## Saturday, April 7, 2018

### The Dimensions of a Ten-Sided Die: Part II

The dimensions of a pentagonal trapezohedron can be expressed with two free variables. One of them determines the overall size of the polyhedron. The other determines the ratio of the distance between the polar vertices relative to the distance of the other vertices to the center.

Is there a canonical value for the ratio? We could insist all vertices lie in a sphere. This gives the die a round shape and is close to the ratio dice manufacturers use.
Since the vertices are now equidistant to the center we can impose the additional constraint $$n + \frac{h}{2} = \sqrt{\frac{h^2}{4} + m^2}$$ We arbitrarily set m to 1 and solve using Mathematica:

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],
   n + h/2 == Sqrt[h^2/4 + m^2],
   m == 1},
   {k, h, w, z, y, f, g, n, m}] // N


The values are $$k \approx 0.809017 \\ h \approx 0.212332 \\ w \approx 0.285586 \\ z \approx 1.49535 \\ y \approx 1.17557 \\ f \approx 1.345 \\ g \approx 0.653491 \\ n \approx 0.899454 \\ m = 1 \\ \alpha \approx 51.8273^\circ \\ \beta = 90^\circ \\ \gamma \approx 128.173^\circ$$