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  • Omar Antolínoantolin@mathstodon.xyz
    May 29, 2026, 6:02 AM

    I learned a cool Euclidean geometry fact this week in our colloquium, in a talk by Benoît Bertrand of the University of Toulouse on joint work of his with my UNAM colleague Lucía López de Medrano. They don't think this geometry fact can be new, but also haven't been able to find a reference. The theorem is a generalization of the fact that the sum of the angles of a triangle is 180° to higher dimensions.

    This generalization is not that the sum of the solid angles of a tetrahedron is constant: that's not true! If you think of tetrahedra that are almost flat, that is, whose vertices are nearly coplanar, there are two types: (1) the vertices are nearly the vertices of a convex planar quadrilateral, or (2) the vertices are nearly a triangle with a point inside it. In case (1) the solid angles are close to 0; in case (2) the one at the central vertex is close to half a sphere (2π steradians), and the others are close to 0. It turns out that those are the extremes and the sum of the solid angles of a tetrahedron is always between 0 and 2π steradians.

    So how does their generalization go? Consider an n-dimensional simplex and select one of its vertices, say P. Now consider all ways coloring the vertices of the simplex with either red or blue in such a way that P is blue (this is half of all colorings). For each coloring look at the vectors of the form R-B where R is some red vertex and B is a blue one, and take the cone they generate (that is, the linear combinations of those vectors with non-negative coefficients). Their theorem says that those cones tile a half-space! (Here "tile" means the interiors are disjoint and the union of the cones is a half-space.)

    Next, a proof. 1/2

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  • Omar Antolínoantolin@mathstodon.xyz
    May 29, 2026, 6:10 AM

    Oh, I almost forgot! Exercise: show that for a triangle this implies the classical theorem on the sum of angles of a triangle.

    Here goes my proof of their result, which is not exactly the same as the proof Bertrand gave in his talk (that one inducts on the dimension):

    First of all, the points of these cones are what are called "affine combinations" of the vertices of the simplex, that is, linear combinations whose coefficents sum to 0. And the vertices of a simplex are affinely independent, which means you can talk about the coefficient of a vertex in such a combination and it is uniquely determined.

    Now those cones have disjoint interiors because, given an interior point X of one of the cones, you can figure out exactly which one it is: the cone corresponding to the coloring whose blue points are the vertices with a negative coefficient in X!

    Finally, the union of the cones consists of all affine combinations of the vertices of the simplex in which the coefficient of the special vertex P is negative. That's a half-space! in the coordinates given by these coefficients it is given by the system x₁ + x₂ + ... + xₙ = 0, x₁ < 0.

    2/2

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