Platonic and
Archimedean Polyhedra

The Platonic Solids, discovered by the Pythagoreans but described by Plato (in the Timaeus) and used by him for his theory of the 4 elements, consist of surfaces of a single kind of regular polygon, with identical vertices. The Archimedean Solids, consist of surfaces of more than a single kind of regular polygon, with identical vertices and identical arrangements of polygons around each polygon. In the following table, the Platonic Solids are indicated in red and the Archimedean Solids in green, blue, and purple. Green is for solids that can be produced by truncating the vertices of either Platonic or the blue Archimedean solids. Blue Archimedean Solids are produced from green ones by continuing the trucation until edges disappear and half the vertices merge. Pairs of Archimedean Solids become identical in that procedure. Purple Archimedean Solids result when, in the five triangles per vertex of the Platonic Icosahedron, one triangle is replaced by either a square (the Snub Cube) or a pentagon (the Snub Dodecahedron). The purple Archimedean solids have the interesting property of having right-handed and left-handed forms.

For information about all this, George W. Hart's "Virual Polyhedra" Site is wonderful; and I first learned about Archimedean solids from The Penguin Dictionary of Curious and Interesting Geometry, by David Wells (Penguin Books, 1991). Now there is a nice little book, Platonic & Archimedean Solids by Daud Sutton (Wooden Books, Walker & Company, New York, 2002), that covers the solids with many related facets of their geometry.

Platonic and Archimedean Polyhedra
P14 vertices4 faces3 faces/
6 edges4 triangles
(3 triangles /vertex)
A112 vertices8 faces3 faces/
18 edges4 hexagons,
4 triangles
(2 hexagons & 1 triangle /vertex)
P26 vertices8 faces4 faces/
12 edges8 triangles
(4 triangles /vertex)
P38 vertices6 faces3 faces/
12 edges6 squares
(3 squares /vertex)
A224 vertices14 faces3 faces/
36 edges8 hexagons,
6 squares
(2 hexagons & 1 square /vertex)
A324 vertices14 faces3 faces/
36 edges8 triangles,
6 octagons
(2 octagons & 1 triangle /vertex)
A412 vertices14 faces4 faces/
24 edges8 triangles,
6 squares
(2 triangles & 2 squares /vertex)
A548 vertices26 faces3 faces/
72 edges6 octagons,
8 hexagons,
12 squares
(1 octagon, 1 hexagon, & 1 square /vertex)
A624 vertices26 faces4 faces/
48 edges18 squares,
8 triangles
(3 squares & 1 triangle /vertex)
A7d24 vertices38 faces5 faces/
60 edges6 squares,
32 triangles
(1 square & 4 triangles /vertex)
A7s24 vertices38 faces5 faces/
60 edges6 squares,
32 triangles
(1 square & 4 triangles /vertex)
P412 vertices20 faces5 faces/
30 edges20 triangles
(5 triangles /vertex)
P520 vertices12 faces3 faces/
30 edges12 pentagons
(3 pentagons /vertex)
A860 vertices32 faces3 faces/
90 edges20 hexagons,
12 pentagons
(2 hexagons & 1 pentagon /vertex)
A960 vertices32 faces3 faces/
90 edges12 decagons,
20 triangles
(2 decagons & 1 triangle /vertex)
A1030 vertices32 faces4 faces/
60 edges12 pentagons,
20 trangles
(2 pentagons & 2 triangles /vertex)
A11120 vertices62 faces3 faces/
180 edges12 decagons,
20 hexagons,
30 squares
(1 decagon, 1 hexagon, & 1 square /vertex)
A1260 vertices62 faces4 faces/
120 edges12 pentagons,
30 squares,
20 triangles
(1 pentagon, 2 squares, & 1 triangle /vertex)
A13d60 vertices92 faces5 faces/
150 edges12 pentagons,
80 triangles
(1 pentagon & 4 triangles /vertex)
A13s60 vertices92 faces5 faces/
150 edges12 pentagons,
80 triangles
(1 pentagon & 4 triangles /vertex)

Johannes Kepler was the first person since antiquity to systematically describe all the Archimedean solids. However, he made one mistake. While the Great Rhombicuboctahedron certainly looks like a Truncated Cuboctahedron, and the Great Rhombicosidodecadhedron a Truncated Icosidodecahedron, which is what Kepler called them, mere truncation does not produce perfectly regular polygons on the surfaces. A little stretching is necessary. I have organized the table above as though Kepler was right, but this ends up being a little deceptive.

Several Archimedean solids can be broken down into parts that can be rotated against each other to produce new polyhedra with less symmetry. All of these rotations will also produce some vertices with different arrangements of the constituent polygons except one, the "pseudo-rhombicuboctohedron," derived from the rhombicubotohedron, where the arrangement of all the vertices is retained (but there are differing arrangements of the polygons around each square).

