A 180-hedron! Is there any limit to the number of congruent faces on a polyhedron?
No. Take a skew prism -- lots of congruent triangular faces -- and put a cap on each end made out of triangles of the same dimensions.
just a minor nit. this cannot always be done; it depends upon the antiprism. er, ... did i interpret this correctly "skew prism" = "(what i've always known as an) antiprism"? anyway, it can be done if the isosceles triangles are sufficiently tall, and that suffices for the "no" answer above. this construction gives a polyhedron with 4n faces. however, the "belt" around the antiprism is not needed! simply take a 2 point suspension of a regular n-gon, which gives a polyhedron with 2n triangular faces. (alternatively, start with a "regular" antiprism, and extend the triangular faces. the planes of the upward-pointing triangles all meet at one point, and the planes of the downward-pointing triangles all meet at another point. this gives a polyhedron with 2n quadrilateral faces. surely there's a picture online ... ) [edit: pictures of stellated "dipyramids" at: http://www.georgehart.com/virtual-polyhedra/pyramids-info.html ] are there examples of polyhedra with all faces congruent, that have an odd number of faces? (i did not see any parity restriction, although the faces must have an even number of sides.) mike