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Yes. For instance, take the octahedron and pick two opposite faces. Add a new vertex over each of the other six faces, dividing each triangle into three triangles (that is, glue a flattened tetrahedron onto each of the six selected faces). Now each of the six old vertices has degree seven, and each of the six new vertices has degree three. (This is like the ...


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This site (to which @Aretino already pointed) lists the dihedral angles as $105.14^\circ$, as illustrated below. (I verified that this angle is same at every edge.) So you should bevel at half this angle, $52.6^\circ$.                     Side view illustrating the dihedral angle.       &...


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It is instructive to work out the vertex coordinates from first principles, then compute dihedral angles from these. 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 ...


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You can construct an infinite class of such polyhedra (all vertices of degree 3 or 7) by starting with the planar tiling by equilateral triangles. Replace one-sixth of the triangular facets by triangular pyramids so that no pyramids share any base vertices; now the vertex of each pyramid has degree 3 and the remaining vertices (three-fourths of the total) ...


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There are many ways to glue the edges you labelled together, depending on whether you preserve or reverse orientation. This has an effect on the resulting manifold. Suppose you glue the edges together in an oriented manner, like this: Then you'll get an orientable surface, with $5$ faces, $10$ edges, and $5$ vertices (note that all "outer" vertices are ...


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The best knowns and easiest such polyhedra are the duals of the prism and of the antiprism with regular heptagonal bases. For other polyhedra with either using regular heptagons or near-miss polyhedra with such faces you might be interested to look into Jim McNeil's Webpage on polyhedra. (Near-miss polyhedra are not a problem as long as your degree 7 ...


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The following is a proof for the only if side of your problem, which requires a good understanding of the duality theorems in linear programming, espeically the complementary slackness thoerem. Proof$\;$ Suppose $F$ is a face of $P$ defined by $c^Tx\le \delta$, i.e., $F=\{x\in P|c^Tx =\delta\}$. Note that $F$ is the set of optimal solutions for linear ...


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