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(Jerry of Nashville, TN. 2000-11-18)
What [polyhedron] has six faces?

A polyhedron with 6 faces is called a hexahedron. The cube is your most obvious hexahedron, however that's certainly not the only one. Disregarding geometrical distortions and considering only the underlying topology, there are 7 distinct hexahedra:

Name of Hexahedron	Edges	Nodes
Triangular Dipyramid	9	5
Pentagonal Pyramid	10	6
Tetragonal Antiwedge	10	6
Hemiobelisk	11	7
Hemicube	11	7
Cube	12	8
Pentagonal Wedge	12	8

The above triangular dipyramid has 5 vertices and 9 edges.  It's the dual of a triangular prism, and looks like two tetrahedra "glued" on a common face.

The pentagonal pyramid has 6 vertices and 10 edges; it's a pyramid whose base is a pentagon. Like all pyramids, the pentagonal pyramid is self-dual.

The above three hexahedra are the only ones which exist in a version where all 6 faces are regular polygons.

The tetragonal antiwedge is the least symmetrical of all hexahedra; its only possible symmetry is a 180° rotation.  This skewed hexahedron has the same number of edges and vertices as the pentagonal pyramid. Its faces consist of 4 triangles and 2 quadrilaterals.  Such a solid is obtained from two quadrilaterals that share an edge [the hinge] but do not form a triangular prism.  After adding two edges to complete the two triangles whose sides are adjacent to the hinge, we are left with a nonplanar quadrilateral and must choose one of its 2 diagonals as the last edge of the polyhedron. Only one choice gives a convex polyhedron.

Loosely speaking, there are two types of tetragonal antiwedges which are mirror images of each other; each is called an enantiomer, or enantiomorph of the other. The tetragonal antiwedge is thus the simplest example of a chiral polyhedron.  In particular, any other hexahedron can be distorted into a shape which is its own mirror image, and the tetragonal antiwedge may thus unambiguously be called the chiral hexahedron.  Each enantiomer is self-dual; a tetragonal antiwedge and its dual have the same chirality.

The other types of hexahedra are more symmetrical and simpler to visualize. One of them may be constructed by cutting off one of the 4 base corners of a square pyramid to create a new triangular face. This hexahedron has 7 vertices and 11 edges.  Its faces include 3 triangles, 2 quadrilaterals and 1 pentagon. It could also be obtained by cutting an elongated square pyramid (the technical name for an obelisk) along a bisecting plane through the apex of the pyramid and the diagonal of the base prism, as pictured at right.  For lack of a better term, we may therefore call this hexahedron an hemiobelisk.

Also with 7 vertices and 11 edges, there's a solid which we may call a hemicube (or square hemiprism), obtained by cutting a cube in half using a plane going through two opposite corners and the midpoints of two edges. Its 6 faces include 2 triangles and 4 quadrilaterals.

With 8 vertices and 12 edges, the cube (possibly distorted into some kind of irregular prism or truncated tetragonal pyramid) is not the only solution: Consider a tetrahedron, truncate two of its corners and you have a pentagonal wedge. It has as many vertices, edges and faces as a cube, but its faces consist of 2 triangles, 2 quadrilaterals and 2 pentagons.

We can build a pentagonal wedge with 2 regular pentagons and 2 equilateral triangles, so that all edges but one are equal. The one "exceptional" edge is the longest side in the two trapezoidal faces. What's its length? Well, look at the wedge "from the side" (so both pentagons project into a line) you see two similar isosceles triangles. The base of the smaller is a regular edge seen perpendicularly (and therefore at its real size), whereas the base of the larger triangle is the length we're after. The ratio of similitude is simply the ratio of the height of a regular pentagon to the distance from a side to an adjacent vertex, namely 1+sin(p/5)/sin(2p/5) = (1+Ö5)/2.  A number known as the golden ratio, which happens to be the ratio of the diagonal to the side in a regular pentagon. The longest edge in our solid is thus 1.6180339887498948482... times the lengt of any other. In other words, both trapezoidal faces are congruent to the diagonal section of a pentagonal face (pictured at right).

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(2007-09-19)   Fattest Tetragonal Antiwedge
The chiral hexahedron of least surface area  (enclosing a given volume).

Let's consider  convex  chiral hexahedra  (see above)  possessing the only allowed symmetry  (namely a  180°  rotation about an axis perpendicular to the hinge where the two tetragonal faces meet).  If the hinge AB has a length of  2 (arbitrary) units, the shape of such a  symmetrical tetragonal antiwedge  is determined by  5  positive parameters  (u,v,x,y,t  with  y>v  for convexity)  giving the coordinates of the  6  vertices in a suitable cartesian system:

A  =

-a 0 0

B  =

a 0 0

C  =

u v -vt

D  =

x y yt

E  =

-x y -yt

F  =

-u v vt

If we call  2q  the  dihedral  angle between the two tetragonal faces, then:

t   =   tan q       and       1 + t ²   =   1 / cos² q

The volume  V  is the sum of  2  pairs of tetrahedra  (each having O as a vertex).

