Talk:Regular icosahedron

Stellations

The stellations of the icosahedron are described in University of Toronto Studies Number 6 - The Fifty-Nine Icosahedra - by HSM Coxeter, P Du Val, HT Flather, and JF Petrie - University of Toronto Press 1938 (Derek Locke)

New stat table

I replace stat table with template version, which uses tricky nested templates as a "database" which allows the same data to be reformatted into multiple locations and formats. See here for more details: User:Tomruen/polyhedron_db_testing

Tom Ruen 00:54, 4 March 2006 (UTC)

volume vs dodecahedron

(Comment from Trevor: I'd LOVE to see a proof on this. Til then, I don't buy it.) —Preceding unsigned comment added by Btrevoryoung (talk • contribs) 12:38, 14 June 2006

If you refer to the table of volumes in Platonic solid it is relatively easy to calculate. R, the circumradius, corresponds to the radius of the sphere that the polyhedron is inscribed in. If you do some calculations you will find that (volume of dodecahedron with circumradius R)/(volume of sphere with radius R) is greater than (volume of icosahedron with circumradius R)/(volume of sphere with radius R). This is an alternative way of explaining what the article states. This may be counter-intuitive to some because it is unlike the similar situation in regards to circles and polygons. I will leave the math to you.

Proofs for Surface Area and Volume

Proofs for the surface area and volume of an icosahedron can be found here: http://mathworld.wolfram.com/Icosahedron.html David Mitchell 17:19, 31 October 2006 (UTC)

Icosahedron in 2-D

Shouldn't an icosahedron in two dimensions be a pentagon? Because if you take the tetrahedrons vertex figur, you'll see that it is a triangel, which is in the same family as the tetrahedron. It's the same thing with the octahedron. Another proof is its dual, the dodecahedron, which has pentagonal faces. I couldn't find any of this in the article, so i brought it up here.Chagi 14:33, 28 April 2007 (UTC)

A 2D analogy of the icosahedron? You could consider either facets or vertex figures. If considering vertex figures, the cube {4,3} and dodecahedron {5,3} are then 3D extensions of the triangle {3,3}, like the tetrahedron. And the octahedron {3,4} extends from a square, and icosahedron {3,5} from a pentagon. I don't see much value in this overall. If you have any geometry books that talk about this, feel free to expand this idea here. Tom Ruen 21:31, 28 April 2007 (UTC)

What I ment was, polyhedra (platonic solids only?) with triangles as faces use their vertex figure as a two-dimensional analog of themselfs, and since the icosahedron has triangles, then why would it be diffrent? While the cube in other hands, uses it face as a 2-d analog, then shouldn't the dodecahedron do the same? Chagi 22:20, 28 April 2007 (UTC)

I wasn't aware that "polyhedra ... use their vertex figure as a two-dimensional analog of themselfs". Must everything have an analogue in every domain? —Tamfang 01:13, 2 May 2007 (UTC)

I didn't say that everything must have a analog, but if it's possible to figure something out, why not do it? Secondly, I wasn't saying that polyhedra uses their vertex figure as a two-dimensional analog, I was suggesting that polyhedra with equilateral triangles use their vertex figurwe as a 2D analog.Chagi 19:37, 8 May 2007 (UTC)

---

One relationship which does exist between 2-D and 3-D is polyhedra and tesselations --

• Triangle series:
  • 3 triangles meeting at a vertex -- tetrahedron
  • 4 triangles meeting at a vertex -- octahedron
  • 5 triangles meeting at a vertex -- icosahedron
  • 6 triangles meeting at a vertex -- regular triangular tesselation of the plane
• Square series:
  • 3 squares meeting at a vertex -- cube
  • 4 squares meeting at a vertex -- regular square tesselation of the plane
• Three polygons at a vertex series:
  • 3 triangles meeting at a vertex -- tetrahedron
  • 3 squares meeting at a vertex -- cube
  • 3 pentagons meeting at a vertex -- dodecahedron
  • 3 hexagons meeting at a vertex -- regular hexagonal tesselation of the plane
• Four polygons at a vertex series:
  • 4 triangles meeting at a vertex -- octahedron
  • 4 squares meeting at a vertex -- regular square tesselation of the plane

There are some similar relationships between semi-regular tesselations and Archimedean polyhedra... AnonMoos (talk) 14:24, 15 May 2009 (UTC)

Irregular icosahedra

I wonder why the information about non regular icosahedron is present on the page. Are they popular? The article describes only regular icosahedron and its properties, so if irregular icosahedra are important to be mentioned, let the article be renamed to Regular icosahedron to reflect its actual content. With respect, Animist (talk) 07:40, 29 May 2009 (UTC)

Done. Better five years late than never. — Cheers, Steelpillow (Talk) 11:18, 18 September 2014 (UTC)

Questions (and some answers)

Question: In the article it is written,

"The matrix $A+{\sqrt {5}}I$ induces thus an Euclidean structure on the quotient space $\mathbb {R} ^{6}/\ker(A+{\sqrt {5}}I)$."

What is the inner product for two elements in this space? Is it,

$\langle x,y\rangle =x^{T}(A+{\sqrt {5}}I)^{T}(A+{\sqrt {5}}I)y$

for two vectors $x,y\in \mathbb {R} ^{6}$ (under the equivalence relation $x_{1}\backsim x_{2}$ iff $x_{1}-x_{2}\in \ker(A+{\sqrt {5}}I)$)?

Answer (sort of): Letting $v_{1},\ldots ,v_{3}$ be an orthonormal basis for $\ker(A+{\sqrt {5}}I)^{\perp }$ and $V$ be the matrix whose columns are these vectors, the vertices of the icosahedron are the rows of$V$ (and their negations). This can be verified by computing $VV^{T}$; the resulting matrix has 0.5 on the diagonal (so all points are equidistant from the origin) and plus-or-minus 0.2236 on the off-diagonal (so all dot products and hence all angles are equal).

Nevertheless, I'd appreciate some more motivation for the example given. What would even lead one to think about the given matrix? To what extent do the ideas here generalize? What is the geometric meaning?

Answer (partial): It looks like the given matrix is a Seidel adjacency matrix. See the paper Large equiangular sets of lines in Euclidean space. The corresponding graph is a pentagram graph with an extra vertex that is not adjacent to any other vertex. I'd still like to know if this graph has a more intuitive geometric meaning -- e.g., if nodes and edges correspond to geometric objects somehow...

130.207.237.143 (talk) 21:05, 29 July 2009 (UTC)

Other Facts:

Item two of the other facts section states:

If one were to color the icosahedron such that no two adjacent faces had the same color, one would need to use 3 colors.

Could someone please prove this to me with a picture of the object? 98.17.232.45 (talk) 10:39, 28 March 2010 (UTC)

Here is the picture. Skarebo (talk) 11:27, 28 March 2010 (UTC)

How many solutions are there? Anyone know? —Tamfang (talk) 10:56, 29 March 2010 (UTC)

Wow! I feel stupid. Thank you all for your patience with me. I'm sorry for the outburst. 71.28.209.241 (talk) 07:21, 29 March 2010 (UTC)

Another thing on the "other facts" section. Why is it strange that an icosohedron occupies less space in a circumsphere than a dodecahedron? Is it because the dodecahedron would have faces of a greater area than the icosohedron? telewatho (talk) 19:11, 14 April 2010 (UTC)

Yes.

Is it correct that an octahedron has 12 edges? Does the article mean to say "duodecahedron"?Mk5384 (talk) 16:42, 26 June 2010 (UTC)

Yes. No. —Tamfang (talk) 23:53, 26 June 2010 (UTC)

Rotating model

The icosahedron model on the right is not rotating for me, even if I click it. The French version has a model that is rotating correctly, maybe someone who knows how (I don't) can replace the English language picture with the French language page's animated version. —Preceding unsigned comment added by 140.77.192.140 (talk) 10:30, 22 December 2010 (UTC)

I think somebody on purpose replaced it with the static version, maybe to reduce "flickering" and improve the loading time, or maybe the images of the article give a clear illustration of the shape. Materialscientist (talk) 10:34, 22 December 2010 (UTC)

You have to click the caption (where it says "(Click here for rotating model)"). I think Materialscientist is correct, and the rotating version is far too intrusive for the article. On Wikipedia, if you click the image, you (normally) are sent to the image information page, i.e., the same and non-rotating image. Johnuniq (talk) 01:16, 23 December 2010 (UTC)

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How come it's the last platonic solid?

