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I don't understand which part of the definition of an archimedian solid excludes prisms from being one. each vertex of a prism has the same polygons around it (4,4,n for a n-gonal prism), and also each vertex is symmetrical to every other vertex, and that isn't even always part of an archimedian solid's definition (see pseudo-rhombicuboctahedron). why are they not classed as archimedian solids?

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  • $\begingroup$ by symmetrical I meant vertex transistive $\endgroup$ – stanley dodds Mar 19 '15 at 7:09
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Prisms and antiprisms, whose symmetry groups are the dihedral groups, are generally not considered to be Archimedean solids, despite meeting the above definition. With this restriction, there are only finitely many Archimedean solids.

From Wikipedia. They're not counted because they're infinite families, and it's the others that are more interesting, I guess. Or maybe people just like having a finite amount of Archimedean solids.

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Prisms don't have 2 or more types of polygons meeting in identical vertices. But other than that your answer should be right and they should be counted.

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  • $\begingroup$ If the end-faces of the prisms are regular $n$ sided polygons and the side faces are squares, then every vertex has an $n$-gon and two squares, and the vertices are essentially identical, in much the same way as a truncated icosahedron is an Archimedean solid. With anti-prisms where the side faces are equilateral triangles, every vertex has an $n$-gon and three triangles, rather like a snub dodecahedron. $\endgroup$ – Henry Mar 19 '15 at 10:08

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