Space Elves

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TSR

The Dimensions of a Ten-Sided Die

TSR introduced the ten-sided die to gamers in 1981 but the idea of using a pentagonal trapezohedron as a die was first patented in 1906.

The pentagonal trapezohedron is face-transitive and each face is parallel to a face on the opposite side. Is the pentagonal trapezohedron the only ten-sided polyhedron with these properties? The pentagonal bipyramid is face-transitive but lacks parallel, opposite sides. The right, regular-octagonal prism has parallel, opposite sides but isn't face-transitive.

The pentagonal trapezohedron isn't vertex-transitive. Two of the vertices, which we can think of as the poles, touch 5 faces each. The remaining 10 vertices are arranged in two parallel pentagons and touch 3 faces each.

In the diagrams above a few labels are provided as an aid to figuring out the dimensions of the pentagonal trapezohedron. In the first diagram the dimensions n and h are along the axis connecting the two poles of the pentagonal trapezohedron. The left diagonal is the central spine of one of the kite-shaped faces, labeled z in the bottom two diagrams. If we rotate the trapezohedron around the axis 36°, we can bring the central spine of an adjacent face into the position of the right diagonal.  The non-polar vertices are in the planes perpendicular to the axis indicated by k or m.

If we rotate the polyhedron so that the axis is perpendicular to the plane of the page, then the second diagram shows how the non-polar vertices are arranged.

By similar triangles in the first diagram this holds $$\frac{k}{n} = \frac{m}{n+h}$$ From the second diagram and trigonometry $$\frac{k}{m} = \cos{\frac{2 \pi}{10}}$$ The remaining equations in the Mathematica code that follows can be justified by similar triangles, trigonomety, or the Pythagorean theorem:

  Solve[{k/n == m/(n + h), k/m == Cos[2*Pi/10], w/z == h/(n + h),
    y/(2*m) == Sin[2*Pi/10], z == Sqrt[m^2 + (n + h)^2],
    f == Sqrt[(z - w)^2 + (y/2)^2], g == Sqrt[w^2 + (y/2)^2]},
    {k, h, w, z, y, f, g}]

$$k = \frac{1 + \sqrt{5}}{4}m \\ h = (2 + \sqrt{5})n \\ w = \frac{3\sqrt{m^2 + (6-2\sqrt{5})n^2}}{4} \\ z = \sqrt{m^2 +(6 - 2\sqrt{5})n^2}  \\ y = \sqrt{\frac{1}{2} \left(5-\sqrt{5}\right)} m \\  f = \sqrt{m^2 + n^2} \\ g = \sqrt{\left(9-4 \sqrt{5}\right) n^2-\frac{1}{2} \left(\sqrt{5}-3\right) m^2} \\ \alpha = 2 \sin ^{-1}\left(\frac{\sqrt{\frac{1}{2} \left(5-\sqrt{5}\right)} m}{2    \sqrt{m^2+n^2}}\right) \\ \gamma = 2 \cos ^{-1}\left(-\frac{\left(\sqrt{5}-3\right) \sqrt{m^2-2 \left(\sqrt{5}-3\right) n^2}}{2    \sqrt{4 \left(9-4 \sqrt{5}\right) n^2-2 \left(\sqrt{5}-3\right) m^2}}\right) \\ \beta = \frac{2 \pi - \alpha - \gamma}{2}$$ We are free to choose m and n to be any positive values.