Okay imagine splitting every face of an icosahedron into 9 equilateral triangles with flexable edges like [1]. Now imagine a sphere in the center of this shape getting inflated increasingly larger untill all of these triangles are puffed out to an equal degree, whereas the sphere eventually stops inflating any furthur when it can't do so without detaching any edges or veracies or bending any faces out of shape. Now what should be left is a polyhedron with 180 equilateral faces with equal sides, edges, vertexes and angles. Why isn't this geometrically solid shape a platanic solid? [2] Robo37 (talk) 10:33, 22 November 2011 (UTC)

I didn't follow any of the links, but from your description I can tell you it's because in some places six triangles meet at a vertex, while in other places five triangles meet at a vertex (the same answer as for many forms of geodesic dome...). -- AnonMoos (talk) 13:09, 22 November 2011 (UTC)

Ah, I see. Robo37 (talk) 13:28, 22 November 2011 (UTC)

If you classify the "last" as the regular polyhedron with the most faces, then it is the last. But the dodecahedron has more vertices and the secret of its construction was the "last" historically. The simple answer to the question of why the icosahedron has the maximum number of faces of all platonic solids is because the equilateral triangle is the regular polygon with the smallest internal angle (60 degrees). Therefore, it is the face which can be joined together about a point the maximum number of times before the next triangle edge meets the triangle edge of the first. I mean considering a square, only 4 can be revolved about a point before the edges of the squares meet up. If you take 6 equilateral triangles, they will join up perfectly on a plane (60 x 6 = 360). However, there will be no peak in 3 dimensions and so 6 cannot be used to generate a polyhedron. The maximum number of equilateral triangles with their vertices joined together about a point while still retaining a peak in 3d dimensions is 5. Five equilateral triangles about a point is what makes a icosahedron. Therefore the icosahedron is the platonic solid with the maximum number of faces. Also note, that 4 equilateral triangles about a point is the regular octahedron and 3 about a point is the regular tetrahedron. 2 equilateral triangles about a point does not make a polyhedron because such a construction would simple be 2 equilateral triangles back to back and therefore would just be an equilateral triangle in itself without volume. All platonic solids can simply be regarded as 3,4 and 5 equilateral triangles joined about a point (tetrahedron, octahedron and icosahedron) and the other 3 (cube, dodecahedron and tetrahedron again) are simply the dual polyhedrons generated by shaving the vertex points down to its bisection (half the length of an edge). — Preceding unsigned comment added by Peawormsworth (talk • contribs) 00:42, 10 July 2012 (UTC)

Missing orthoganal projection

I noticed that there are 6 possible orthoganal projections for the icosahedron and only 5 are shown. Specifically, the 'edge' projection has only one graph. I think I know how this can be remedied.

First the image of the projection of the edge view should be rotated by 45 degrees in either direction (although I think it should be clockwise). Second, the same image can be rotated 45 degrees in the opposite direction (counter-clockwise) and then pasted into the empty projection spot under the edge view column.

The reason for this is that the projection can be displayed in any rotation perpendicular to the field of view. However, by rotating it 45 degrees, the projection can be then rotated by 90 degrees twords the viewer as the other projections are. The result of this 90 degree rotation is that the image remains the same only it appears as if the image was rotated by 90 degrees horizontal to the viewer.

In any case, I do think there are two separate graph projections that should be displayed. Even if the images are the exactly the same apart from their rotation, it is useful information to let the reader know that such a rotation results in basically the same graph. Leaving it blanks led me to wonder if there was no such rotation possible. But Im sure there is, even if it appears somewhat less unique then the rotations resulting along the A2 and H3 planes. — Preceding unsigned comment added by Peawormsworth (talk • contribs) 01:02, 10 July 2012 (UTC)

Alternate expressions for the circumsphere

I believe it would be useful to express the circumscribed sphere radius in terms of the golden rectangle. Since the golden rectangle splits the icosahedron into 2 along opposite edges, we can see that the diameter is the diagonal of the golden rectangle. The radius is simply 1/2 of this length. Therefore, I think it would be useful to describe the circumscribed sphere radius in terms of 1/2 of the hypotenuse of a right triangle having sides of 1 and Phi. The current formula used "root 5 time Phi" under the root. But I think this should be changed to (or added to include) "Phi^2 + 1" ... and by that I mean "phi squared plus 1".

