[Math] Dihedral angles of a pentakis dodecahedron

geometrypolyhedra

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.

Best Answer

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.

Related Question