Talk:Great retrosnub icosidodecahedron

Vertex figure confusion
I keep seeing conflicting vertex figures for this one. Which is right? Professor M. Fiendish, Esq. 12:45, 3 September 2009 (UTC)
 * 1) Some sources say the vertfig is 3.3/2.3.5/3.3.
 * 2) Other sources say it is 3/2.3/2.3/2.5/3.3/2.
 * 3) And still others say it is (3.3.3.3. 5/2) /2.


 * That's a mess. The image File:Great retrosnub icosidodecahedron vertfig.png shows the last verf. Tom Ruen (talk) 18:22, 3 September 2009 (UTC)


 * Uh oh. By Wythoff_symbol, and |3/2 5/3 2, verf is 3.3.q.3.p = 3.3.3/2.3.5/3 - first one! Tom Ruen (talk) 18:25, 3 September 2009 (UTC)


 * shows 3.3.3/2.3.5/3 AND my vertex figure diagram which looks like (3.3.3.3.5/2)/2. So it may be correct, BUT something needs explaining. At least 3/2="backwards triangle", 5/3="backwards pentagram", but what effect that has, I can't see now. Tom Ruen (talk) 18:34, 3 September 2009 (UTC)


 * Wenninger's book also shows the same verf image that looks like (3.3.3.3.5/2)/2. It's worth trying to explain, but I can't do that now. Great snub icosidodecahedron has a similar verf 3.3.3.3.5/2, with Wythoff symbol |3 5/2 2, but without double wrapping, so that'a clue! Tom Ruen (talk) 18:54, 3 September 2009 (UTC)


 * My hunch is there's something sinister about Schwarz triangles with retrograde (p/(p-1)) reflection orders. Tom Ruen (talk) 20:07, 3 September 2009 (UTC)


 * Wait a minute, there's a problem with the first one. If you have 3.3.3/2.3.5/3, then the second face is a triangle, and the third one is a retrogade triangle that would coincide with the second one! Professor M. Fiendish, Esq. 02:47, 4 September 2009 (UTC)


 * That interpretation comes from the Wythoff symbol, apparently insufficient to deal with retrograde faces. Tom Ruen (talk) 03:00, 4 September 2009 (UTC)


 * For the second one, I don't think making all the faces retrogade makes them surround the vertex twice. All that does is turn your polyhedron inside-out. You need to have that "/2" at the end. Professor M. Fiendish, Esq. 03:39, 4 September 2009 (UTC)


 * For any Wythoff symbol of the form |p q r, the vertex figure would be of the form p.3.q.3.r.3, which would result in $$\frac32.3.\frac53.3.2.3$$, with the 2 (= digon) collapsing to end up with the equivalent to the first one. DBoffey (talk) 20:00, 18 August 2012 (UTC)
 * The problem with that has already been pointed out above by me (with my old account): {3/2} is a retrograde triangle, which would coincide with the next {3} prograde triangle. So the first vertex figure cannot be right. The vertex figure (3.3.3.3.5/2)/2, however, seems to generate the correct polyhedron (it works in Stella, which generates the uniform polyhedra by their vertex figures). Double sharp (talk) 04:21, 19 August 2012 (UTC)

Zvi Har'El's paper Uniform Solution for Uniform Polyhedra has this sentence: "Let O be the Fermat point of the Schwarz triangle PQR, i.e, the point where the sides subtend equal angles (assuming it exists4)." He then uses this to construct the snub polyhedra. But the note 4 referenced says "The existence of O is verified by a continuity argument, at least for the case where all the angles of PQR are smaller than 2π/3 (this is not the case for (3/2 3/2 5/2) and (3/2 5/3 2))." These two are the Schwarz triangles that generate the small retrosnub icosicosidodecahedron (sirsid, yog-sothoth) and great retrosnub icosidodecahedron (girsid, azathoth) respectively – the very polyhedra where the general vertex configuration derived from the Wythoff construction disagrees with the actual vertex configuration.

When a Euclidean triangle has an angle greater than 2π/3, the Fermat point is situated at the obtuse-angled vertex instead of at the first isogonic center. (Schwarz triangles are spherical, but evidently something also happens to the Fermat point of a spherical triangle once at least one of its angles passes 2π/3.) Har'El neglects this case, that is only useful for sirsid and girsid (the other snub polyhedra that may be so derived are degenerate): is this the reason why the vertex configurations derived for sirsid and girsid from their Wythoff construction don't make sense? What happens when we take this into account? Double sharp (talk) 06:31, 30 March 2014 (UTC)
 * P.S. It's also useful for gike, | 2 3/2 3/2. Double sharp (talk) 15:48, 16 March 2015 (UTC)

See Talk:List of uniform polyhedra for more. Double sharp (talk) 20:12, 16 March 2015 (UTC)

General
It would be nice if the explicit formula for &xi; could be transformed so as not to include any calculations on the complex plane. DBoffey (talk) 20:00, 18 August 2012 (UTC)
 * IIRC, this can't be done in general, but maybe it's possible for this special case. Double sharp (talk) 04:23, 19 August 2012 (UTC)
 * ...nope, it can't be done in this case. Double sharp (talk) 12:47, 15 March 2014 (UTC)
 * Out of curiosity, based on the timestamps, did you really spend 18 months seeing if the roots could be expressed only in terms of real numbers? I can see how the roots are derived from the cubic polynomial equation, but I have no idea how to prove that that's the only possible expression for the roots. -- 66.87.79.145 (talk) 17:28, 28 June 2015 (UTC)
 * No, I forgot about it and came back later. ;-) The equation has three real roots, but none are rational (obviously, since the constant term is 1/τ), so this is the casus irreducibilis. That article I just linked gives a proof that in such a case, though the roots are real, they cannot be expressed using only real radicals, but only complex radicals. You can however express them using real numbers only if you allow the use of the cosine and arccosine functions. But I don't think that's really a standard way to present them, and in my opinion it's even uglier than the complex-radicals presentation as it requires nesting the functions. Double sharp (talk) 19:27, 28 June 2015 (UTC)

Schläfli symbol of this polyhedron
Schläfli symbol of Great retrosnub icosidodecahedron is written s{3/2,5/3} on this page now, but it looks like that it doesn't add up, while Snub dodecahedron is sr{5,3}, and Great snub icosidodecahedron is sr{5/2,3}, Great inverted snub icosidodecahedron is sr{5/3,3}. The correct is $$s \scriptstyle\begin{Bmatrix} 3/2 \\ 5/3 \end{Bmatrix}$$ or sr{3/2,5/3} or sr{5/3,3/2}, isn't it?--円口類 (talk) 05:44, 9 September 2018 (UTC)
 * You're right; I've changed to to sr{3/2,5/3}. This and all the polyhedra you mention are different geometric realisations of the same abstract polytope. Double sharp (talk) 06:24, 9 September 2018 (UTC)
 * Thanks!--円口類 (talk) 06:44, 9 September 2018 (UTC)

"Azathoth (geometry)" listed at Redirects for discussion
An editor has identified a potential problem with the redirect Azathoth (geometry) and has thus listed it for discussion. This discussion will occur at Redirects for discussion/Log/2022 March 31 until a consensus is reached, and readers of this page are welcome to contribute to the discussion. Steel1943 (talk) 19:00, 31 March 2022 (UTC)