# Compute face angles for a snub disphenoid (Johnson solid J84)

What I understand so far:

This shape has 12 triangular faces (all equilateral triangles since this is a Johnson solid) and 8 vertices. We have 4 faces meeting at 4 of the vertices and 5 faces meeting at the other 4 vertices.

The two vertices where 4 faces meet can be split into 2 groups, each group containing 2 such vertices that are connected between them and not connected to any of the 2 vertices in the other group. This means we can get this shape by taking two regular octahedra, removing 2 adjacent faces from each and then joining them.

We could also build it starting from a pentagonal bipyramid as shown here.

What can be seen in the above figure is that at each vertex where 5 faces meet, we have 2 groups of 2 faces that are almost in the same plane.

What I want to know

How can I compute the angles of the faces?

I want to get these angles without starting from the vertex coordinates which I can find online because that's just starting from the solution.

• Three comments : 1) You should have said that the faces are all equilateral triangles ; 2) I suppose that the coordinates you mention are those given in the excellent : en.wikipedia.org/wiki/Snub_disphenoid. 3) Why would it be a shame to use these coordinates ? Using "pure geometry" in this case looks to me too complicated... – Jean Marie Aug 20 '19 at 10:25
• "pure geometry" won't be able to solve this at all - you need to solve a cubic to get the answers, and that's bad news for actually constructing this object with euclidean tools. – Dan Uznanski Aug 20 '19 at 14:31

Orient the polyhedron so that the two reflection planes of symmetry coincide with the $$xz$$- and $$yz$$-planes; that is to say, the twofold rotational axis of symmetry is the $$z$$-axis. Consequently, without loss of generality, for a unit edge length, the vertices are of the form $$(\pm a, 0, -b), \quad (\pm 1/2, 0, c), \quad (0, \pm a, b), \quad (0, \pm 1/2, -c)$$ for $$a, b, c > 0$$. This gives rise to the system of equations \begin{align*} \left(a-\frac{1}{2}\right)^2 + (b+c)^2 &= 1, \\ a^2 + \frac{1}{4} + (b-c)^2 &= 1, \\ 2a^2 + 4b^2 &= 1. \end{align*} The unique nontrivial positive solution to this system results in roots of a cubic; specifically, $$a$$ satisfies $$2 - 2 a - 3 a^2 + 2 a^3 = 0$$ and $$b = \frac{1}{2} \sqrt{1 - 2a^2}, \quad c = \sqrt{\frac{1 + a - a^2}{2}}.$$ The numeric values are approximately \begin{align*} a &\approx 0.64458427322415498454 \\ b &\approx 0.20556156585325953808 \\ c &\approx 0.78393092423256364992. \end{align*} The computation of the dihedral angles is tedious but straightforward. These are, in radians, approximately $$1.6789775749362604980, 2.1248189366147042378, 2.9049356464412552144.$$