4 Dimensional "Platonic" Polytopes
1. 5-cell, Pentatope or Simplextetrahedra51010self-dual
2. 8-cell, Tesseract or Hypercubecubes16322416-cell
3. 16-celltetrahedra824328-cell
4. 24-celloctahedra249696self-dual
5. 120-celldodecahedra6001200720600-cell
6. 600-celltetrahedra1207201200l20-cell
In four dimensions, the five Platonic Solids have six analogues. Polyhedra become "polytopes." The Pentatope and the Tesseract are relatively easy to understand, and to illustrate with projections, as analogues of the Tetrahedron and the Cube.

Robert Heinlein wrote a science fiction story, "--And He Built a Crooked House--" [1941], based on the Tesseract. He imagined a man building a house that would be an unfolded tesseract, with five cubes arranged in a cross on the main floor, a single cube below, and a tower of two cubes above. At the time he wrote the story, Heinlein and his second wife lived in a house at 8777 Lookout Mountain Avenue, in the hills above Hollywood; and the house of Heinlein's fictional architect is said to be at 8775 Lookout Mountain, which would be next door to the Heinleins, except that this address doesn't exist -- the house next door is 8773.

The story begins by zeroing in on the location, first with the statement that Americans are considered crazy around the world ("You Yanks"), but they tend to blame it on California, while Californians blame it on Los Angeles. But Angeleños attribute all this to Hollywood, with the "violent cases" residing up in Laurel Canyon. Lookout Mountain Ave is in a side canyon above Laurel. Heinlein says, "The other Canyonites donít like to have it mentioned."

In the 1960's, Joni Mitchell and Graham Nash lived in a house in Laurel Canyon, the very house of the song "Our House," performed by Crosby, Stills, Nash, and Young. Indeed, the actual house was not on Laurel Canyon Blvd. It was 8217 Lookout Mountain Avenue, just down the street from the "Crooked House." Nash and Mitchell had had breakfast at Art's Deli in Studio City, and Nash wrote the song when they got home, and Mitchell actually was putting flowers in the vase she had just bought on Ventura Blvd.

I have found that the web of roads in that area of the Santa Monica Mountains, many of them dirt, made for enjoyable drives, both during the day and at night, with scenic views of the city lights below. Lookout Mountain Avenue itself eventually led into ways that led down to Sunset Boulevard. Since maps of the area were often incomplete and inaccurate, part of the charm was exploration. The roads and turnouts also attracted others, who might consume intoxicants or engage in sex, the evidence of which was often left on the ground. The neighbors evidently tired of this, and access began to be restricted by gates, usually just to prevent use of the roads as thoroughfares. Since the roads, especially the dirt ones, were very narrow, this made it difficult to turn around, which discouraged use of them altogether. It all seemed like the passing of a pleasant and more innocent time.

At the Google Maps streetview screens, it looks like many of these roads are now paved and the gates seem to be gone. Some of the shoulders also seem to have collapsed; and places were people used to turn out and park have now vanished, with fences protecting people from going off the road.

In the story, a minor earthquake causes the tesseract house to collapse into a true tesseract, with only one cube remaining visible on the street, and the other seven cubes presumably falling into the fourth dimension. Inside the house, all the cubes are still accessible, but the windows look out on strange scenes, perhaps on different worlds, or on no world. As an afterstock from the earthquake begins to shake the house, the architect and his companion leap out of a window, despite their fear that the strange flora they see may not be on Earth. Actually, the things are simply the eponymous plants, stange enough in their own right, at Joshua Tree National Monument in the California desert. They return to Los Angeles to discover that the house has disappeared completely.

A model or projection of a tesseract can be constructed in a very intuitive way by analogy with the construction of the cube. For a cube, we begin with a square. A square has four sides. On each side we attach another square. We can do all this on a plane. Given the third dimension, we can then fold up the outside squares until they meet. When they do meet, we discover that they have formed a new square, on a parallel plane to the original square. This construction can be projected back onto the original plane in an infinite number of ways, depending on how the cube is rotated in 3 dimensional space. In the projections, however, at least four squares, and perhaps all six, will be distorted in shape and/or size.

For the tesseract, we begin with the cube. It has six faces. On each face we attach another cube. Assuming a fourth dimension, the outside cubes can be "rotated" until they meet. In meeting, they will form an eighth cube, in a parallel space to that of the original cube. This construction can be projected back into three dimensional space, giving us 3-d models of a tesseract, or onto a plane, giving us projections of a tesseract, as here. In 3-d, two of the cubes may be undistorted; but the other six will be deformed. In the 2-d projections, of course, none of the cubes will be undistorted. In "folded tesseract 1," the rhombic shape of some of the distorted cubes can be made out easily. In "folded tesseract 2," one cube is smaller than the other (from perspective on its distance in the 4th dimension), while the cubes that attach them seem to take the form of trucated pyramids.

