Hemiobelisk

A hemiobelisk is one half of an obelisk, or elongated square pyramid. It is obtained by cutting an equilateral obelisk with unit edge lengths along a plane passing through opposite corners of its square base and the apex of the pyramid.

It is implemented in the Wolfram Language as PolyhedronData["Hemiobelisk"].

This solid is one of two having 7 vertices, 11 edges, and 6 faces (the other being the hemicube) and is an example of one of the 7 topological classes of convex hexahedron.

When constructed from a unit elongated square pyramid, the solid consists of two unit squares, an isosceles right triangle with unit base lengths, two unit equilateral triangles, and an isosceles right pentagon with base length and all other sides of length 1.

Pegg (2026) gave a particular hexahedron with the same arrangement of faces as a hemiobelisk and Heesch number at least 1 in three dimensions. His construction has 24 surrounding congruent copies. The vertices of this hexahedron can be taken as

This hexahedron is one of three congruent pieces in a polyhedron dissection of a heptahedron. The heptahedron is obtained by attaching a regular tetrahedron to a triangular face of the space-filling polyhedron formed by truncating a regular tetrahedron at each vertex, one quarter of the way along each edge (Pegg 2026). In the coordinates above, each hexahedron has volume .

This hexahedron has nonzero Dehn invariant, so it is not a space-filling polyhedron. The edges having dihedral angles that are rational multiples of contribute zero, while the remaining edges give

where denotes the tensor product over the rationals, denotes the equivalence class of the dihedral angle modulo rational multiples of , and is a basis angle used in the Dehn invariant entry. Since is not an allowed rational cosine at a rational multiple of by Niven's theorem, is irrational, so and therefore are nonzero.

Other hexahedra with the face arrangement of a hemiobelisk are space-filling hexahedra. Examples arise by dissecting the quarter-edge truncation of a regular tetrahedron into four or twelve congruent pieces (E. Pegg, Jr., pers. comm., Oct. 23, 2026). Their ability to tile space therefore differs from that of the particular hexahedron above.

The skeleton of the hemiobelisk is the hemiobelisk graph.