Talk:Rhombic triacontahedron

Is there a name for the 60-sided polyhedron formed when you cut each rhombus into two roughly equilateral triangles? I was trying to find a polyhedron to best describe Spaceship Earth (Disney), and this was the best I could find. But if that other one (equivalent to replacing each side of a dodecahedron with five triangles) has a name, I should call it one of those. --SPUI 18:27, 25 Jan 2005 (UTC)

Found it - pentakis dodecahedron. --SPUI (talk) 01:24, 7 Apr 2005 (UTC)

6-dimensional hypercube project?

I removed this statement until someone can verify it. I added vertex-first orthogonal projection, but not sure about that even. SockPuppetForTomruen (talk) 05:40, 23 May 2008 (UTC) The rhombic triacontahedron forms the (hull of) a [vertex-first orthogonal projection] of a 6-dimensional hypercube to 3 dimensions.

Wayback Machine (archive.or

"The image of a unit cube in 6-space, parallel to the cubes of the lattice of points with integer coordinates, is a rhombic triacontahedron in p-space!" Mekomancer (talk) 16:03, 16 November 2022 (UTC)

Tom, I just verified that indeed the 6-cube projected using the Golden ratio produces a 3D object with an outer shell of the Rhombic triacontahedron.

The basis vectors are: x = {1, φ, 0, -1, φ, 0} y = {φ, 0, 1, φ, 0, -1} z = {0, 1, φ, 0, -1, φ} There are 64 vertices and 192 unit length edges forming pentagonal symmetry along specific axis (as well as hexagonal symmetries on other axis).

This one has the inner edges removed preserving the vertices).

Jgmoxness (talk) 23:39, 2 January 2012 (UTC)

Very nice! Tom Ruen (talk) 23:59, 2 January 2012 (UTC)

Dihedral angle

Dihedral angle is said to be 144 degrees. Mathworld says it is either pi/5 for faces sharing an edge or pi/3 for faces sharing only one point. Who is right ? —Preceding unsigned comment added by Fdecomite (talk • contribs) 12:53, 15 December 2009 (UTC)

Edit : solved : just a different way to count angle (144+36=180), dihedral angle between non-adjacent faces does not seem to be widely used.

Rhombic Triacontahedron TX

Rhombic Triacontahedron TX formed by dodecahedrons, heureka by Jorge Taxa, in 29.3.2003

File:///D:/Documents and Settings/Vanessa Taxa/Desktop/15. diamantetx.jpg

File:///D:/Documents and Settings/Vanessa Taxa/Desktop/18. diamantetx1.jpg Jorgetaxa--Jorgetaxa (talk) 05:38, 6 March 2011 (UTC)--Jorgetaxa (talk) 04:48, 6 March 2011 (UTC) — Preceding unsigned comment added by Jorgetaxa (talk • contribs) 04:32, 6 March 2011 (UTC)

Ball of Whacks

Pretty cool toy. We should have an article on it. It's worth mentioning that five adjacent faces form what Roger von Oech calls a "bow tie", and six of those comprise the "whole ball of wax". That's the "less than obvious relationship between the rhombic triacontahedron and the cube," that it's got octahedral symmetry (citation needed). 68.173.113.106 (talk) 20:33, 20 November 2011 (UTC)

Rhombic Tricontahedron in Nature

Linus Pauling in his largely ignored Spheron Theory of the atomic nucleus posited the rhombic tricontahedron composed of spherons as one of the structures marking the completion of a full shell there. A spheron is NOT a sphere, but a grouping of nucleons. When close packed, these form an approximation of the solid in question. Whether he was correct or not has not been thoroughly examined. 67.81.236.32 (talk) 10:17, 23 November 2011 (UTC)

inclusion, arrangement, contents

The rhombic triacontahedron is also interesting in that its vertices includes the arrangmenmt of all the platonic solids. It contains ten tetrahedrons, five hexahedrons, five octahedrons, an icosahedron and a dodecahedron.

"The arrangement of all the platonic solids" is an obscure phrasing. How about this:

The vertices of the rhombic triacontahedron include those of the compound of ten tetrahedra, the compound of five cubes, the icosahedron and the dodecahedron.

The compound of five octahedra has 30 vertices, which are unlikely to fit neatly into the 32 of the R30, so I'm skeptical of the assertion that five octahedra are "contained". And the other compounds mentioned share their vertices with the dodecahedron, so it's looking less remarkable. —Tamfang (talk) 22:13, 9 September 2014 (UTC)