodecagonal prisms without theirtops and bottoms. This makes a single layer with 6 squares around a point. • Squares–8 around a point. (1) Start with the semi-regular pseudopolyhedron shown by Wachman, Burt, andKleinmann (WBK) on page 9 at the top. It looks like triangular prisms joined by cubes to form a surface with4 squares and one triangle around each vertex (designated 4 × squares–8 around a point . With triangles gone there are 4 squares around eachpoint, with two sides each which the ant traverses circling the vertex. (2) On the same page at the bottom arehexagonal prisms joined by cubes in a similar manner (4 × squares–8 arounda point . (3) On page 15, is a multilayer structure (4 ). Instead of having two planes of squares connected bytunnels as in (Gott 1967), this has planes of squares connected by tunnels going down and columns going upto adjoining punctuated planes of squares. Remove selected squares on the planes of squares (those touching attheir corners 2 columns and 2 tunnels) to create an envelope polyhedron squares–8 around a point . (4) A similar4 structure on page 16 has a different arrangement of columns and tunnels. Remove selected squares (on theplanes of squares, those touching at their corners 2 columns and 1 tunnel on one plane and 1 column and 2tunnels on the next and repeat) to create an envelope polyhedron squares–8 around a point . (5) Make a singlelayer of cubes that cover a plane. From the top they look like a checkerboard. Delete the tops and bottoms ofthe cubes, to leave a checkerboard shaped grid of cells. Four cubes join at a vertex, so an ant circling the vertex,traverses two interior squares of cube 1, then two interior squares of cube 2, then two interior squares of cube3, and finally two interior squares of cube 4 giving 8 squares around a point. (6) Take the structure 4 × Gott and bottoms off. This gives squares–8 around a point . (7) There is a biplane version of the previous structurewhere the two planes are connected by triangular tunnels. (8) Take the filmstrip, triangles–6 around a point (2,listed above) and build triangular prisms on top of it. Now delete the triangular faces. This leaves an infiniterow of interlaced triangular prisms without their tops and bottoms, giving squares–8 around a point . (9) Takethe structure 4 on page 50 of WBK, Delete the squares in the planes of squares whose four corners touch twocolumns and two tunnels. (10) Take the structure 3 × on page 71 of WBK and remove the triangles to make squares–8 around a point . This structure has hollow shells like that shown in Fig. 2, missing their red squaresjoined by cubes missing two opposite squares attaching them where their two missing red squares used to be.(11) Start with the structure 3 × on page 78 of WBK, remove the triangles to produce envelope polyhedron4 . An ant at each vertex navigates both sides of 4 squares as it circles the vertex. This has a diamond structurewith a hexagonal ring of squares circling each carbon bond as an axis. (12) Start with the Wells (1977) structure4 on page 87 of WBK and delete squares which are at the center of a flat cross shaped pattern. These squareshave dihedral angles on all their four edges equal to 180 ◦ . (13) Start with the Wells (1977) structure 4 onpage 88 of WBK and delete the squares that have 4 dihedral angles of 215 ◦ (cid:48) at their edges, as indicated inWBK (square COB in their diagram). These deleted squares sit on the surfaces of truncated octahedrons, thesquares that remain lie on surfaces of hexagonal prisms. (14) Squares and octagons (4 × ) tile the plane, to usethe nomenclature of WBK. Make them into octagonal prisms and square prisms (cubes) without their tops andbottoms. Add fins (as in Fig. 5) to the tops of the octagonal prism sides where two octagonal prisms meet. Thenrepeat vertically to make alternate layers of prisms and fins. Gives 8 squares around a point. (15) Trianglesand dodecagons (3 × ) also tile the plane. Similarly, make them into a single planar layer of triangular anddodecagonal prisms without their tops and bottoms. Add fins (as in Fig. 5) to each dodecagonal prism side thatattaches to another dodecagonal prism. This likewise gives 8 squares around a point. (16) Triangles, squares,and hexagons tile the plane (3 × × × × . Turn them into prisms without their topsand bottoms to make a single planar layer. An ant circling a vertex would visit 2 interior squares of each of 4prisms, giving 8 squares around a point. • Squares–10 around a point . (1) Start with the structure 3 × on page 99 of WBK and delete the triangles.This takes envelope polyhedra squares–6 around a point (as shown in Fig. 2) and replaces the yellow squareswith cubes without ends linking the envelope polyhedron to similar copies of