Geometry of a d10

Question

A d10, in roleplayers' lingo, is a 10-sided dice.

Several different shapes for this item exist, but the most common is that of two 5-sided pyramids with their equilateral bases lying on the same plane and offset by 36° (1/10, or half a side of a pentagon), as shown in the lower part of the image:

Both pyramids have their apex on the vertical of A, and the upper part of the drawing (shown at a different zoom below, so that all letters can be read) is built upon points B and C for the upper pyramid (projected: H and J) and I and F for the lower one (projectet: K and G).

It is trivial to see that the x component of AC is ABcos(36°)

Now, the planes defined by the 10 side faces of the pyramids meet on a jagged line that goes up and down the plane where the original bases lie. I suppose the meeting point between a face and the edge of an opposite pyramid are N and P but I'm not sure I'm right.

Suppose that ∠HAJ (and therefore also ∠KLG) is 90° (then, because of simmetry, ∠LPM and ∠LNM are 90° too), what's the LN/LP ratio?

I'm sure I can calculate ∠JHL and ∠HJL from the ratio between the sides (which is the same as the ratio between the x component of AC and and ABcos(36°)) and that I could get JL and HL from there but I can't think about a way to find NJ and PH.

Images created with geogebra

Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. — José Carlos Santos, Feb 02 '18 at 18:57
@JoséCarlosSantos I'm currently making the effort. I think I'll be able to post a partial process in a day or so. With pictures. — Zachiel, Feb 02 '18 at 19:35
I think it should be straightforward, though perhaps tedious, to deal with the jagged seam at the “equator” by examining the intersections of the planes of an upper face and its two neighboring lower faces. — amd, Feb 02 '18 at 21:06
And as conjectured here for this ratio the kite indeed has a right angle. — WimC, Feb 02 '18 at 21:38

score 2 · Answer 1 · answered Feb 03 '18 at 12:17

Not a very geometric solution but it gets the job done. Take three consecutive points on the jagged equator: $$(\cos\frac{\pi}5, -\sin\frac{\pi}5, x), (1,0,-x), (\cos\frac{\pi}5, \sin\frac{\pi}5, x)$$ where $x$ is some still to be determined deviation. The plane through these points determines the apex at $(0,0,(5+2\sqrt 5) x)$. Now $x$ needs to be taken such that the three points $$(-1, 0, x), (0,0,(5+2\sqrt 5) x), (1, 0, -x)$$ in the $XZ$ plane (vertex apex vertex) form a right angle. This gives $$(4+2\sqrt 5)(6+2\sqrt 5)x^2 = (44 + 20 \sqrt 5)x^2=1.$$ For this value of $x$ the length of the long edge is $\sqrt{1+(4+2\sqrt 5)^2x^2}=\tfrac12\sqrt{5+\sqrt5}$ and of the long diagonal $\sqrt{1+(6+2\sqrt 5)^2x^2} = \sqrt{\sqrt 5}$. The ratio long diagonal to long edge is therefore $\sqrt{\sqrt 5 -1}$.

Geometry of a d10

1 Answers1