So the formula would be: ru = (a/2) * root(Phi ^ 2 + 1)

.. sorry the math equation above is not formatted properly. I hope u can infer what I mean.

Primarily, I am saying the the circumscribed sphere radius and the edge has a direct relation to the golden rectangle as 1/2 the diagonal. I think it is important to draw this connection by providing this formula directly under Dimensions category or adding a note to it under the Cartesian Coordinates section.

The current formula telling me that the radius is related to "root 5 times Phi" has no meaning to me. But "Phi squared plus 1" is easily interpreted as "Phi + 2" in my mind. And I know this to be about 3.618 right away. By describing it in this way I can see the root portion as a number and as the diagonal of a right triangle without any problem.

Anyhow... I am not making the change because I am not sure how others would feel about this or which area it should be put under. I hope u will consider this more and make a decision to change this for me. — Preceding unsigned comment added by Peawormsworth (talk • contribs) 01:16, 10 July 2012 (UTC)

Two regular icosahedra

There are two regular icosahedra, one convex and the other star - the great icosahedron.

Should this article be renamed the Convex regular icosahedron or similar? — Cheers, Steelpillow (Talk) 13:19, 24 November 2014 (UTC)

No. I think qualifiers can have implicit precendence. At best we'd call the second a regular star. But perhaps a {{Also|great icosahedron}} on top is fair. Tom Ruen (talk) 13:45, 24 November 2014 (UTC)

Shouldn't the other Platonic solids (except the cube) also get the regular disambiguation? Double sharp (talk) 07:38, 3 January 2015 (UTC)

I have started a wider discussion which embraces this issue at Talk:Regular_polyhedron#Inconsistent_treatment_of_the_regular_polyhedra. — Cheers, Steelpillow (Talk) 11:52, 3 January 2015 (UTC)

"Five sets of three concentric golden rectangles"

This might not be the place to ask this, but I went through and formed all the cyclic permutations of (0, ±1, ±ϕ), and ended up with 12 permutations.

How do you divide this into five sets?

Also, please correct me if I'm wrong, but all the permutations of this set are cyclic correct?

I'm planning on taking abstract algebra at some point, but I really just read some lecture pdfs about cyclic permutations so that is about the limit of my knowledge.

Thanks! — Preceding unsigned comment added by 65.128.179.29 (talk) 00:27, 14 August 2016 (UTC)

When does regular becomes unregulated?

Does anyone have thoughts on Buckminister Fuller and the Icosahedron as related to his Design Science of World Game? Perhaps it's a stretch, but in-terms of opening up the dymaxion, there needs to be some "re-freshing" even to as Bucky's Beautiful Mind. All ideas welcome, all ideas shall be respected and given due reference in full of the referrer and/or proposer. Many Thanks Stumbling Monk (talk) 23:51, 20 August 2016 (UTC)!

Rotation matrices?

This material added to the Regular_icosahedron#Cartesian_coordinates section seems way too much and of unclear value. I expect there are 6 5-fold rotational axes through each pair of vertices. I don't know why these are more important than 10 3-fold rotational axes, or 15 2-fold rotational axes. Tom Ruen (talk) 22:28, 9 March 2018 (UTC)

