The maximum volume polytope with nine vertices inscribed in the sphere
A classical problem in convex and discrete geometry asks for the convex polyhedron of greatest volume whose vertices are chosen from the unit sphere $\mathbb{S}^2$. For a prescribed number $N$ of vertices, the problem is known only in a small number of cases. In this paper we resolve the next outstanding case, $N=9$. We prove that every convex polyhedron with at most nine vertices on $\mathbb{S}^2$ has volume at most $3\sqrt{2\sqrt{3}-3}$, with equality, up to rotation, precisely for a triaugmented triangular prism of an explicitly determined shape. The proof combines combinatorial and geometric reductions with sharp volume estimates. By a theorem of Berman and Hanes (Mathematische Annalen, 1970), a volume maximizer must be simplicial, reducing the $2,606$ combinatorial types of $9$-vertex polyhedra to $50$. We prove that a maximizer cannot have a trivalent vertex, leaving only five combinatorial types, which are treated using geometric and combinatorial arguments. In particular, we determine the exact maximizer within the triaugmented triangular prism class, and characterize the equality case.