I'm new to the world of mathematical descriptions of polyhedra, and I'm wondering if, for a Pentakis Dodecahedron, the dihedral angles are uniform at each vertex. The visualization of the P.D. on the wikipedia page compels me to believe that, if they are uniform, the dihedral angles at both the longer edges of each triangle are convex and that the dihedral angles at the shorter edges are concave. My intuition is that face-transitivity should imply that the dihedral angles between any faces ought to be the same. If that is correct, then it stands to follow that uniformity of dihedral angle says nothing of concavity or convexity in a polyhedron.

I ask this question because I'm trying to build a pentakis dodecahedron out of wood and I need to know angle at which to cut the beveled edges.

  • $\begingroup$ Since you reference the "longer" and "shorter" edges of the triangular faces, I take it you're specifically referring to Catalon Solid version of the PD; the one that's dual to the truncated icosahedron? (The one Wikipedia uses for the main picture, in green) $\endgroup$
    – pjs36
    Commented Apr 19, 2015 at 18:43
  • $\begingroup$ Yes! I'm referring to the Catalan Solid version. $\endgroup$ Commented Apr 19, 2015 at 18:44
  • $\begingroup$ ahh -- I think I mistook the red shape under the main description as another representation of the pentakis dodecahedron, when it is, in fact, a deltahedron. $\endgroup$ Commented Apr 19, 2015 at 18:49
  • $\begingroup$ It actually is "another" PD: The name just refers to adding pentagonal pyramids to all faces of a regular dodecahedron; both the red and green do that. The heights of the pyramids differ though. So geometrically, there is no such thing as "the" PD, but a whole class of PD's that 'look' the same (combinatorially equivalent), but aren't geometrically the same (not scaled versions of each other). $\endgroup$
    – pjs36
    Commented Apr 19, 2015 at 18:54
  • $\begingroup$ The question actually still stands -- does having a dihedral angle imply uniform convexity and uniform angle between any two planes that share an edge in the polyhedron? $\endgroup$ Commented Apr 19, 2015 at 18:54

2 Answers 2


This is really just a comment that's too long. Hopefully, once I understand fully what you're asking and we have uniform terminology, someone will be able to answer your (at least specific) question.

The dihedral angle is angle at which two planes meet; for polyhedra, it's the angle at which two faces meet. All polyhedra, convex or not, have dihedral angles.

However, due to the way it's measured, the dihedral angle can tell convex from non-convex polyhedra: We always measure the dihedral angle as the angle inside the polygon.

enter image description here

In the picture above, imagine the lines are altitudes of triangle faces (and we can't see the faces; they're perpendicular to our line of sight). Then the angle $\varphi$ corresponds to a dihedral angle of a convex portion of our polyhedron; the angles all "bend toward" the interior of the polyhedron. This would be a convex meeting of faces, as we have $\varphi < \pi$.

However, for a non-convex polyhedron, the angles may "bend away" from the interior of the polytope. In this case, it would actually be $2\pi - \varphi$ that's the interior dihedral angle, and we'd have $\varphi > \pi$.

Thus, we can characterize convex polytopes based on their interior dihedral angles: They're all less than $\pi$.

There's another angle measure for polyhedra that can (only sometimes) detect a lack of convexity: The angular defect. To calculate the angular defect at a vertex, we add up all the angles of faces that make up the vertex, and subtract this sum from $2\pi$. For example, in a regular tetrahedron, three faces meet at each vertex, and the angles of each face are $\pi/3$; thus our angular defect is $2\pi - 3(\pi/3) = \pi$. In a regular dodecahedron, the angular defect is $2\pi - 3(3\pi/5) = 2\pi - 9\pi/5 = \pi/5$.

I only bring 'angular defect' up because I can't quite parse your phrase that "the dihedral angles are uniform at each vertex." To my knowledge, in $3$-space, dihedral angles really only apply to edges (where just two faces meet), and not to vertices.

For the Catalan PD, apparently the dihedral angle is constant for all edges. I suspect that the angular defect is not constant, and can be one of two different things, but I haven't done any calculations.

  • $\begingroup$ I understand why my terminology was incorrect -- I meant to say that "the dihedral angles are uniform at each edge" -- which means that the dihedral angle would be constant for all edges. $\endgroup$ Commented Apr 20, 2015 at 21:37
  • $\begingroup$ Also -- thank you so much for the explanation! I now understand what an angular defect is. What is an example of a polytope with constant angular defect at each vertex, constant dihedral angles at each edge and face-transitivity? It can't be an Archimedean solid because they are not necessarily face-transitive. $\endgroup$ Commented Apr 20, 2015 at 21:41
  • $\begingroup$ @NikhilShinday You're quite welcome! I suspect that "face transitivity" alone forces us to use a small class of polyhedra, and that we would be able to "inspect those by hand" to find that only the Platonic Solids (completely regular) would satisfy those criteria. That could probably be a question of its own though! :) $\endgroup$
    – pjs36
    Commented Apr 20, 2015 at 21:54

There is some useful information (as well as a beautiful image) at this webpage that may help:



You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .