# congruent regular faces imply congruent dihedral angles?

I have seen definitions of regular polyhedra where it is assumed only that the faces are congruent and regular (equilateral). I have seen also definitions where two conditions are assumed:

1. Congruent regular faces.
2. Congruent Dihedral angles.

I belive that the condition (1) above implies the condition (2) above.

Is this right?

Update:

The following link: A congruence problem for polyhedra

Has the following :

Theorem 1.2 (Cauchy, 1839). Two convex polyhedra with corresponding congruent and similarly situated faces have equal corresponding dihedral angles.

I believe this is the line of search that I would go for.

Final Update: At this point the answers with counter-examples are good. I wanted to point out to the book in this link Polyhedra

A set of counter-examples is given in Figure 2.18. But most important is the following theorem:

Theorem: Let $P$ be a convex polyhedron whose faces are congruent regular polygons. Then the following statements about $P$ are equivalent:

1. The vertices of $P$ all lie on a sphere.
2. All the dihedral angles of $P$ are equal.
3. All the vertex figures are regular polygons.
4. All the solid angles are congruent.
5. All the vertices are surrounded by the same number of faces.

Again, thanks to all who helped.

## 2 Answers

Perhaps the simplest counterexample is the bipyramid formed by sticking together two equal regular tetrahedra so that a face on each of them coincides.

It depends what else is given about the polyhedron.

Namely, there exist a moderate number of Johnson Solids, like the gyroelongated square dipyramid below which is obtained from a triangular antiprism by attaching two square pyramids. The Johnson solids have regular faces with equal edge lengths, but by definition are not regular. In lieu of saying that all dihedral angles are equal, a common component of the definition of a Platonic Solid is that an equal number of faces must meet at each vertex -- this should be another way to force regularity. In the image above, we have vertices of degree $4$ and $5$.

• There are two things that I did not put in my question. (i) convexity, (ii) equal distance to its center of mass (all vertices in a sphere). However I haven ot the second constraint in any of the defintions. Convexity implies Platonic Solids. I do not want to prove this by exhausting posibilities such as show that there are only 5 Plantonic solids and in those all dihedral angles within are congruent. – Herman Jaramillo Dec 10 '15 at 14:05
• I appreciate you correct and prompt answer. So, it is safe to say that if (i) all vertices are on the surface of a sphere. (ii) all the faces are equilateral/equiangle (regular) surfaces. Then the polyhedron is regular. Right? – Herman Jaramillo Dec 10 '15 at 15:33