V   =   ¹/₃ [ det (B,C,D)  +  det (D,E,F) ]   =   ²/₃ yt  ( av + uy + vx )

The above computation can be done mentally, using the preceeding layout which features the relevant triplets of columns adjacent to each other (and in the right order).  The first determinant  (2avyt)  is trivial.  The second one is quite simple too, when computed as  det (D,D+E,F).

Expressing the areas of triangles and quadrilaterals as cross-products  (and using pairwise equalities)  we obtain the total surface area  S  of the hexahedron:

S   =	|| BC ´ BD ||  +  || DE ´ DF ||  +  || AD ´ BF ||
Where :
|| BC ´ BD || 2 =	(1 + t 2 ) [ (u-a) y + (a-x) v ] 2  +  4 (vyt) 2
|| DE ´ DF || 2 =	4 (y-v) 2 [ x 2  +  t 2 y 2 ]  +  4 t 2 (uy + xv) 2
|| AD ´ BF || 2 =	(1 + t 2 ) [ (u+a) y + (a+x) v ] 2

The maximum volume for a given surface  (or, equivalently, the minimum surface for a given volume)  is obtained when the differential forms  dV  amd  dS  are  proportional  (the coefficient of proportionality is the Lagrange multiplier associated to whichever quantity is considered a  constraint  under which the other is to be optimized).

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(2000-11-28)   Enumeration of Polyhedra
How many polyhedra have a given number of faces and/or edges?

We maintain on this site an extensive table with the latest data. Let's just say it's not easy to count these things.

Polyhedra which are mirror images of each other are not counted as distinct. In the above article, we counted 7 types of hexahedra, but there would be 8 if we tallied separately both chiralities of the tetragonal antiwedge.

There's only one tetrahedron:

There are two types of pentahedra: the triangular prism and the square pyramid.

There are 7 hexahedra (see previous article), 34 heptahedra, 257 octahedra, 2606 enneahedra, 32300 decahedra, 440564 hendecahedra, 6384634 dodecahedra, 96262938 tridecahedra, 1496225352 tetradecahedra, etc.

Duality :

The  dual  of a polyhedron is the polyhedron obtained by switching the roles of vertices and faces:  Edges of the dual connects nodes associated with adjacent faces of the original polyhedron  (dual polyhedroa have the same number of edges).  Duality is a  topological  relation which is geometrically ambiguous...

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BenO (Ben Ocean, 2001-07-28, via e-mail)   Platonic Solids  
What are the x,y,z coordinates of the vertices of the 5 Platonic Solids?

Only 5 solids exist (up to rotation and/or scaling) whose faces are congruent regular polygons.  These very special polyhedra are known as  Platonic solids.  (See below for a generalization to n dimensions.)

Regular Tetrahedron A B C D x = 2 2 -4 0 y = 2Ö3 -2Ö3 0 0 z = -Ö2 -Ö2 -Ö2 3Ö2

Shown at right are the Cartesian coordinates of the vertices of a regular tetrahedron ABCD centered at the origin. These may be scaled and/or rotated. As given, this tetrahedron has:

• A side equal to 4Ö3.
• A height equal to 4Ö2.
• A circumscribed sphere of radius 3Ö2.

An alternate set of Cartesian coordinates for a smaller regular tetrahedron (of side 2Ö2, inscribed in a sphere of radius Ö3) would consist of every other vertex in a cube of side 2 (see below) centered at the origin. This highlights some geometrical relationships and dims others:

A = [+1,+1,+1],   B = [+1,-1,-1],   C = [-1,+1, -1],   D = [-1,-1,+1]

In such a regular tetrahedron, two vertices are seen from the center at an angle known as the tetrahedral angle (which is very familiar to chemists) whose cosine is -1/3 and whose value is 109.47122°... The dihedral angle between a pair of faces is supplementary to that angle; its cosine is 1/3 and its value is about 70.52878° (this may be called a cubic angle, for a reason which follows). Expressed in radians, three times this angle minus a flat angle (p) gives the value [in steradians, sr] of the solid angle at each corner of the tetrahedron, namely 0.55128559843... This is about 4.387 % of the solid angle of a whole sphere (4p). Some astronomers prefer to use the square degree as a unit of solid angle (a square degree equals p²/180² sr); the solid angle at the corner of a regular tetrahedron is 1809.7638632... square degrees (or 6515150 square minutes).

We may choose 3 coordinates from the set {-1,+1} in 8 different ways and these correspond to the coordinates of the 8 vertices of a cube of side 2 centered at the origin (and inscribed in a sphere of radius Ö3). Seen from the center of a cube, the angular separation between corners is either a flat angle (180° between diametrically opposed vertices), a tetrahedral angle of cosine -1/3 (about 109.47° between the opposite corners in a face), or a cubic angle whose cosine is 1/3 and which is supplementary to a tetrahedral angle (about 70.53° between adjacent corners). The solid angle at each corner of a cube is clearly p/2, namely 1/8 of a whole sphere (4p).

