Rich writes: << There's a plane geometry theorem that the sum of the vertex angles of a polygon is pi*(#vertices-2). Is there a 3D analog?

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In short, the external spherical angles at all vertices of a polyhedron K must add up to 4*pi. (As in 2D this can be thought of as a limiting case of the embedded Gauss-Bonnet theorem for smooth surfaces in 3D (resp. smooth curves in 2D). Approximate a polyhedron K by an (internal) smooth surface M. M will have a Gauss map G: M -> S^2 -- G(x) = the point of S^2 having its outward unit normal = the same vector as the outward unit normal of M at x. Let JDG: M -> R be the Jacobian determinant, i.e. the factor by which G magnifies area at each point of. Then Gauss-Bonnet tells us that the (integral over M of JDG dA) = 2pi*X(M), where X(M) = its Euler characteristic = 2 for K a topological sphere. Back to K: As M approaches K smoothly (at least in the C^3 topology) the Gauss map is no longer well-defined at edges or vertices of M. But its image can be replaced by the set of outward unit normals to all planes P in space that (assuming K convex) keep K inside of one of the closed half-spaces defined by P. The area of the spherical image of all the unit normals must still be 2pi*X(K) = 4*pi. This is the sum over all vertices of the external spherical angle. It's a nice little exercise to calculate the formula for the spherical angle at a vertex where 3 edges meet, whose unit edge vectors (i.e., vectors leaving the vertex in the directions of the edges it's in) are say u,v,w. --Dan