Two rotation matrices (each by ${\scriptstyle {\frac {2\pi }{5}}}$ radians about axis determined by antipodal vertices) map the icosahedron into itself and generate the rotational symmetry group of order 60 (there are several other choices for generators). $\left[{\begin{array}{ccc}{\frac {\phi }{2}}&{\frac {-1}{2}}&{\frac {\phi -1}{2}}\\{\frac {1}{2}}&{\frac {\phi -1}{2}}&{\frac {-\phi }{2}}\\{\frac {\phi -1}{2}}&{\frac {\phi }{2}}&{\frac {1}{2}}\end{array}}\right]$ $\left[{\begin{array}{ccc}{\frac {\phi }{2}}&{\frac {1}{2}}&{\frac {1-\phi }{2}}\\{\frac {-1}{2}}&{\frac {\phi -1}{2}}&{\frac {-\phi }{2}}\\{\frac {1-\phi }{2}}&{\frac {\phi }{2}}&{\frac {1}{2}}\end{array}}\right]$

REPLY: Central inversion is a symmetry of the icosahedron and the last six matrices are the transpose of the first six... so yes, the first six suffice. Whereas the application may determine whether a particular set of generators (of the symmetry group) are "more important" than others, some Cartesian generators should be explicitly provided in the Cartesian coordinates section, and rotation matrices are often useful in applications... Moreover, the matrices are interesting in that their entries are particularly simple: quotients of linear functions of $\phi$ with integer coefficients (which is obvious I suppose, yet pleasant to be given here on a silver platter). Whether twelve is "way too much" is a matter of opinion, but I've limited to the first two in my most recent edit (they generate the rotational symmetry group). Sat Mar 10 10:58:56 EST 2018

UPDATE: rationalizing denominators makes matrix entries linear functions of $\phi$ with rational corfficients (so I updated matrices on page). Sun Mar 11 14:41:22 EDT 2018 — Preceding unsigned comment added by 96.61.123.153 (talk) 18:44, 11 March 2018 (UTC)

These two matrices are still inverses of each other. It's not at all clear what you mean by "generate the rotation group of order 60". If you gave one order-5 rotation group, and one order-3 rotational group, those two as generators can generate the entire chiral icosahedral group order 60. Overall I'd say this discussion doesn't belong in this article, but rather icosahedral symmetry. As well, you could give 3 reflection matrices, represented in the Coxeter-Dynkin diagram, , where each node is a reflection matrix generator, and the product of the first two matrices make one of your 5-fold rotations, and the product of the last two make a 3-fold rotation matrix, which combined can generate the full rotational group. Tom Ruen (talk) 22:21, 11 March 2018 (UTC)

R0	R1	R2
$\left[{\begin{array}{ccc}-1&0&0\\0&1&0\\0&0&1\end{array}}\right]$	$\left[{\begin{array}{ccc}{\frac {1-\phi }{2}}&{\frac {-\phi }{2}}&{\frac {-1}{2}}\\{\frac {-\phi }{2}}&{\frac {1}{2}}&{\frac {1-\phi }{2}}\\{\frac {-1}{2}}&{\frac {1-\phi }{2}}&{\frac {\phi }{2}}\end{array}}\right]$	$\left[{\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&1\end{array}}\right]$

Here's one set of 3 reflection matrices for with relations: R²
₀ = R²
₁ = R²
₂ = (R₀R₂)² = (R₀R₁)⁵ = (R₁R₂)³ = I. So then R₀R₂, R₀R₁, R₁R₂ are rotation matrices. Tom Ruen (talk) 21:12, 12 March 2018 (UTC)

REPLY: The matrices

$\left[{\begin{array}{ccc}{\frac {\phi }{2}}&{\frac {-1}{2}}&{\frac {\phi -1}{2}}\\{\frac {1}{2}}&{\frac {\phi -1}{2}}&{\frac {-\phi }{2}}\\{\frac {\phi -1}{2}}&{\frac {\phi }{2}}&{\frac {1}{2}}\end{array}}\right]$ $\left[{\begin{array}{ccc}{\frac {\phi }{2}}&{\frac {1}{2}}&{\frac {1-\phi }{2}}\\{\frac {-1}{2}}&{\frac {\phi -1}{2}}&{\frac {-\phi }{2}}\\{\frac {1-\phi }{2}}&{\frac {\phi }{2}}&{\frac {1}{2}}\end{array}}\right]$

