An enneahedron for Herschel
Christian Perfect
Dr Michael White ↓
It doesn't have $D_6$ symmetry!
Maths!
\begin{align} \mathbf{h}_1 &= \left(0,0,h_1\right) & \mathbf{h}_2 &= \left(0,0,h_2\right) \\ \mathbf{v}_1 &= \left(0,0,0\right) & \mathbf{v}_3 &= \left(1,0,0\right) \\ \mathbf{v}_0 &= \frac{\left(\mathbf{v}_1+\mathbf{v}_3\right)}{2} – \mathbf{h}_1 & \mathbf{v}_2 &= \frac{\left(\mathbf{v}_1+\mathbf{v}_3\right)}{2} + \mathbf{h}_1 \\ &= \left(\frac{1}{2},0,-h_1\right) & &= \left(\frac{1}{2},0,h_1\right) \\ \mathbf{v}_4 &= \frac{\left(\mathbf{v}_1+\mathbf{v}_3+\mathbf{v}_{10}\right)}{3} – \mathbf{h}_2 & \mathbf{v}_5 &= \frac{\left(\mathbf{v}_{10}+\mathbf{v}_3\right)}{2} – \mathbf{h}_1 \\ &= \left(\frac{1}{2}, \frac{\sqrt{3}}{6},-h_2\right) & &= \left(\frac{3}{4},\frac{\sqrt{3}}{4},-h_1\right) \\ \mathbf{v}_6 &= \frac{\left(\mathbf{v}_{10}+\mathbf{v}_3\right)}{2} + \mathbf{h}_1 & \mathbf{v}_7 &= \frac{\left(\mathbf{v}_1+\mathbf{v}_3+\mathbf{v}_{10}\right)}{3} + \mathbf{h}_2 \\ &= \left(\frac{3}{4},\frac{\sqrt{3}}{4},h_1\right) & &= \left(\frac{1}{2}, \frac{\sqrt{3}}{6},h_2\right) \\ \mathbf{v}_8 &= \frac{\left(\mathbf{v}_1+\mathbf{v}_{10}\right)}{2} + \mathbf{h}_1 & \mathbf{v}_9 &= \frac{\left(\mathbf{v}_1+\mathbf{v}_{10}\right)}{2} – \mathbf{h}_1\\ &= \left(\frac{1}{4}, \frac{\sqrt{3}}{4},h_1\right) & &= \left(\frac{1}{4}, \frac{\sqrt{3}}{4},-h_1\right) \\ \mathbf{v}_{10} &= \left(\frac{1}{2},\frac{\sqrt{3}}{2},0\right) \end{align}
It lives!
Alistair, finish your bees post!
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