## Saturday, March 31, 2018

## Thursday, March 22, 2018

## Saturday, March 17, 2018

## Sunday, March 11, 2018

### 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

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:

$$ 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

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.
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