Icosidodecahedra, Icosahedra, and Dodecahedra

Posted on 13 November 2014 by RobertLovesPi

If one starts with a single icosidodecahedron, and then augments its pentagonal faces with dodecahedra, and its trianguar faces with icosahedra, this is the result.

This figure has gaps in it where two pentagons and two triangles meet around a vertex. If one puts icosidodecahedra in those gaps, this is the resulting figure.

Next, once again, the pentagonal faces are augmented with dodecahedra, and the triangular faces with icosahedra.

These virtual polyhedral models were all built using Stella 4d, available at http://www.software3d.com/Stella.php.

A Dozen Triangula

Posted on 25 October 2014 by RobertLovesPi

This dodecahedron is adorned with images of the Triangulum Galaxy. The plural of “Triangulum” is “Triangula,” is it not?

Software credit: this rotating image was created using Stella 4d: Polyhedron Navigator, which is available at http://www.software3d.com/Stella.php.

Reaugmenting the Dodecahedron

Posted on 21 September 2014 by RobertLovesPi

Suppose you take a central dodecahedron, and then augment each of its faces with a dodecahedron. That would be an augmented dodecahedron. If you augment this figure with another layer of dodecahedra, then you have the reaugmented dodecahedron:

After another level of such augmentation, you get this — the metareaugmented dodecahedron:

Both these images were created using Stella 4d, which you can try here.

An Unsolved Problem Involving the Icosahedron and the Dodecahedron, and Their Circumscribed Spheres

Posted on 21 September 2014 by RobertLovesPi

This is apparently a problem, posed by Gregory Galperin, which went unsolved at the Bay Area Math Olympiad in 2005. I haven’t solved it yet, but I’m going to try, as I work on this blog-post. My 2010 source is a paper about Zome which may be read, as a .pdf, at bact.mathcircles.org/files/Summer2010/zomes-6-2010.pdf. The problem involves a dodecahedron and an icosahedron, each inscribed inside the same sphere of radius r, and asks which has the greater volume. At the time the authors wrote this paper, they knew of no solution, and I know of none now, but I do like a challenge.

My idea for solving this begins with Zome (info on Zome: see http://www.zometool.com, as well as other sites you can find by googling “Zome”). In the Zome geometry system, using B1 struts for the edges of both a dodecahedron and an icosahedron, R1 struts are the radii of the circumscribed sphere for the icosahedron, and Y2 struts are the radii for the circumscribed sphere of the dodecahedron. Since volume formula for polyhedra are generally given in term of edge-length, I need to find B1 in terms of R1 for the icosahedron, and find B1 in terms of Y2 for the dodecahedron.

Icosahedron: find B1, in terms of R1.

There exists a right triangle which can be built in Zome which has a hypotenuse equal to 2R1, and legs epqual to B1 and B2. B2 = φB1, so, by the Pythagorean Theorem, (2R1)^2 = (B1)^ + φ²(B1)², which simplifies to 4(R1)^2 = (1 + φ²)(B1)^2, which can then be solved for B1 as B1 = sqrt[4(R1)^2/(1 + φ²)]. B1 here is the icosahedron’s edge-length, while R1 is the radius of its circumscribed sphere.

Dodecahedron: find B1, in terms of Y2.

In the Zome system, Y2 = φY1, and Y1 = [sqrt(3)/2]B1. Rearrangement of the first of these equations yields Y1 = Y2/φ, and substitution then yields [sqrt(3)/2]B1 = Y2/φ, which then can be rearranged to yield B1 = 2Y2/[φsqrt(3)]. B1 here is the dodecahedron’s edge-length, while Y2 is the radius of its circumscribed sphere.

Next, find the volume of the icosahedron inscribed inside a sphere, in terms of that sphere’s radius.

According to http://mathworld.wolfram.com/Icosahedron.html, the volume of an icosahedron is given by V = (5/12)[3 + sqrt(5)]a³, where a is the edge length, or B1 in the first indented section, between the two images, above. Then, by substitution, V = (5/12)[3 + sqrt(5)]{sqrt[4(R1)^2/(1 + φ²)]}³, which then becomes (with “r” being the radius of the circumscribed sphere) V = (5/12)[3 + sqrt(5)][2r/sqrt(1 + φ²)]³ = (40/12)[3 + sqrt(5)][1/sqrt(1 + φ²)]³r³ = (10/3)[3 + sqrt(5)][1/sqrt(1 + φ²)]³r³. Then, using the identity φ² = φ + 1, this can be further simplified to V = (10/3)[3 + sqrt(5)][1/sqrt(2 + φ)]³r³.

Next, find the volume of the dodecahedron inscribed inside the same sphere, in terms of that sphere’s radius, r.

