Hemisphere

A hemisphere is either half of a sphere cut by a plane through its center. Its curved surface has area $2\pi r^2$. Closing that surface with its equatorial disk encloses a half-ball of volume $2\pi r^3/3$. The curved surface and enclosed solid have distinct roles in surface-cost and volume calculations.

For a closed hemisphere of radius $a>0$, take a rectangular parallelepiped with a face on the flat base, base half-widths $x,y>0$ and height $z>0$. Its curved-boundary constraint is $x^2+y^2+z^2=a^2$ and its volume is $4xyz$. A Lagrange multiplier gives $x=y=z=a/\sqrt{3}$, yielding the displayed maximum. The full edge lengths are $2a/\sqrt{3},2a/\sqrt{3},a/\sqrt{3}$.

This bound remains valid when arbitrary orientations are allowed. Write a cuboid's half-edge lengths as $u_1,u_2,u_3$, its center components in its orthonormal edge basis as $v_i$, and the vertical unit-vector components as $n_i$. Containment above the base plane requires $\sum n_i{v}_i\ge \sum |n_i|u_i$. Hence $|v_j|\ge u_j$ for at least one $j$: otherwise every contributing term would be strictly smaller than its required bound. Its farthest vertex has squared distanceContainment in the sphere bounds the right side by $a^2$. The arithmetic-geometric mean inequality then gives $2u_1{u}_2{u}_3\le a^3/(3\sqrt{3})$, so the full volume $8u_1{u}_2{u}_3$ cannot exceed the displayed value. The constructed cuboid attains it. With strict containment and no boundary contact, the same value is a supremum.

For a nonzero vector $u$, the open hemisphere of $S^d$ with direction $u$ is $\{x\in S^d:u\cdot x>0\}$. Its boundary is excluded, so opposite open hemispheres are disjoint.

Choose $N$ points of $S^d$ such that every $d+1$ of their position vectors are linearly independent. Any equatorial hyperplane then contains at most $d$ points. If $N-d\ge 2r-1$, at least one of each opposite pair of open hemispheres contains $r$ points. Such configurations exist: take distinct real parameters $t_i$ and normalize $(1,t_i,\dots ,t_i^d)$; the Vandermonde determinant proves the required independence. This weaker one-of-two guarantee already suffices for the topological lower bound in Lovász theorem on Kneser graphs.

For integers $n=2k+d$ with $k\ge 1$ and $d\ge 0$, there are $n$ labelled points of $S^d$ such that every open hemisphere contains at least $k$ of them. The normalized vectors $(-1{)}^i(1,t_i,\dots ,t_i^d)$ for increasing $t_i$ give a signed moment curve construction. A nonzero polynomial of polynomial degree at most $d$ cannot have enough sign changes for fewer than $k$ positive alternating sample values. Zero sample values can be perturbed negatively by a Lagrange interpolation polynomial. Labels allow repeated positions when $d=0$.