Similarly, there are 6 ways of choosing 3 coordinates from the set {-1,0,+1} so that only one of them is nonzero. These correspond to the coordinates of the 6 vertices in a regular octahedron of side Ö2 centered at the origin (and inscribed in a sphere of radius 1). From the center of a regular octahedron, two vertices are seen separated either by a right angle (90°) or by a flat angle (180°).

The volume of a regular dodecahedron is (15+7Ö5)/4 times the cube of its side. The dihedral angle has a cosine of -1/Ö5 and a value of about 116.565°

Topical pages with other relevant data :   Paul Bourke, Ron Knott, VB Helper, etc.

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Archimedean solids  
Catalan solids are the duals of Archimedean polyhedra.

Two vertices of a polyhedron are called  equivalent  if there's a rotation leaving the polyhedron unchanged which transforms one of these vertices into the other.

All the vertices in a platonic solid are equivalent.  This is also the case for all  equilateral  prisms and antiprisms.  Other convex polyhedra whose vertices are all equivalent are called  Archimedean solids.

For a solid to be Archimedean, all its vertices must have the same arrangement of faces around them, but this is not sufficient. A counterexample is the elongated square gyrobicupola (J37), also called pseudo-rhombicuboctahedron (pictured at left), which is not Archimedean.

There are 13 Archimedean solids (two of which are chiral, the snub cube and snub dodecahedron).  Archimedes of Syracuse (c.287-212 BC) may have discovered them all, but only 12 of them were known during the Renaissance.  Kepler (1571-1630) added the  snub dodecahedron  when he reconstructed the whole set in 1619.

The truncated icosahedron (the shape of a traditional soccer ball) is now more commonly known as a buckyball ever since it was found to be the structure of a wonderful new molecule, now called fullerene (C60) in honor of the famous American architect R. Buckminster ("Bucky") Fuller (1895-1983), who created and advocated geodesic domes in the late 1940s.

The buckyball is one of 4 Archimedean solids without triangular faces.  The other three are the truncated octahedron (at left), the great rhombicosidodecahedron (at right) and the great rhombicuboctahedron.

The 4 Archimedean polyhedra illustrated so far are simplicial (i.e., only 3 edges meet at each vertex).  There are 3 others such simplicial polyhedra, illustrated next, which happen to be obtained
(like the buckyball and truncated octahedron above) by truncating a Platonic solid.

Finally, 5 edges meet at every vertex of the two chiral Archimedean polyhedra:

The  snub dodecahedron  (above right) was forgotten until Kepler rediscovered it!

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(2002-06-14)
Is there a systematic way to name polyhedra?

Only up to a point.  The most "generic" way is to use for polyhedra the same naming scheme as for polygons, by counting the number or their faces:  Thus, a tetrahedron has 4 faces, a pentahedron has 5, a dotriacontahedron (also called triacontakaidihedron) has 32 faces.  However, counting just faces is not nearly enough to describe a polyhedron, even from a topological standpoint.  In some cases, a nonstandard counting prefix is traditionally used for certain very specific polyhedra.  For example, the dotriacontahedron shown at right is an Archimedean solid unambiguously known as an icosidodecahedron (literally, a polyhedron with 20+12 faces) because it includes 20 triangular faces and 12 pentagonal ones.  The icosidodecahedron could also be called a pentagonal gyrobirotunda but that name would masks its much greater symmetry compared to the related pentagonal orthobirotunda (J34).  For the same reason, a special name has been given to the cuboctahedron, which might otherwise be called a triangular gyrobicupola.

If the icosidodecahedron had not claimed the title for the above reason, the name could have been given to another Archimedean solid with 32 faces, the so-called truncated dodecahedron (which has 20 triangular faces and 12 decagonal ones).  It wasn't...
      The notoriety of the icosidodecahedron has made it tempting for some (knowledgeable) people to use the nonstandard icosidodeca prefix (instead of dotriaconta or triacontakaidi ) to name other unrelated things (like a 32-sided polygon).  Resist this temptation...

The general situation is similar to the naming of chemical compounds.  Certain families can be identified and a systematic naming can be introduced among such families.  The next article gives the most common such examples.

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(2000-11-19)   Types of polyhedra named after a polygon :

Take a regular polygon (an hexagon, say) and construct a polyhedron by considering an identical copy of that hexagon in a parallel plane.  Join each vertex of the hexagon to the corresponding vertex in its copy and you obtain what's called an hexagonal prism.  Instead, you may twist the copy slightly and join each vertex to the two nearest vertices of the copy. What you obtain is an hexagonal antiprism. In such families, the polyhedron is named using the adjective corresponding to the name of the polygon it's built on (e.g., "hexagonal").

There are several other families besides  prisms  and  antiprisms  for which this pattern applies. For example, if you cut a prism with a plane containing some edge of either base polygon (but not intersecting the other), this "half" prism is called a  wedge  (it includes the base polygon and its featured edge).

Alternately, if the cutting plane contains only a single vertex, instead of a whole edge, the polyhedron we obtain by cutting a prism is best called an hemiprism.

