# Three Polyhedra with Only Pentagonal Faces The polyhedron above has 72 pentagonal faces of two types. The next one below has three different types of pentagon for its 132 faces. After that is a polyhedron with sixty faces, all of which are non-convex pentagons.  All three of these all-pentagon polyhedra were created using Stella 4d: Polyhedron Navigator. This program is available here.

## 6 thoughts on “Three Polyhedra with Only Pentagonal Faces”

1. teachezwell on said:

They are all spectacular! Wish I could have them in hand.

Liked by 1 person

2. I love pentagons, although my preference is convex.
Question, is it possible to have a polyhedron of 240 convex pentagons?
I think it is, but I don’t have enough math background to determine if it is so.

Liked by 1 person

• I must admit that I simply do not know. If I find one, ever, I’ll post it on this blog.

Like

• steve v on said:

Trivially, a 120 sided antiprism capped with 120 sided pyramids has 240 convex 5-fold vertices so its dual would be a polyhedron of 240 convex pentagons. Some 10 sided dice are in the same family, so think of one of these with vastly more faces. A more interesting example is formed by taking the rhombic icosidodecahedron and performing the Snub* operation on it so that each of its 60 vertices is replaced by a new quadrilateral with 4 vertices (total 240) and each of its edges is replaced by a couple of triangles. The resulting polyhedron has 240 convex 5-fold vertices so its dual would be a polyhedron of 240 convex pentagons (and icosahedral symmetry.) *The snub operation is ambiguous; the terms (a) snub cube and (b) snub cuboctahedron refer to the same solid. I’m using it in the former sense, where the vertices of the original cube generate the triangular faces of the cuboctahedron, rather than the latter where they are already present. In this case the 4 fold vertices of the rhombic icosahedron generate new quadrilaterals.

Liked by 1 person