itself. The envelope polyhedron squares–6 around a point shown in Fig. 2 fits in an imaginary cube, the added cubes without ends link it alongthe imaginary cubes 12 diagonals to similar envelope polyhedra in nearby cubes. Circling a vertex the ant willtraverse an outside red square of Fig. 2, two outside squares of a cube without ends, it will then go throughthe triangular hole, visit the two inside squares of that same cube without ends, then the inside red square, theinside two squares of another cube without ends, before coming out of the triangular tunnel and traversing thecorresponding two outside squares of that cube without ends before returning to where it started: 10 squaresaround the vertex. (2) Start with a single layer of cubes tiling a plane, remove the tops and bottoms of the cubes,then add fins to every other east-west square (similar to Fig. 5). Then repeat vertically to create alternatinglayers of planes of open cubes and fins. An ant crawling around each vertex will visit two interior squares of eachof 4 cubes joining each other, plus the two sides of one fin, giving squares–10 around a point . (3) Triangles andhexagons tile the plane (3 × squares–10around a point . (4) Triangles and squares tile the plane (3 × ). Turn these into prisms without tops andbottoms to make a single layer. Each vertex is surrounded by 5 prisms, each with two interior faces the ant mustvisit so it is also squares–10 around a point . (5) Triangles and squares tile the plane (3 × × × squares–10 around a point . • Squares–12 around a point. (1) Triangles and squares tile the plane (3 × ). Turn these into prisms withouttops and bottoms to make a single layer. Add fins to squares joining two open cubes. Then repeat vertically tomake alternating layers of prisms and fins. An ant will traverse two sides of a fin, then interior pairs of facesof 5 prisms as it circles each vertex, giving squares–12 around a point . (2) Triangles and squares tile the plane(3 × × × nvelope Polyhedra squares–12 around a point . • Hexagons–4 around a point. (1) Start with a hollow tetrahedron and cut its corners off. Cutting the cornersoff each triangular face creates 4 hexagonal exterior faces and 4 hexagonal interior faces, with 4 triangular holesconnecting the interior and exterior. Each vertex connects 2 exterior and 2 interior hexagons. (2) Start with ahollow icosahedron and cut its corners off. This creates 20 exterior hexagonal faces, 20 interior hexagonal facesand 12 pentagonal holes connecting the interior and exterior. (3) Start with a plane tessellation of hexagons. Itis possible to remove every third hexagon in such a way that hexagonal holes are created in the plane of hexagonssuch that every vertex borders one of the hexagonal holes. These holes connect to the other side of the plane,so an ant circling the vertex will visit 2 hexagons on the top of the plane, go through the hole and traverse 2hexagons on the bottom of the plane before returning to where it started. • Hexagons–8 around a point.
Tessellate the entire three dimensional space with octahedrons and tetrahedrons.This tessellates space into cells with all triangular faces. Now cut the corners off each of the triangular facesturning each of them into regular hexagons. Each edge in the original tessellation is truncated to 1/3 its formerlength as the corners of the triangular faces are cut off. At each end of one of these truncated original edges is avertex of the envelope polyhedron. Each original edge is bordered by two octahedral and two tetrahedral volumes,so four back to back hexagons join at a truncated original edge, and intersect at the vertex at the end of thatedge. As an ant circles this vertex he will traverse 2 interior hexagons of an octahedral volume, then 2 interiorhexagons of a tetrahedral volume, then 2 interior hexagons of an octahedral volume, then 2 interior hexagons ofa tetrahedral volume–8 hexagons around a point. This envelope polyhedron is reminiscent of octagons–8 arounda point which was based on a tessellation of space by cubes. • Octagons—4 around a point. (1) Start with the regular skew polyhedron squares–6 around a point , and cut thecorners off all the squares, making all of them into octagons. Where two squares formerly met at an edge, twooctagons from one side of the surface would meet at a shortened edge, and each vertex at the end of each ofthese shortened edges will now connect these two octagons with two more from the other side of the surface ofthe original regular skew polyhedron. (2) Start with a plane tessellation of squares, truncate all the squares tocreate octagons and leave square holes in the plane. Vertices connect 2 octagons on the top side of the planewith 2 octagons on the bottom side. • Decagons–4 around a point.