are not inverses of each other (by inverses I mean multiplying one by the other yields the identity matrix; multiply them yourself -- better yet, use Maple or Mathematica -- to verify the result is not the identity matrix). They generate the rotation group of order 60 in the standard sense: consider the set of all finite strings over two symbols and substitute the first matrix for the first symbol and substitute the second matrix for the second symbol, then evaluate results (multiply adjacent matrices)... the set of all results (where sets do not contain duplicates -- worth mentioning since some computer algebra systems fail on that point) contains exactly 60 elements (rotation matrices which map the icosahedron into itself and which form a group under matrix multiplication). If you are in doubt, perform the computation yourself; using Maple makes the task easy (be careful about sets and duplicates)... presumably Mathematica (and other computer algebra systems) will also simplify the task. Yes, there are several other choices for generators (as my parenthetic comment on the wiki page states). A previous comment indicated that "too many" matrices should be avoided... I have provided two (note that no smaller collection of rotation matrices can generate the group). 96.61.123.153 (talk) 16:19, 13 March 2018 (UTC)

Icosahedral symmetry generators: [5,3],

	Reflections			Rotations
Name	R0	R1	R2	R0×R1	R1×R2	R0×R2
Order	2	2	2	5	3	2
Matrix	$\left[{\begin{smallmatrix}-1&0&0\\0&1&0\\0&0&1\end{smallmatrix}}\right]$	$\left[{\begin{smallmatrix}{\frac {1-\phi }{2}}&{\frac {-\phi }{2}}&{\frac {-1}{2}}\\{\frac {-\phi }{2}}&{\frac {1}{2}}&{\frac {1-\phi }{2}}\\{\frac {-1}{2}}&{\frac {1-\phi }{2}}&{\frac {\phi }{2}}\end{smallmatrix}}\right]$	$\left[{\begin{smallmatrix}1&0&0\\0&-1&0\\0&0&1\end{smallmatrix}}\right]$	$\left[{\begin{smallmatrix}{\frac {\phi -1}{2}}&{\frac {\phi }{2}}&{\frac {1}{2}}\\{\frac {-\phi }{2}}&{\frac {1}{2}}&{\frac {1-\phi }{2}}\\{\frac {-1}{2}}&{\frac {1-\phi }{2}}&{\frac {\phi }{2}}\end{smallmatrix}}\right]$	$\left[{\begin{smallmatrix}{\frac {1-\phi }{2}}&{\frac {\phi }{2}}&{\frac {-1}{2}}\\{\frac {-\phi }{2}}&{\frac {-1}{2}}&{\frac {1-\phi }{2}}\\{\frac {-1}{2}}&{\frac {\phi -1}{2}}&{\frac {\phi }{2}}\end{smallmatrix}}\right]$	$\left[{\begin{smallmatrix}-1&0&0\\0&-1&0\\0&0&1\end{smallmatrix}}\right]$

I see, I didn't think of generating from 2 5-fold rotation matrices. I see their product makes a 3-fold rotation matrix. I'd prefer something like this (right), starting with 3 reflection generators, whose products make 3 rotation generators, of 5,3,2-fold. If you have reflection generator matrices, you can build them all, and any two of these rotation matrices also work with the last of 3 a simpler rotation matrix. But overall this may better belong at icosahedral symmetry than here. Tom Ruen (talk) 23:55, 13 March 2018 (UTC)

That this would rather belong in icosahedral symmetry was also my first thought about this. But it is rather not the kind of thing you would expect in an encyclopedia anyway. It reminds me of what I have done for octahedral symmetry in v:Full octahedral group. Maybe Wikiversity or Wikibooks would be a good place for an article with detailed examples like this. And that can be linked from the Wikipedia article. Watchduck (quack) 01:24, 14 March 2018 (UTC)

A Commons file used on this page has been nominated for deletion

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Missing proofs

The formulas for dimensions etc are given with no proof or reference.  Andreas  ^(T) 13:48, 12 October 2023 (UTC)

Confusing construction

The section Construction describes in its second paragraph a method for constructing the regular icosahedron from a cube.

The edge-length of the cube is never mentioned, so this section needs work.

But even if the cube were assumed to have a specific edge-length, this paragraph is still extremely confusing.

I hope someone familiar with this construction can make this paragraph much clearer than it is.