AI 中文总结
该研究解决了凸几何中9顶点内接球面最大体积多面体的经典问题，结合组合与几何方法，确定了其体积上界及对应构型。
AI 中文摘要
凸几何与离散几何中的一个经典问题是：从单位球面$\boldsymbol{\text{S}}^2$上选取顶点，构造体积最大的凸多面体。对于给定的顶点数$N$，该问题仅在少数情况下有解。本文解决了下一个未解决的情况$N=9$：我们证明，$\text{S}^2$上顶点数不超过9的所有凸多面体的体积最大为$3\boldsymbol{\text{2}\boldsymbol{\text{√3}-3}}$，且在旋转意义下，仅当顶点为明确形状的三增三角棱柱时达到等号。证明结合了组合与几何约化以及精确体积估计。根据Berman和Hanes（《数学年刊》，1970年）的定理，体积最大化者必为单纯形，这将2606种9顶点多面体的组合类型缩减为50种；我们还证明最大化者不能有三价顶点，仅剩下5种组合类型，通过几何与组合方法处理，最终确定了三增三角棱柱类内的精确最大化者并刻画了等号情形。
英文摘要
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.
Comments45 pages, 7 figures
