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The regular $d$-simplex has dihedral angle $\arccos(1/d)<90^\circ$, and the $d$-cube has dihedral angle exactly $90^\circ$. The maximal dihedral angle of a prism over a $(d-1)$-simplex is also $90^\circ$. For each $d\ge 2$, these are the polytopes with the smallest maximal dihedral angle among all convex $d$-dimensional polytopes.

Question: What are the $d$-polytopes with the next smallest maximal dihedral angle?

I am more interested in the value of the angle than the polyope, and I am okay with bounds rather than exact values.

More precisely, I wonder for which $d\ge 2$ there is a $d$-polytope $P\subset\Bbb R^d$ (not the simplex, the cube or a prism) with all dihedral angles smaller than

  • $\arccos(-\frac1d)>90^\circ$, or even
  • $\frac13(\pi + \arccos(-\frac1d))>90^\circ$.
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  • $\begingroup$ For reference when $d=3$: en.wikipedia.org/wiki/Table_of_polyhedron_dihedral_angles. Is the answer known for small $d$? $\endgroup$
    – Harry Richman
    Commented May 9, 2020 at 18:51
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    $\begingroup$ As stated I think the answer is yes for all $d$: just take the cube and "squash" it slightly so some dihedral angles increase and some decrease. $\endgroup$
    – Harry Richman
    Commented May 9, 2020 at 18:54
  • $\begingroup$ @HarryRichman I see, that was unexpected. In my application I have additional assumptions on the polytope, namely, that its $(d-2)$-faces are simplices, so I suppose this is an essential assumption. Anyway, I will need to think about this and maybe edit the question. $\endgroup$
    – M. Winter
    Commented May 9, 2020 at 18:59

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