# Talk:Archimedean solid

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Field: Geometry

## Untitled

"semi-regular" should be defined; right now it is not clear what the difference between Johnson and Archimedian solids is. Furthermore, I take it that the platonic solids also count as Archimedian? --Axelboldt

I don't think so. semi-regular, i think has to do with the fact that multiple kinds polygons can meet at a vertex And platonic solids are definitely not Archimedean. (since there are five platonic solids, and thirteen archimedean ones....)

I think not being face-uniform just means there are more than one kind of regular polygon used. And I guess that being vertex-uniform means that there is a rotation that can move any one chosen vertex to the place of any other chosen vertex, while also mapping all other verticies to the place of either itself or another vertex. I hope so, because if I guessed wrongly, then my pictures and coordinates probably aren't of Archimedean solids. Ксйп Cyp 20:24, 31 Jul 2003 (UTC)
Hey, that was a reply in less than 2 years... How's that for wiki-fast replies? Ксйп Cyp 23:16, 31 Jul 2003 (UTC)
Suppose to the above comment on rotation, mirroring also counts, since otherwise I can't see how a truncated cuboctahedron or truncated icosidodecahedron would count. Ксйп Cyp 10:13, 1 Aug 2003 (UTC)

A platonic solid is made up of only one type of regular polygon, and is vertex transitive. An Archimedian solid is any vertex transitive solid and is made up of two or more regular polygons. Obviously, these two can't be the same. And a Johnson solid is any solid that is made up of two or more regular polygons, that aren't vertex transitive. Vertex transitive means that any vertex, all the same polygons meet, in the same order and number. Hope that clears it up. Timeroot (talk) 00:33, 24 June 2008 (UTC)

Not quite. Vertex-transitivity means that the solid has a symmetry group under which any vertex maps to any other. The congruency of vertex figures follows from this, but not the other way around. The elongated square gyrobicupola (aka "false rhombicuboctahedron") has a triangle and three squares at each vertex, but the sixteen vertices of its octagonal prism are not in the same symmetry class as the eight vertices of the 'tops' of the cupolae. —Tamfang (talk) 04:02, 9 July 2008 (UTC)

The last sentence uses the term "regular vertex" without defining it. AxelBoldt 09:50, 2 Oct 2003 (UTC)

I worked a bit on the definition of Archimedean solids, but it's still not satisfactory. Right now, it's not clear why the Elongated Square Gyrobicupola is excluded. AxelBoldt 10:13, 2 Oct 2003 (UTC)

Contains a lot of sections, seems to be in the format of mentioning an Archimedean solid site, someone reviewing it, and selecting their favourite solid. There's at least 2 about this page. I hope copying the comments about this site to here would be considered fair use or something... Κσυπ Cyp   19:15, 18 Jan 2004 (UTC)

Jennifer Brosten reviews http://en2.wikipedia.org/wiki/Archimedean_solid

This website was nice because it has the general idea of the Archimedean solids in a rather concise manor though it also offers more in depth information on polyhedras and all of the different Archimedean Polyhedra. The table showing the different Archimedean Polyhedra was well done, because it illustrated the figures while also giving useful information about the vertices, faces and how they meet. The illustrations make it so that you can see the 3D aspect, whereas many sites only show the front without being able to see what is happening at the back of the object as well. The web site does not stop at giving the general information of the different solids. If you click on the names of the solids, you are taken to a new web page which is devoted strictly to that solid.

My favorite Archimedean polyhedra would have to be the Icosidodecahedron. What made me pick this shape was first it's name, because it's kinda fun to try to say. The Icosidodecahedron is made up of 20 triangular faces and 12 pentagon faces. There are a total of 60 edges and 30 vertices. At each vertex, there are 2 triangles and 2 pentagons meeting. They go triangle-pentagon-triangle-pentagon.

Mine would have to be one with both icosahedral-group and octahedral-group symmetry. Does any such polyhedron, Archimedean or not, exist? —The Doctahedron, 68.173.113.106 (talk) 21:50, 22 November 2011 (UTC)

Mary Moser reviews http://www.ezresult.com/article/Archimedean_solid (Note: That is a not so good mirror of Wikipedia, without working pictures... Ironically, the only complaint about the article is that the pictures don't work.)

I was able to find the above website and saw that it had all the basic information presented clearly as well as some interesting history. I really appreciated that throughout the sight vocab words are linked to further definitions and explenations. There is a lot of potential in this sight unfortunately it seems the pages that are supposed to provide images of the polyhedra are not working, (at least I was unable to view them). Also I would like to see descriptions relating them to the platonic solids (essentially how we get the Archimedean polyhedron by truncating the platonic polyhedron).

I think a good example of the pictures and information my initial sight is missing can be found at http://www.ul.ie/~cahird/polyhedronmode/favorite.htm. I especially liked the animation showing the truncations for some of the polyhedra.