A  deltohedron  is what a regular n-gonal dipyramid becomes if we twist its upper pyramidal cone 1/2n of a turn with respect to the lower one: The intersection of the two cones becomes a solid whose faces are quadrilaterals [see figure at left].  Do not confuse this with the deltahedra defined below !

Topologically, we obtain what's called an  n-gonal antiwedge  by starting with an  n-gonal wedge  (as described above)  and splitting each of its n-3 lateral tetragonal faces into two triangles  (by introducing just one diagonal of each such quadrilateral as a new edge).  An equivalent  geometrical  construction starts with two  (non-coplanar)  n-gons  sharing an edge  (the so-called  hinge)  and the two triangular faces formed by an extremity of that hinge together with the adjacent vertices found on each n-gon.  We're then left with n-3 lateral tetragons (which are not, in general, planar quadrilaterals) from which we build 2n-6 additional triangular faces  (for a grand total of 2n-2 faces, including the two n-gons).  Note that there's only one way to split a  nonplanar  tetragon into two triangular faces to form a  convex  polyhedron.

A priori, the above constructions yield  2^n-3  types of n-gonal antiwedges  which may differ either by their topology or their chirality.  However, some pairs of such configurations may be obtained from each other by a 180° rotation about an axis perpendicular to the  hinge.

Besides the trivial case of the "trigonal antiwedge"  (which is just a fancy name for an ordinary tetrahedron)  the simplest such polyhedron is the tetragonal antiwedge,  a remarkable hexahedron which stands out as the simplest example of a chiral polyhedron  (the two possible tetragonal antiwedges are mirror images of each other).

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What are deltahedra? [Not to be confused with  the above deltohedra.]

Deltahedra are simply polyhedra whose faces are all equilateral triangles (a polyhedron whose faces are triangles which are not all equilateral is best called an irregular deltahedron). A deltahedron [or an irregular deltahedron] has necessarily an even number of faces (2n faces, 3n edges, and n+2 vertices).

A noteworthy fact is that there are only 8 convex deltahedra (disallowing coplanar adjacent faces).  Namely:

4 faces	6 faces	8 faces	10 faces

1. Regular tetrahedron (4 faces).
2. Triangular Dipyramid (6 faces).
3. Regular octahedron or "Square Dipyramid" (8 faces).
4. Pentagonal Dipyramid (10 faces).
5. Snub Disphenoid (12 faces), J84.
   [The Snub Disphenoid was originally called "Siamese dodecahedron" by Freudenthal and van der Waerden, who first described it in 1947.]
6. Triaugmented Triangular Prism (14 faces), J51.
7. Gyroelongated Square Dipyramid (16 faces), J17.
8. Icosahedron or "Gyroelongated Pentagonal Dipyramid" (20 faces).

12 faces	14 faces	16 faces	20 faces

All told, the convex deltahedra include

• 3 Platonic solids (tetrahedron, octahedron and icosahedron).
• 3 dipyramids (triangular, square and pentagonal).
• 3 Johnson solids (J17, J51, and J84).

This adds up to 8, instead of 9, because the regular octahedron happens to be counted twice (as a Platonic solid and a square dipyramid)...

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Johnson Solids and Polyhedral Nomenclature

It's quite a challenge to enumerate all  convex  polyhedra whose faces are regular polygons.  Besides infinitely many prisms and antiprism, this includes only:

•   5   Platonic polyhedra
• 13   Archimedean solids
• 92   Johnson solids.

Norman W. Johnson gave the full classification in 1966, by adding the 92 polyhedra now collectively named after him.

Only 5 of the 92 Johnson solids are chiral, namely:

• J44 :   gyroelongated triangular bicupola.
• J45 :   gyroelongated square bicupola.
• J46 :   gyroelongated pentagonal bicupola.
• J47 :   gyroelongated pentagonal cupolarotunda.
• J48 :   gyroelongated pentagonal birotunda.

Nomenclature

To describe these and other common polyhedra, some systematic nomenclature is useful.  In particular, any polyhedron gives rise to many other types whose names include one or more of the following adjectives:

• Elongated: By replacing (one of) the largest m-sided polygon, with an m-gonal prism (that polygon may not be a face of the polyhedron, but an "internal" polygon with apparent edges).  This adds m vertices, 2m edges, and m faces.  The simplest example, shown at right, is the elongated tetrahedron (J7), which is an heptahedron. 
• Gyroelongated: By replacing (one of) the largest m-sided polygon, with an m-gonal antiprism (that polygon is usually not a face of the polyhedron, but an "internal" polygon with apparent edges). This adds m vertices, 3m edges, and 2m faces. Gyroelongation can be performed in two different ways (often leading to different chiral versions of the same polyhedron). 
• Snub: Snubbing is an interesting chiral process which, roughly speaking, amounts to loosening all faces of a polyhedron and rotating them all slightly in the same direction (clockwise or counterclockwise), creating 2 triangles for each edge and one m-sided polygon for each vertex of degree m. A polyhedron and its dual have the same snub(s)!  If a polyhedron has k edges, its snub has 5k edges, 2k vertices and 3k+2 faces. 
• Truncated: By cutting off an m-gonal pyramid at one or more (usually all) of the vertices. This add (m-1) vertices, m edges and 1 face for each truncated vertex. 
• Augmented: By replacing one or more of the m-sided faces with an m-gonal pyramid, cupola, or rotunda. 
• etc.

