Hexagonal ditetragoltriate

The hexagonal ditetragoltriate or hiddet is a convex isogonal polychoron and the fourth member of the ditetragoltriate family. It consists of 12 hexagonal prisms and 36 rectangular trapezoprisms. 2 hexagonal prisms and 4 rectangular trapeozpirsms join at each vertex. However, it cannot be made uniform. It is the first in an infinite family of isogonal hexagonal prismatic swirlchora.

This polychoron can be alternated into a triangular double antiprismoid, which is also nonuniform.

It can be obtained as the convex hull of 2 similarly oriented semi-uniform hexagonal duoprisms, one with a larger xy hexagon and the other with a larger zw hexagon.

Using the ratio method, the lowest possible ratio between the longest and shortest edges is 1:$$\frac{2+\sqrt2}{2}$$ ≈ 1:1.70711. This value is also the ratio between the two squares of the two semi-uniform duoprisms.

Vertex coordinates
The vertices of a hexagonal ditetragoltriate, assuming that the trapezoids have three equal edges of length 1, centered at the origin, are given by:
 * $$\left(0,\,±1,\,0,\,±\frac{2+\sqrt2}{2}\right),$$
 * $$\left(0,\,±1,\,±\frac{2\sqrt3+\sqrt6}{4},\,±\frac{2+|sqrt2}{4}\right),$$
 * $$\left(±\frac{\sqrt3}{2},\,±\frac12,\,0,\,±\frac{2+\sqrt2}{2}\right),$$
 * $$\left(±\frac{\sqrt3}{2},\,±\frac12,\,±\frac{2\sqrt3+\sqrt6}{4},\,±\frac{2+\sqrt2}{4}\right),$$
 * $$\left(0,\,±\frac{2+\sqrt2}{2},\,0,\,±1\right),$$
 * $$\left(0,\,±\frac{2+\sqrt2}{2},\,±\frac{\sqrt3}{2},\,±\frac12\right),$$
 * $$\left(±\frac{2\sqrt3+\sqrt6}{4],\,±\frac{2+\sqrt2}{4},\,0,\,±1\right),$$
 * $$\left(±\frac{2\sqrt3+\sqrt6}{4},\,±\frac{2+\sqrt2}{4},\,±\frac{\sqrt3}{2},\,±\frac12\right).$$