# Talk:Disdyakis triacontahedron

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Field:  Geometry

Remove calculation comment: Tom Ruen 08:40, 29 July 2006 (UTC)

• Side lengths for r=12:

s1=4.35411996 s2=6.55839670 s3=7.69022184 (Calculated by Garrith McLean)

## Hexakis icosahedron?

Why's it also called a hexakis icosahedron? Professor M. Fiendish, Esq. 02:53, 2 September 2009 (UTC)
That's what Williams calls it, I guess each triangular face of the icosahedron is divided into 6 faces. Tom Ruen (talk) 03:03, 2 September 2009 (UTC)
Holden (Shapes, space, and symmetry, 1971) calls it that too. [1] Tom Ruen (talk) 03:08, 2 September 2009 (UTC)

## Sides?

Something is wrong with the edges. In the pictures, you clearly see that the sides are right-angled. In the infobox however; It says that the edges are 4, 6 and 10. Whith the pythagorean theorem we see:

${\displaystyle a^{2}+b^{2}=c^{2}={\sqrt {a^{2}+b^{2}}}=c={\sqrt {4^{2}+6^{2}}}\approx 7.21110255093}$

That means that it is not 4, 6, 10. Does anyone know what it is? --Berntisso (talk) 19:39, 7 September 2011 (UTC)

First, they are not right-angled. They are only right-angled if you project them at a right angle. Secondly, the face configuration (4,6,10) is not referring to edge lengths; they are referring to how many faces surround each type of vertex.—Tetracube (talk) 20:09, 7 September 2011 (UTC)
Hmmm... the faces have right angles only as a spherical tiling with spherical triangle faces. There you can extract the angles from the V4.6.10 notation; internal angles are 360/4, 360/6, 360/10 or 90,60,36. The polyhedral face angles are a bit smaller - specifically 88d58'31", 58d14'17",32d46'12" (adding to 180 degrees) by Robert Williams. SockPuppetForTomruen (talk) 22:01, 7 September 2011 (UTC)
So, which are the proportions of the lenghts of the triangle? If the ratio is irrational, is there a good integer approximation? --RokerHRO (talk) 15:41, 10 February 2015 (UTC)

## Twisty puzzles

Whether this figure can be a puzzle mechanism is currently the biggest unsolved problem in the area of mechanical puzzles. In the realm of twisty puzzles, it's known as "big chop". EdPeggJr (talk) 18:44, 24 August 2015 (UTC)

I moved the statement into a new Uses section. It would be good to get more sources for the claims. Tom Ruen (talk) 19:54, 24 August 2015 (UTC)
The outer shape for a twisty puzzle is very flexible. If a good mechanism was found, many variations of this would be made. It's not particularly the dodecahedral shape that's wanted, it's the big chop mechanism. I'd want the sphere version myself. But currently it's an unsolved problem. Tens of thousands of twisty puzzle mechanisms have been developed, but this unsolved one is the biggie. 98.212.151.160 (talk) 01:46, 25 August 2015 (UTC)

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