Other terms are available to describe certain interesting special cases:

• Cingulum (Latin: girdle; cingere to gird).
• Fastigium (Latin: apex, height).
• Sphenoid (Greek: wedge).
• etc.

The rest of the nomenclature used in the context of Johnson solids, is best described in the words of Norman W. Johnson himself :

"  If we define a lune as a complex of two triangles attached to opposite sides of a square, the prefix spheno- refers to a wedgelike complex formed by two adjacent lunes. The prefix dispheno- denotes two such complexes, while hebespheno- indicates a blunter complex of two lunes separated by a third lune. The suffix -corona refers to a crownlike complex of eight triangles, and -megacorona, to a larger such complex of 12 triangles. The suffix -cingulum indicates a belt of 12 triangles. "

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How do polyhedra generalize to 4 dimensions or more?

The equivalent of a polyhedron in dimension 4 is called a polychoron (plural polychora). Polychora are discussed extensively on beautifully illustrated pages proposed by George Olshevsky and Jonathan Bowers.

Although the introduction of the term polychoron is fairly recent, it seems now generally accepted, as there's no serious competition (the etymology of "polyhedroid" is poor and misleading). The term was coined in the 1990's by George Olshevsky, whose earlier proposal of "polychorema" (plural: "polychoremata") was unsuccessful. Olshevsky's new proposal had the early support of Norman W. Johnson, after whom the 92 "convex regular-faced solids" are named (Johnson solids).

A polychoron is bounded by 3-dimensional faces, called cells. The four-dimensional equivalent of the Euler-Descartes formula is a topological relation which relates the number of vertices (V), edges (E), faces (F), and cells (C) in any polychoron enclosing a portion of hyperspace homeomorphic to a 4D open ball (provided edges, faces and cells are homeomorphic, respectively, to 1D, 2D and 3D open balls):

V - E + F - C   =   0

In an unspecified number of dimensions, the counterpart of a 2D polygon, a 3D polyhedron, or a 4D polychoron is called a polytope. The boundary of an n-dimensional polytope consists of hyperfaces which are (n-1)-dimensional polytopes, joining at hyperedges, which are (n-2)-dimensional polytopes. (All the hyperfaces of an hyperface are thus hyperedges.) This vocabulary is consistent with the well-established term hyperplane to designate a vector space of codimension 1 (in a hyperspace with a finite number n of dimensions, a hyperplane is therefore a linear space of dimension n-1). We also suggest the term hyperline for a linear space of codimension 2 (and, lastly, hyperpoint to designate a space of codimension 3).

To denote the p-dimensional polytopes within a polytope of dimension n, the following terms may be used: vertex (p=0; plural "vertices"), edge (p=1), face (p=2), cell or triface (p=3), tetraface (p=4), pentaface (p=5), hexaface (p=6), ... hypervertex (p=n-3), hyperedge (p=n-2), hyperface (p=n-1), hypercell (p=n).

To establish and/or memorize the n-dimensional equivalent of the Euler-Descartes formula for "ordinary" polytopes in n dimensions, it's probably best to characterize each such polytope by the open region it encloses (boundary excluded), except in dimension zero (the 0-polytope is a single point). For the formula to apply, each such region should be homeomorphic [i.e., topologically equivalent] to the entire Euclidean space of the same dimension, or equivalently to an open ball of that dimension. An edge is an open segment, a face is an open disk, a cell is an open ball, etc. [for example, the ring between two concentric circles is not allowed as a face, and the inside of a torus is disallowed as a cell]. Then we notice that a number can be assigned to any polytope (and a number of other things) called its Euler characteristic (c), which is additive for disjoint sets, equal to 1 for a point and invariant in a topological homeomorphism (so that topologically equivalent things have the same c). For our purposes, this may be considered an axiomatic definition of c. It may be used to establish (by induction) that the c of n-dimensional Euclidean space is (-1)ⁿ, which is therefore equal to the c of our "ordinary" open polytopes (HINT: A hyperplane separates hyperspace into three parts; itself and 2 parts homeomorphic to the whole hyperspace). The c of all "ordinary" closed polytopes in n dimensions is the c of a closed n-dimensional ball and it may be obtained by inspecting the components of the boundary of any easy n-dimensional polytope like the hypercube or the simplex polytope discussed below. It turns out to be equal to 1 in any dimension n. If the hypercell itself (the polytope's interior) is excluded from the count, as it is in the traditional 3-dimensional Euler-Descartes formula, the RHS of the formula will therefore be 2 in an odd number of dimensions and zero in an even number of dimensions. For example, in 7 dimensions, if we denote by T the number of tetrafaces, by P the number of pentafaces (hyperedges), and by H the number of hexafaces (hyperfaces), we have:

V - E + F - C + T - P + H   =   2

We may focus on the n-dimensional equivalent of the Platonic solids, namely the regular convex polytopes, whose hyperfaces are regular convex polytopes of a lower dimension, given the fact that the concept reduces to that of a regular polygon [equiangular and equilateral] in dimension 2. In dimension 3, this gives the 5 regular polyhedra. In dimension 4, we have 6 regular polychora. In dimension 5 or more, only 3 regular polytopes exists which belong to one of the following three universal families (also present in spaces of lower dimensions):

Family	n-Polytope	dim.	V 0	E 1	F 2	C 3	T 4	p-Faces p
Simplex		n	C(n+1,p+1)
	Point	0	1
	Segment	1	2	1
	Triangle	2	3	3	1
	Tetrahedron	3	4	6	4	1
	Pentachoron	4	5	10	10	5	1
Cross Polytope		n	C(n,p+1) 2p+1
	Point	0	1
	Segment	1	2	1
	Square	2	4	4	1
	Octahedron	3	6	12	8	1
	Hexadecachoron	4	8	24	32	16	1
Hypercube		n	C(n,p) 2n-p
	Point	0	1
	Segment	1	2	1
	Square	2	4	4	1
	Cube	3	8	12	6	1
	Tesseract	4	16	32	24	8	1

The regular simplex polytope is obtained by considering n+1 vertices in dimension n, so that each one is at the same distance from any other (its hyperfaces are simplex polytopes of a lower dimension). Choosing as vertices all points whose Cartesian coordinates are from the set {-1,+1}, we obtain an n-dimensional hypercube (of side 2). The hyperfaces of an hypercube are hypercubes of a lower dimension. The dual of the above hypercube is the regular cross polytope whose vertices have a single nonzero coordinate, taken from the set {-1,+1}.

The  interactive hypercube  at right is from Kurt Brauchli (details here).  Click and drag with the mouse to turn the cube aound the chosen axes (H and V) indicated in the menu bar.  This 5D cube projects like a 3D cube if you rotate only around axes 0, 1 or 2.  The fourth and fifth dimensions appear with axes 3 and 4.   With the bottom cursor, you may choose a distant (left) or close-up perspective (right).

Your browser is unable to run a Java "applet" and cannot display this interactive picture...   Sorry.

The word  tesseract  was coined by  Charles Howard Hinton  (1853-1907)  the son-in-law of the logician George Boole (1815-1864).  C. Howard Hinton was obsessed with visualizations of the fourth dimension involving the tesseract, which he presented in his 1904 book entitled  The Fourth Dimension,  whose posthumous third edition (1912) has been made available online by rkumar, thanks to Banubula.

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EnolaStraight (2002-05-07)
What's the radius of the circle touching 3 touching unit circles?
What's the radius of the sphere touching 4 touching unit spheres?
[In this (edited) question, "touching" means "externally tangent (to)".]

The generalization of this question to any number of dimensions is a classic demonstration that whatever geometrical intuition we may have developed in two or three dimensions may not be trusted in a space of more dimensions. The two-dimensional case [pictured at right] shows 3 congruent circles, centered on the vertices of an equilateral triangle, touching each other and a much smaller circle [pictured as a red disk] whose radius has to be determined. Based on this 2-D case [and, to a lesser extent, on the 3-D case] it would seem that such an inner ball [disk, sphere, or n-dimensional hypersphere] would always be small enough to fit inside the simplex [equilateral triangle, regular tetrahedron, n-dimensional regular simplex] formed by the centers of the congruent balls. This happens to be true for a dimension equal to 4 or less, but fails for a dimension of 5 or more. In a very large number of dimension, the (linear) size of the inner ball is about 41% the size of the outer ones. More precisely, Ö2-1 = 0.41421356... is the limit of that ratio when the number of dimensions tends to infinity. Read on...

Consider the center O of the n-dimensional simplex formed by the n+1 centers of the congruent balls [each of radius 1]. The critical quantity is the distance D(n) from the center O to any vertex; the radius of the inner ball is simply D(n)-1. Well, because O is the center of gravity of n+1 vertices, it is on the line that goes from a vertex to the center of gravity of the n others. It divides that line in a 1 to n ratio. The length of that line is therefore (1+1/n)D(n) and it is also one side of a right triangle whose other side is of length D(n-1) and whose hypotenuse is of length 2 (it's the side of the simplex, the distance between the centers of two balls). In other words, D(n) is given recursively by the relations:

D(1) = 1     and     D(n) = Ö[4-D(n-1)²] / (1+1/n)

This recursion can be used to prove by induction the following simple formula:

D(n)   =   Ö[ 2n / (n+1) ]

The simplicity of this result is a hint that there might be a more direct way to obtain it. D(2) = 2/Ö3 says that the radius of the inner circle in the above figure is 2/Ö3-1 [about 15.47 %] of the radius of any outer circle. Similarly, the corresponding ratio for spheres is ½Ö6-1 [about 22.4745 %]. The limit of D(n) is Ö2: In a space with a very large number of dimensions, the ratio of the radius of the inner hypersphere to the radius of any outer hyperspheres is thus slightly less than Ö2-1 [about 41.42 %].