According to https://en.wikipedia.org/wiki/Dodecahedron, the volume of an icosahedron is given by V = (1/4)[15 + 7sqrt(5)]a³, where a is the edge length, or B1 in the second indented section, below the second image, above. Then, by substitution, V = (1/4)[15 + 7sqrt(5)]{2Y2/[φsqrt(3)]}³, which then becomes (with “r” being the radius of the circumscribed sphere) V = (8/4)[15 + 7sqrt(5)]{1/[φsqrt(3)]}³r³ = 2[15 + 7sqrt(5)]{1/[3sqrt(3)]}(1/φ³)r³ = (2/3)[15 + 7sqrt(5)][sqrt(3)/3](1/φ³)r³ = [2sqrt(3)/9][15 + 7sqrt(5)](1/φ³)r³.

So, with the “r” in each case being the same, the icosahedron is larger than the dodecahedron iff (10/3)[3 + sqrt(5)][1/sqrt(2 + φ)]³ > [2sqrt(3)/9][15 + 7sqrt(5)](1/φ³), which simplifies to (5)[3 + sqrt(5)][1/sqrt(2 + φ)]³ > [2sqrt(3)/3][15 + 7sqrt(5)](1/φ³), which simplifies further to {5/[sqrt(2 + φ)]³}[3 + sqrt(5)] > [2sqrt(3)/3φ³][15 + 7sqrt(5)], which is, as a decimal approximation, is (0.726542528)(5.2360679774998) > (3.464101615/12.708203932)(30.6524758), or 3.804226 > 8.355492, which is false, meaning that the dodecahedron is larger, not the icosahedron.

Now for the bad part: I think I’m wrong, but I don’t know where the error lies. I’m also tired. If any of you see the mistake, please point it out in a comment, and I’ll try to fix this after I’ve rested.

Update: if the websites http://rechneronline.de/pi/icosahedron.php and http://rechneronline.de/pi/dodecahedron.php work correctly, then the dodecahedron is larger. Evidence:

This does not, however, mean that I did the problem correctly. I merely stumbled upon the correct answer. How do I know this? Simple: the ratio I obtained was too far off. Therefore, I would still welcome help clearing up the mystery of where my error(s) is/are, in the calculations shown above.

Two Dodecahedra with Varying Rotation-Types of the Same Design Shown On Their Faces

Posted on 15 September 2014 by RobertLovesPi

The images shown on the faces of these dodecahedra appeared in the last post, and were made using Geometer’s Sketchpad and MS-Paint.

Assembling the polyhedral images and creating these rotating .gif files required another program, Stella 4d, which is available at http://www.software3d.com/Stella.php.

The Rhombic Dodecahedron / Rhombic Triacontahedron Compound

Posted on 12 September 2014 by RobertLovesPi

Stella 4d has this in its built-in library. You can try it at http://www.software3d.com/Stella.php.

The Dodecahedron / Octahedron Compound

Posted on 12 September 2014 by RobertLovesPi

This rotating image was assembled using Stella 4d, available at http://www.software3d.com/Stella.php.

A Collection of Four Polyhedra Decorated with Mandalas

Posted on 11 September 2014 by RobertLovesPi

First, a cuboctahedron.

Next, its dual, the rhombic dodecahedron.

And, after that, the icosidodecahedron.

And finally, its dual, the rhombic triacontahedron.

All of these rotating images were assembled using Stella 4d, available at http://www.software3d.com/Stella.php.

Music Video: Murder By Death’s “Those Who Stayed” & “I’m Afraid of Who’s Afraid of Virginia Woolf”

Posted on 3 August 2014 by RobertLovesPi

Music: the first two tracks from the Murder By Death album Like the Excorcist, But More Breakdancing. Please visit their website, http://www.murderbydeath.com, to buy this band’s music and merchandise. While you’re there, I recommend checking their concert calendar, to see if they may be playing near you soon. Murder By Death concerts, which I’ve seen six times now, are not to be missed!

Visuals: rotating polyhedra, all with icosidodecahedral symmetry, generated using Stella 4d: Polyhedron Navigator, which you can try for yourself at http://www.software3d.com/Stella.php. The polyhedra shown are, in order of appearance:

The icosahedron
The compound of the icosahedron and its dual, the dodecahedron
The dodecahedron, with all faces the same color
The small stellated dodecahedron, or first stellation of the dodecahedron, in a single color
The small stellated dodecahedron, with only parallel faces having the same color (six-color arrangement)
The great dodecahedron, or second stellation of the dodecahedron, six-color arrangement
The great stellated dodecahedron, or third stellation of the dodecahedron, six-color arrangement
Stellating the dodecahedron a fourth time, to return it to its original form, but in the six-color arrangement this time
The icosidodecahedron, with triangular faces invisible, and pentagonal faces shown using the six-color arrangement
The icosidodecahedron, all faces visible now, and colored by face type
The fourth stellation of the icosidodecahedron (its first stellation is the dodecahedron, the second is the icosahedron, and the third is the compound of the first two, all of which have already been seen)
The fifth stellation of the icosidodecahedron
The convex hull of the fifth stellation of the icosidodecahedron, which is a slightly-truncated icosahedron
The truncated icosahedron which is a true Archimedean Solid, since all its faces are regular
The truncated icosahedron’s second stellation (the first is the already-seen icosahedron)