Start with a hollow dodecahedron and cut off the corners. This turns each pentagonalface into a decagon with 10 sides. The cut off corners become 20 triangular holes, connecting 12 interior decagonswith 12 exterior decagons. This has the topology of a sphere with 19 handles, the most complicated multiply-connected topology of any finite envelope polyhedra. • Dodecagons–4 around a point. (1) Start with the regular skew polyhedron, hexagons–4 around a point. Cutoff the corners of all the hexagons turning them into dodecagons (with 12 sides each). Where two hexagonsmet on an edge, the edge will now be shortened, and a new vertex will be created at each end point of eachshortened original edge. This vertex will now connect two dodecagons from one side of the original regular skewpolyhedron with two dodecagons from the other side. That makes dodecagons–4 around a point . (2) Do the sameoperation starting with the regular skew polyhedron hexagons–6 around a point . (3) Start with the semi-regularpseudopolyhedron 3 × on page 30 of WBK and remove the triangles. This creates an envelope polyhedron dodecagons–4 around a point where a single plane of decagons is punctuated by triangular holes. (Dodecagonsand triangles tessellate a plane with one triangle and two dodecagons around each vertex, once the triangles areholes, there are two dodecagons with top and bottom faces left around each vertex to give 4 around each vertex). B. ADDITIONAL ENVELOPE POLYHEDRA WITH MIRROR VERTICESBelow, in the WBK nomenclature are additional envelope polyhedra containing mirror image vertices. Page refer-ences are from WBK. : (1) From 3 × × shown on page 85 of WBK at bottom right. These are icosahedrons connected to each other byoctahedral tunnels in a structure where the icosahedrons resemble carbon atoms in a diamond structure and the four4 Gott octahedral tunnels originating from each resemble the carbon bonds in the diamond structure. Remove four trianglesfrom each icosahedron, which are opposite the four octahedral tunnels. This creates a 3 envelope polyhedron withmirror image vertices where there are three exterior and three interior octahedral tunnel triangles, and three exteriorand three interior icosahedral triangles around each vertex with a triangular hole connecting the interior and exteriortriangles around each vertex. (4) From 3 × : (1) From 4 page VIII, 1, by deleting a square from each vertex of an infinitely tall cylinder with 2 n sides ( n ≥ squares–6 around a point (1)described in the previous section of the Appendix. (2) From 4 × ×
12 page 10, by deleting the dodecagons. (4) from 4 × × page 13, by deleting the triangles to make a ladder made of squares with holes in the side rails and cubic rungs withopen ends. This is a vertical stack of cubes with alternately open east-west ends and open north-south ends. (6) From4 × ×
12 on page 41, by deleting the dodecagons. (8) From 3 × page 48, by deleting the triangles. (9) From 3 × on page 61 delete the triangles. (10) Tessellate the plane withdodecagons, squares and hexagons, build dodecagonal prisms, cubes and hexagonal prisms on these. Delete their topsand bottoms. (11) From 4 × × × × × onpage 97, deleted the octagons and add a square at each vertex connecting the remaining two squares to make a bent L shaped pattern of 3 squares (where the two legs of the L are bent in opposite directions at the edges of the squaresby 45 ◦ each). This gets traversed twice by an ant circling the vertex to give 4 . The structure consists of octagonalprisms, missing their octagonal tops and bottoms, pasted together at right angles at square sides—octagonal ringsof squares meeting at right angles in a three-dimensional structure. (This can also be constructed from squares–10around a point mentioned in the main body of this paper by deletion of the appropriate squares). : (1) From 4 page 17, by deleting squares. Bottom right picture shows some squares face on; these representtowers of open cubes seen from the top, eliminate squares from these towers seen edge on in the bottom right picture,so as to leave the squares seen face on in the bottom right picture as fins connecting the other 2 by 2 boxlike structures.(2, 3, 4, & 5) From 4 on page 20, delete the horizontal squares in the right hand picture. Or from 4 on page 20,instead delete alternate squares in the filmstrips of squares connecting octagonal columns of squares. There are twoways to do this. Or delete the squares on the vertical sides of half of the open cubes connecting the octagonal columns.(6) From 4 × × on page 46, delete the squares seen face-on in the upper right hand picture. Or from 4 on page 46instead delete alternate squares in the filmstrips of squares connecting hexagonal and octagonal columns of squares.There are two ways to do this. Or delete squares on the vertical sides of the open cubes connecting the hexagonalcolumns in the bottom right figure. (12) From 4 × × × × × ×
