Hendecagonal-snub dodecahedral duoprism

The hendecagonal-snub dodecahedral duoprism or hensnid is a convex uniform duoprism that consists of 11 snub dodecahedral prisms, 12 pentagonal-hendecagonal duoprisms, and 80 triangular-hendecagonal duoprisms of two kinds. Each vertex joins 2 snub dodecahedral prisms, 4 triangular-hendecagonal duoprisms, and 1 pentagonal-hendecagonal duoprism.

Vertex coordinates
The vertices of a hendecagonal-snub dodecahedral duoprism of edge length 2sin(π/11) are given by all even permutations with an odd number of sign changes of the last three coordinates of: as well as all even permutations with an even number of sign changes of the last three coordinates of: where
 * $$\left(1,\,0,\,\phi\sqrt{\phi\left(\xi-1-\frac1\xi\right)}\sin\frac\pi{11},\,\xi\phi\sqrt{3-\xi^2}\sin\frac\pi{11},\,\phi\sqrt{\xi(\xi+\phi)+1}\sin\frac\pi{11}\right),$$
 * $$\left(\cos\frac{j\pi}{11},\,±\sin\frac{j\pi}{11},\,\phi\sqrt{\phi\left(\xi-1-\frac1\xi\right)}\sin\frac\pi{11},\,\xi\phi\sqrt{3-\xi^2}\sin\frac\pi{11},\,\phi\sqrt{\xi(\xi+\phi)+1}\sin\frac\pi{11}\right),$$
 * $$\left(1,\,0,\,\phi\sqrt{3-\xi^2}\sin\frac\pi{11},\,\xi\phi\sqrt{1-\xi+\frac{1+\phi}\xi}\sin\frac\pi{11},\,\phi\sqrt{\xi(\xi+1)}\sin\frac\pi{11}\right),$$
 * $$\left(\cos\frac{j\pi}{11},\,±\sin\frac{j\pi}{11},\,\phi\sqrt{3-\xi^2}\sin\frac\pi{11},\,\xi\phi\sqrt{1-\xi+\frac{1+\phi}\xi}\sin\frac\pi{11},\,\phi\sqrt{\xi(\xi+1)}\sin\frac\pi{11}\right),$$
 * $$\left(1,\,0,\,\xi^2\phi\sqrt{\phi\left(\xi-1-\frac1\xi\right)}\sin\frac\pi{11},\,\phi\sqrt{\xi+1-\phi}\sin\frac\pi{11},\,\sqrt{\xi^2(1+2\phi)-\phi}\sin\frac\pi{11}\right),$$
 * $$\left(\cos\frac{j\pi}{11},\,±\sin\frac{j\pi}{11},\,\xi^2\phi\sqrt{\phi\left(\xi-1-\frac1\xi\right)}\sin\frac\pi{11},\,\phi\sqrt{\xi+1-\phi}\sin\frac\pi{11},\,\sqrt{\xi^2(1+2\phi)-\phi}\sin\frac\pi{11}\right),$$
 * $$\left(1,\,0,\,\xi^2\phi\sqrt{3-\xi^2}\sin\frac\pi{11},\,\xi\phi\sqrt{\phi(\xi-1-\frac1\xi)}\sin\frac\pi{11},\,\frac{\phi^2\sqrt{\xi(\xi+\phi)+1}\sin\frac\pi{11}}\xi\right),$$
 * $$\left(\cos\frac{j\pi}{11},\,±\sin\frac{j\pi}{11},\,\xi^2\phi\sqrt{3-\xi^2}\sin\frac\pi{11},\,\xi\phi\sqrt{\phi(\xi-1-\frac1\xi)}\sin\frac\pi{11},\,\frac{\phi^2\sqrt{\xi(\xi+\phi)+1}\sin\frac\pi{11}}\xi\right),$$
 * $$\left(1,\,0,\,\sqrt{\phi(\xi+2)+2}\sin\frac\pi{11},\,\phi\sqrt{1-\xi+\frac{1+\phi}\xi}\sin\frac\pi{11},\,\xi\sqrt{\xi(1+\phi)-\phi}\sin\frac\pi{11}\right),$$
 * $$\left(\cos\frac{j\pi}{11},\,±\sin\frac{j\pi}{11},\,\sqrt{\phi(\xi+2)+2}\sin\frac\pi{11},\,\phi\sqrt{1-\xi+\frac{1+\phi}\xi}\sin\frac\pi{11},\,\xi\sqrt{\xi(1+\phi)-\phi}\sin\frac\pi{11}\right),$$
 * j = 2, 4, 6, 8, 10,
 * $$\phi = \frac{1+\sqrt5}2,$$
 * $$\xi = \sqrt[3]{\frac{\phi+\sqrt{\phi-\frac5{27}}}2}+\sqrt[3]{\frac{\phi-\sqrt{\phi-\frac5{27}}}2}.$$