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The Wikipedia page for the snub dodecahedron provides explicit coordinates for its vertices, from which one can of course by brute force calculate that there are only two dihedral angles. Can anyone give a (hopefully insightful/more intuitive) shorter proof that there must be only two such angles? The symmetries of the object immediately imply that there can be at most three such angles: the triangle-pentagon one, the angle between two triangles each of which is adjacent to a pentagon, and the angle between a triangle not adjacent to any pentagon and one that is. But is there any "quick" way to see that the latter two angles must be equal?

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Focus on a pair of adjacent triangles $T_1$ and $T_2$. Their vertices lie on the circumscribing sphere $S$. Their dihedral angle is supplementary to the angle $\angle P_1OP_2$ where $O$ is the centre of $S$ and $P_i$ is the centroid of $T_i$. But this angle only depends on the radius of the sphere and the edge lengths of the equilateral triangles $T_i$. (There's a rotation of the sphere taking $T_1$ and $T_2$ to any configuration of two adjacent triangles with the same sidelengths with vertices on $S$.)

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    $\begingroup$ Aha! I was not recognizing the fact that a symmetry of the circumsphere demonstrating the equality of two triangle-triangle dihedrals does not have to be a symmetry of the full snub dodecahedron. Thanks! $\endgroup$ – Glen Whitney Jul 10 '18 at 9:15

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