The famliar Pentagram is, strangely enough, a two dimensional (2-d) projection of the Pentatope. Since a Pentatope contains five Tetrahedra, it should be possible to find five distinct two dimensional projections of a Tetrahedron in the projection of the Pentatope. In the diagram at right this can be seen. Highlighted in red are each of the five Tetrahedra, with an independent red Tetrahedron for comparison. While it seems like this should be excellent fuel for fantasy or science-fiction connections between higher dimensional reality and occult practices, I have not noticed any such use of it that way. Even better, if the red lines are taken to be the projection of a Square with two diagonals, then the black lines can make each drawing the projection of a Pyramid [note].

A model or projection of a pentatope can be constructed in an intuitive way by analogy with the construction of the tetrahedron. For a tetrahedron, we begin with a triangle. A triangle has three sides. On each side we attach another triangle. We can do all this on a plane. Given the third dimension, we can then fold up the outside triangles until they met. They meet at a single vertex, so this completes the solid. No additional figure is generated. The construction can be projected back onto the original plane in an infinite number of ways, depending on how the tetrahedron is rotated in 3 dimensional space. In the projections, however, at least three triangles, and perhaps all four, will be distorted in shape and/or size.

For the pentatope, we begin with the tetrahedron. It has four faces. On each face we attach another tetrahedron. Assuming a fourth dimension, the outside tetrahedra can be "rotated" until they meet. They will meet at a single vertex, completing the polytope. Again, no additional form is generated. This construction can be projected back into three dimensional space, giving us 3-d models of a pentatope, or, as illustrated above, onto a plane, giving us projections of the pentatope. In "folded pentatope 2," four distorted tetrahedra appear to occupy the center of the base tetrahedron.

n-Dimensional "Platonic" Polytopes, n > 4
Polytopenumber of (n-1) D cellsverticesduals3-d analogue
1. (n + 1) celln + 1 n-cellsn + 1self-dualTetrahedron
2. 2n-cell2n (2n-2)-cells2n2n-cellCube
3. 2n-cell2n n-cells2n2n-cellOctahedron
Unlike the match of cube with tesseract and tetrahedron with pentatope, there is no 3-d analogue to the 24-cell.

In all dimensions greater than four, there are exactly three analogues to the Platonic Solids. This is, curiously, exactly half the forms we find in 4 dimensions.

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Platonic and Archimedean Polyhedra, Note;
The Lizard-Spock Expansion

The full pentagram provides a diagram for the suggested "Lizard-Spock Expansion" of the traditional "rock, paper, scissors" game. This was publicized in The Big Bang Theory in 2008 [episode 2:8]. Later the show acknowledged that this was developed by "internet pioneer" Sam Kass, who himself acknowledges the collaboration of Karen Bryla.

Ordinarily, (1) scissors cut paper, (2) paper covers rock, and (3) rock breaks scissors. With the addition of the lizard and (Mr.) Spock (from Star Trek), (4) rock crushes lizard, (5) lizard poisons Spock, and (6) Spock smashes scissors. However, we now have the occurrence of additional matches. Rock still breaks scissors, but now (7) scissors also decapitate lizard, (8) Spock vaporizes rock, (9) lizard eats paper, and (10) paper disproves Spock. The three actions of "rock, paper, scissors" expand to ten actions with the addition of two new moves.

One might ask, "Why expand the system by two instead of just one?" One reason would be the asymmetry of the result. If we only add "lizard," which gives us the square diagram at left, we see that two of the items defeat two others but that the other two only defeat one. This means that in playing the game it would be a very poor strategy ever to choose paper or lizard, which will be defeated twice as often as they can ever defeat another themselves. This asymmetry is inevitable given that each vertex of the square is the intersection of three lines. So each vertex must either defeat two or be defeated by two. Adding both lizard and Spock means that each vertex is the intersection of four lines, which can be divided evenly. Kass and Bryla devoted some thought to the Lizard-Spock Expansion.

The premise on The Big Bang Theory was that "rock, paper, scissors" did not provide enough alternative choices, resulting in too many ties; but then, when used, the "Expansion" resulted in everyone always choosing Spock. Given the preference of players for Spock, for symbolic and personal reasons, the best strategy would be to choose lizard or paper, both of which defeat Spock. However, since another player with similar understanding might choose lizard or paper, lizard is the best choice, since it defeats both Spock and paper.

Now we get further explansions of the game, for instance with the addition of the Zombie and the LHC (i.e. Large Hadron Collider -- the particle accelerator at Cern in Switzerland/France). This allows for three victories and three defeats for each gesture. The Zombie hand is a "limp-wristed gesture, with all fingers opened," while the LHC is a "fist with the thumb and index finger extended," like the hand shape for a gun, since the collider uses a "particle gun" to inject particles into the machine. Another large body of rules must be added, which I will not detail here. This is intriguing but all begins to seem like too much.

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The Chinese Elements and Associations, Note 1