12 on page 59 delete the dodecagons.(18) From 3 × on page 60 at left delete the triangles, leaving two parallel layers of checkerboard pattern of justthe white squares connected by filmstrips of squares. (19) From 3 × on page 62 delete the triangles. (20, 21, 22)Attach square fins (one per vertex) to the single layer of 4 (10) described above to attach to layers above to makea vertical stack of layers. There are three ways to do this, as three double-sided squares meet at a vertex in a singlelayer, and so we have three different places to attach a fin. (23) Octagons and squares tile the plane. Turn them intoprisms and make a planar layer of octagonal prisms and cubes without their tops and bottoms. Add a series of parallelsquare fins above opposite sides of each cube. Repeat so that these parallel fins connect to a similar layer above withmirror image vertices. : (1) From 4 on page 45, delete every other square forming the hexagonal rings. (2) From 4 on page 53, deletehorizontal squares from North-South ties in the lower left figure. (3) Delete similar squares from 4 on page 54 atupper right. (4) Delete horizontal squares on North-South tunnels from Petrie 4 on page 67, upper right figure. (5)Take squares–8 around a point (8) described above and add fins connecting it to other copies of squares–8 arounda point . This can also be obtained from Fig. 5 by deleting the 12 squares at the back of Fig. 5, leaving one row ofopen ended triangular prisms with fins sticking up. This structure repeats vertically. It may also be viewed as twovertical checkerboards punctuated with holes [ squares–6 around a point (1) described above] connected by horizontaleast-west zig-zag filmstrips [ squares–4 around a point ] giving 4 . The holes in the squares–6 around a point punctured nvelope Polyhedra squares–10 around a point (2). Thishas vertical checkerboards punctuated by square holes, connected by horizontal sequences of squares which are sidesof cubes. Now between checkerboards rotate by 90 ◦ to produce vertical sequences of squares that are sides of cubes.We now have punctuated checkerboards connected by alternating horizontal and vertical sequences of squares that aresides of cubes. : (1 & 2) Start with squares–10 around a point (3) and add fins to the top of the single layer of open endedtriangular and hexagonal prisms tiling a plane. There are two ways to do this. Add vertical fins to alternate verticalsquare sides of the hexagonal prisms, or add vertical fins to square sides between two triangle prisms where the finis in the same plane as sides of the two hexagonal prisms that it touches at vertices. Now repeat vertically to createalternate layers of open prisms and fins. These of course have mirror vertices. (3, 4) Start with the plane tessellation3 × and construct prisms over it. Now take the tops and bottoms off these prisms. These are cubes and triangularprisms. Add square fins to this to connect to the next vertical layer of open prisms. There are two ways to do thisthat produce mirror vertices. Place the fins in parallel above alternate squares joining a cube and a triangular prism,or place the fins instead above squares connecting triangular prisms : In Fig. 5 the repeated fins and rows of squares repeated vertically produce vertical checkerboards punctuatedwith holes [ squares–6 around a point (1)] connected by horizontal zig-zag filmstrips of squares. Now rotate the set of12 squares at the back of the figure by 90 ◦ in a vertical plane, so that the zig-zag filmstrip of squares stands vertically.We now have a series of punctuated checkerboards connected to adjacent checkerboards by horizontal rows of zig-zagfilmstrips of squares–4 around a point on one side and vertical rows of zig-zag filmstrips of squares–4 around a point on the other side. This gives 4 + 6 + 4 = 14 squares around a point: 4 .Interestingly, squares, which have 4 sides, seem to produce the greatest variety of structures in 3D, just as carbon,which has four bonds with adjacent atoms, produces the richest chemistry.REFERENCES Coexter, H. S. M., 1937, “Regular Skew Polyhedra in Threeand Four Dimensions, and Their Topological Analogues,”
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