In dimension n, the distance from the center O to any of the hyperplanes (of dimension n-1) of the "faces" is D(n)/n.  Therefore, if the radius D(n)-1 is greater than is D(n)/n., the inner hypersphere bulges outside of the n-dimensional regular simplex formed by the centers of the outer hyperspheres. This happens as soon as n²-5n+2 > 0, which is the case when n is at least equal to 5. This higher-dimensional configuration is contrary to the intuition we would form by looking only at the two-dimensional and/or three-dimensional cases...

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Dr. Murali V.R. (2004-02-25; e-mail)
What is the volume of a regular antiprism?

A regular antiprism is a polyhedron whose faces are two parallel n-gonal bases  [regular polygons with n sides]  and 2n equilateral triangles called lateral faces.

Look at the outline of such a solid from above, and what you see is a regular polygon with 2n sides (every other vertex is on the top base, and every other one is on the bottom).  The angle at each vertex of this outline is thus  q = p-p/n.

Now, each lateral face is seen as an isosceles triangle having an angle q at the top and featuring a base observed at its real size  a  (as the direction of observation is perpendicular to it).  The height of such an isosceles triangle is thus:

½ a cotan (q/2)   =   ½ a tan (p/2n)

This quantity is also equal to the length of a side of a right triangle whose hypotenuse is the true height of a lateral face (namely ½ aÖ3) and whose other side is the height  h  of the antiprism  (namely, the distance between its bases).  This gives the height  h  of the antiprism in terms of the length  a  of its edges:

h   =   ½ a	Ö	3 - tan2(p/2n)

Consider the circumscribed prism of height h  whose base is the 2n-gonal outline.  Each side of this outline is equal to  ½ a / cos(p/2n).  Its surface area is therefore:   (n a²/4) / sin(p/n)   and the volume of the prism is h times that.

The  antiprism  is obtained from this prism by removing 2n triangular pyramids of height h whose bases are all congruent to the above isosceles triangle, for a combined base area of  (n a²/2) tan(p/2n)   and a total volume h/3 times that.

The volume V of the antiprism is the difference between these two volumes:

V   =   (n a3 / 24)  [ 3 / sin(p/n) - 2 tan (p/2n) ]	Ö	3 - tan2(p/2n)

This can be rewritten in a much more palatable form, using   t = tan (p/2n)  :

V   =   n a³ ( 3 - t ² ) ^3/2  /  48t

h   =   ½  a Ö 3 - tan2(p/2n)   V   =   n h 3 / 6 tan(p/2n)

Two noteworthy special cases (for a = 1):

• V = 1 / Ö72   when n=2.  A regular tetrahedron! [A degenerate but valid case.] 
• V = ½ Ö3   when n=3.  A regular octahedron...

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(2006-01-18)   Szilassi Polyhedron  &  Császár Polyhedron
Two polyhedra of genus 1 [topology of a torus] dual of each other.

 

Author	Date	Faces	Edges	Vertices
Lajos Szilassi   (1942-)	1977	7  hexagons	21	14
Ákos Császár  (1924-)	1949	14  triangles	21	7

In a polyhedron, a line between two nonadjacent vertices is called a diagonal.  When every pair of vertices is connected by an edge, the polyhedron has thus  no diagonals.  The tetrahedron is an example of a polyhedron without diagonals, so is  Császár's polyhedron.  By duality, this means that every face of the  Szilassi heptahedron  has an edge in common with each of the other 6 faces...

Geometrically, the Szilassi polyhedron has an axis of 180° symmetry: 3 pairs of congruent faces and a symmetrical face (darkest in the picture).  This symmetry allows one to build a full mental picture of the polyhedron from the image at right  (obtained from David Eppstein's Geometry Junkyard).

Beyond Császár and Szilassi...

The Descartes-Euler formula for a polyhedral surface of  genus  G  is:

V - E + F   =   2 - 2G

In a polyhedron where all pairs of faces share an edge, E = (F-1)F/2.  Also, we have V = 2E/3, since any vertex must belong to only 3 faces.  Eliminating E and V using these two relations makes the above Euler formula boil down to:

G   =   (F-4)(F-3) / 12

As G is an integer, F must have definite values modulo 12, namely 0, 3, 4 or 7.  Beyond F=4 (the tetrahedron of genus 0) and F=7 (the Szilassi heptahedron of genus 1) the next possibility would thus be F=12, a dodecahedron of genus 6... 

Curved toroidal maps which can't be straightened :

The number of colors required to color a map (or a graph) so that no two adjacent patches (or nodes) are of the same color is called its  chromatic number.  For any given surface other than the  Klein bottle  (a surface of genus 1 on which any map can be colored with only 6 colors)  the maximal chromatic number of all maps drawn on it depends only on its geometric genus G, according to the following formula, proposed by P.J. Heawood in 1890  (A000934).

Maximal Chromatic Number   =   ë ½ ( 7  +  ( 48G + 1 ) ^½ ) û

For genus 0, this amounts to the celebrated  4-color theorem  for planar or spherical maps, as proved by Appel and Haken in 1977.  Otherwise, the formula was shown to be an upper bound by Heawood in 1890.  It was found to be  exact  (except for the Klein bottle)  in 1968, when Ringel and Youngs showed how a map with this many countries could always be drawn on a surface of genus G so that two countries always have a common boundary.