While each of the polyhedra we are discussing is really interesting and fun to explore, the assignment is to choose one favorite so I choose truncated cuboctahedron. It has 26 faces (12 squares, 8 hexagons and 6 octagons), 72 edges, and 48 vertices.

## Educational toy

Jovo [[1]] is a toy that is ideal for constructing Archimedean solids. Can such a link be in the Wikipedia, or is it too commercial? --80.162.63.207 17:01, 5 Feb 2005 (UTC)

## Inscribed Archimedean polyhedra

NOTE: This text for a new and uncompleted section was removed from article!

Edit here as you like and move back when it's done!
Tom Ruen 21:43, 14 January 2006 (UTC)

When the Archimedean polyhedra are inscribed in a sphere, they occupy the following percentages of that sphere's volume:

• truncated tetrahedron 40.134%
• cuboctahedron 56.270%
• truncated cube 57.682%
• truncated octahedron XX%
• rhombicuboctahedron XX%
• truncated cuboctahedron XX%
• snub cube XX%
• icosidodecahedron XX%
• truncated dodecahedron XX%
• truncated icosahedron XX%
• rhombicosidodecahedron XX%
• truncated icosidodecahedron XX%
• snub dodecahderon XX%
I hope someone who has the percentages will add them here. I was attempting to make the Archimedean article include some of the same things that are in the Platonic article. Unfortunately, removing the list to the discussion page instead of leaving it on the article page means that it will probably languish here, and not get finished, but we'll see. Jimaginator 13:06, 19 January 2006 (UTC)

Here is how you could calculate those percentages: Go onto the page for that solid, and scroll to the part about the cartesian co-ordinates. Take the three numbers, square each of them, and take their sum (for the truncated tetrahedron, these would yield 11, for example). Now take the square root of that. Now, in that same section, it will say "For a side length of --." The truncated tetrahedron, for example, will have "the square root of 8". Divide this side length number by the previous number you got (the square root of 11), and put that into the formula for volume, which will be else where on the page. Now, divide that result of that formula by 4/3pi, and multiply by 100. This is your percentage, and it will work for any Archimedean solid. I already did a few. Timeroot (talk) 17:57, 24 June 2008 (UTC)

## Wythoff construction

TEST TABLE

All the Archimedean solids can be constructed by Wythoff construction of a spherical tiling.

Parent Truncated Rectified Bitruncated
(truncated dual)
Birectified
(dual)
Cantellated Omnitruncated
(Cantitruncated)
Snub
Wythoff symbol
p-q-2
q | p 2 2 q | p 2 | p q 2 p | q p | q 2 p q | 2 p q 2 | | p q 2
Coxeter-Dynkin diagram
Vertex figure pq (q.2p.2p) (p.q.p.q) (p.2q.2q) qp (p.4.q.4) (4.2p.2q) (3.3.p.3.q)
Tetrahedral
(3 3 2)

{3,3}

(3.6.6)

(3.3a.3.3a)

(3.6.6)

{3,3}

(3a.4.3b.4)

(4.6a.6b)

(3.3.3a.3.3b)
Tetrahedral
3-3-2

{3,3}

(3.6.6)

(3.3.3.3)

(3.6.6)

{3,3}

(3.4.3.4)

(4.6.6)

(3.3.3.3.3)
Octahedral
(4 3 2)

{4,3}

(3.8.8)

(3.4.3.4)

(4.6.6)

{3,4}

(3.4.4a.4)

(4.6.8)

(3.3.3a.3.4)
Octahedral
4-3-2

{4,3}

(3.8.8)

(3.4.3.4)

(4.6.6)

{3,4}

(3.4.4.4)

(4.6.8)

(3.3.3.3.4)
Icosahedral
(5 3 2)

{5,3}

(3.10.10)

(3.5.3.5)

(5.6.6)

{3,5}

(3.4.5.4)

(4.6.10)

(3.3.3a.3.5)
Icosahedral
5-3-2

{5,3}

(3.10.10)

(3.5.3.5)

(5.6.6)

{3,5}

(3.4.5.4)

(4.6.10)

(3.3.3.3.5)

## Blue and yellow

Hi,
I've just created the following table, and I think the bottom line would look much more logical with blue and yellow exchanged. The trigons and hexagons, which are blue at the moment, correspond to the yellow trigons in the two big files above. The squares (or trigon pairs in the snubs), yellow at the moment, have no corresponding faces in all the files above. They just correspond to the vertices in the two big files. Thus I think, these "new" faces should have the "new" color. Any thoughts? Lipedia (talk) 23:33, 7 August 2010 (UTC)

Yes. —Tamfang (talk) 03:01, 9 August 2010 (UTC)

That would make sense. Who made these pictures, anyway? —The Doctahedron, 68.173.113.106 (talk) 21:45, 22 November 2011 (UTC)

## Why are there only 13 Archimedean solids?

Could someone add a section on why there are only 13 Archimedean solids please, and explain it well? Thanks, please do it asap Akhi666 (talk) 18:55, 11 October 2014 (UTC)