When  G  is  (F-4)(F-3) / 12,  the Heawood formula gives precisely a maximal chromatic number equal to F  [check it].  The challenge met by Szilassi for genus 1  (a topological torus)  was to draw a  Ringel-Youngs map with 7  flat countries.  Could the same feat be possible for the next cases,  starting with a genus-6 surface consisting of 12 planar faces?

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Miro Dabrowski (of Bunbury, Australia. 2006-07-21)
What are the dihedral angles in a  buckyball ?
How do I taper 32 ready-to-glue wooden faces ?

First, let's consider how the overall size of the buckyball relates to the length of its edges  (a).  One easy way to do so is to consider the  equator  of the ball if the  polar axis  goes through the centers of two opposite pentagonal faces  (for example, let the polar axis be vertical in the above picture).  This equator is a  regular decagon  whose sides are of length  1.5 a  (namely, the distance between the middles of two nonadjacent edges in a  regular hexagon  of side  a).  The radius of the circle  circumscribed  to that decagon is the so-called  midradius  (r)  of the buckyball  (i.e., the distance from the ball's center to the middle of any  edge).  Thus,  3a/ 4r  is the  sine  of  p/10  (or 18°)  which is  1/2f.  This yields the first relation below.  The other equation gives the  circumradius  (R)  of the buckyball  (i.e., the distance  from the center to any  vertex )  obtained from the Pythagorean theorem  (r² + a²/4  =  R² ).

r / a   =	3/4  (1+Ö5)   =   3 f/2		=   2.42705098312484227230688025...
R / a   =	1/4  Ö	58  +  18 Ö5	=   2.47801865906761553756640791...

We may now envision a  solid  buckyball  (a truncated regular icosahedron)  as consisting of 32 straight regular pyramids  (12 pentagonal and 20 hexagonal ones)  with a common apex  at the center O.  All the lateral faces of those pyramids are congruent to an isosceles triangle of base  a  and sides  R.

A thick buckyball "shell" may be assembled from wooden faces (hexagonal or pentagonal) whose sides are tapered at an angle equal to the angle between the base and any lateral face of such a pyramid.  This is the only way to have the same taper for all hexagonal sides, in order to simplify tooling and final assembly. 

The interior shape will not be a perfect hexacontahedron if boards of the same thickness are used for both types of faces.  If needed, this flaw (which is normally hidden) could be eliminated by shaving about 5.1% off the boards used for the hexagonal blocks  (since the height of an hexagonal pyramid is about 94.9% of the height of a pentagonal one).

Each n-gonal pyramid consists of n slices  (as pictured at left)  sharing an edge from O to the center  M  of the n-gon  (n is 5 or 6). a  is the length of every edge. R  is the  circumradius  (distance from the center O to any vertex). r  is the  midradius  (distance from O to [the middle of] any edge). bn  =  ½ a / tan(p/n)   is the inscribed radius of the n-gonal face.

The aforementioned taper angles  (q = 0  would be a straight cut)  are obtained from the formula   q_n  =  arcsin ( b_n / r ).  Numerically, this boils down to:

q5   =   arcsin Ö	(5+Ö5)  /  90	=   16.4722106927603966099...°
q6   =   arcsin [	(Ö5 - 1)Ö3 / 6 ]	=   20.9051574478892990329...°

Between an hexagon and an adjacent pentagon, the dihedral angle is  q₅+q₆.  The dihedral angle between two adjacent hexagons is 2q₆.

Angular distances seen from the center of a buckyball :

Angle...	Symbol	Expression	Value in degrees
subtended by an edge	a	2 arcsin  a 2R	23.28144627 °
from face center to vertex	b5	arcsin  a 2R sin p/5	20.07675127 °
	b6	arcsin  a 2R sin p/6	23.80018260 °
from face center to [middle of] edge	q5	arcsin  a 2r tan p/5	16.47221069 °
	q6	arcsin  a 2r tan p/6	20.90515745 °
between middles of edges, 1/3 of the way around an hexagon		p/5	36 °

A useful check is obtained by considering any journey along half a  great circle  (180°) which includes an edge between hexagons:

180°   =   p   =   a  +  2 (b₅  +  q₅ )  +  4 q₆

A similar "trick" was used to obtain trivially the last entry of the above table  (actually, this was our starting point to obtain the midradius-to-edge ratio).

The  deficiency  at each of the 60 buckyball vertices is 12°, namely  4p/60 = p/15.  It may also be computed directly as  2p-2(2p/3)-3p/5.

The  deficiency  at a vertex of a polyhedron is defined to be what's missing from a full turn  (2p  or 360°)  when you add up all the angles of the faces which meet at that vertex.  The name is historically linked to Descartes'  deficiency theorem  which states that, in any polyhedron homeomorphic to a sphere, the sum of the deficiencies at all vertices is equal to an angle of 4p  (or 720°).