A Cluster of Nine Octahedra, and Related Polyhedra

Posted on 22 November 2014 by RobertLovesPi

If one starts with a central octahedron, then augments each of its eight triangular faces with identical octahedra, this is the result.

It is then possible to augment each visible triangle of this cluster with yet more octahedra, which produces this result, in which some octahedra overlap each other.

After making this, I wanted to see its convex hull: the smallest, tightest-fitting convex polyhedron which can contain a given non-convex polyhedron. (I use Stella 4d: Polyhedron Navigator to perform these manipulations of polyhedra, and this program makes this a fast and easy process. If you’d like to try this software, even as a free trial download, the website to visit is http://www.software3d.com/Stella.php.) Here’s what this convex hull, which bears a resemblance to the rhombcuboctahedron, looks like.

Looking for previously-unseen, and interesting, polyhedra, I then starting stellating this convex hull. I did find something interesting — to me, anyway — after only two stellations.

That concluded my latest polyhedral investigation, but I certainly don’t intend it to be my last.

Two Different Clusters of Thirty-One Rhombicosidodecahedra

Posted on 16 November 2014 by RobertLovesPi

The cluster above was formed by starting with one rhombicosidodecahedron, and then augmenting each of its thirty square faces with another rhombicosidodecahedron. If you examine the single rhombicosidodecahedron below, though, you’ll see that this can be done in two different ways, each of which produces a cluster with the same degree of symmetry as the original. (If less symmetry is acceptable, there are far more than two different ways to obtain such clusters, but that does not interest me.)

When new rhombicosidodecahedra are attached to the square faces of this central rhombicosidodecahedron, they new ones can be oriented such that pentagons are placed above pentagons, and triangles above triangles — or the new ones can be oriented the other way, so that pentagons are placed above triangles, and triangles above pentagons. One of these possibilities produces the cluster at the top of this post. The other possibility produces the cluster shown below.

All of these polyhedral manipulations were performed using a program called Stella 4d: Polyhedron Navigator, which may be purchased (or tried for free, as a trial download) at http://www.software3d.com/Stella.php.

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 Cuboctahedral Cluster of Rhombic Dodecahedra

Posted on 12 November 2014 by RobertLovesPi

It is well-known that the cuboctahedron and the rhombic dodecahedron are dual polyhedra. However, until I stumbled upon this, I was unaware that rhombic dodecahedra could actually be arranged into a cluster with the overall shape of a cuboctahedron.

[Software credit: see http://www.software3d.com/Stella.php for more information about Stella 4d, the program I use to make these rotating images. A free trial download is available at that website.]

A Cluster of Thirteen Rhombic Dodecahedra, and Three Other Related Polyhedra

Posted on 12 November 2014 by RobertLovesPi

One of the thirteen rhombic dodecahedra in this cluster cannot be seen, for it is hidden in the middle. The other twelve are each attached to a face of the central rhombic dodecahedron.

If one then creates the convex hull of this cluster — the smallest convex polyhedron which can contain it — this is the result:

This polyhedron has fifty faces: the six square faces of a cube, the eight triangular faces of an octahedron, the twelve rhombic faces of a rhombic dodecahedron, and twenty-four rectangles to fill the gaps between the other faces.

This fifty-faced polyhedron also has an interesting dual, with 48 faces, all of which are kites. Half of these 48 kites are of one type, and arranged into eight panels of three kites each, while the other half are arranged into six panels of four kites each:

Returning to the fifty-faced polyhedron, two images above, here is what happens if one tries to make each face as regular as possible:

In this polyhedron, the six squares are still squares, the eight triangles are still regular, and the twelve rhombi are still rhombi, although these rhombi are wider than before. The 24 rectangles, however, have now been transformed into isosceles trapezoids.

Building, and Analyzing, Octahedral Lattices of Rhombic Triacontahedra

Posted on 8 November 2014 by RobertLovesPi

To build an octahedral lattice of rhombic triacontahedra, start with a single rhombic triacontahedron.

Next, augment its faces with more rhombic triacontahedra — but not all thirty faces. Instead, only augment six of them — the six which lie along mutually-perpendicular x-, y-, and z-axes. Another way to look at this is that you only augment the North, South, East, West, top, and bottom faces.

There are now seven rhombic triacontahedra in this small lattice, and one can begin to see the octahedral structure which is forming. The next step is to perform the same type of augmentation on each of the rhombic triacontahedra which exist at this point.

The overall octahedral shape of this lattice is now quite obvious. Also, there is one rhombic triacontahedron in the center, six in the layer next to the center, and eighteen in the outer layer, for a total of 1 + 6 + 18 = 25 rhombic triacontahedra, in the third of these figures, immediately above.

This latest augmentation increases the number of rhombic triacontahedra in the cluster by 38, for a total of 25 + 38 = 63 rhombic triacontahedra in the lattice shown immediately above. This pattern can, of course, be continued indefinitely — and, as it increases, the overall octahedral shape of the lattice becomes progressively more clear.

At this point, the 5th in the sequence, the number of new triacontahedra added, in the outermost layer, becomes more difficult to count, but it is certainly possible. In the middle level of this outermost layer, there are sixteen new triacontahedra. In the levels above and below that, there are twelve new rhombic triacontahedra, each. The next levels up and down contain eight more, each. Above and below those two levels, there are four each — and going one more step up and down takes one to the top, with one rhombic triacontahedron at the top, plus one more at the bottom. The number of new rhombic triacontahedra is, therefore, 16 + 12 + 12 + 8 + 8 + 4 + 4 + 1 + 1, which equals 66. Add 66 more (on the outside) to the 63 already inside, and you have the total number of rhombic triacontahedra in this latest lattice: 129.

The number of rhombic triacontahedra at each point in this series of geometric shapes is, itself, interesting. Here’s what we have so far.

When n = 1, there is 1, or n, rhombic triacontahedral “cell” in the structure.
When n = 2, there are (1) + (4 + 1 + 1) = 1 + 6 = 7 cells. This is also equal to 3(1) + (4).
When n = 3, there are (1) + (4 + 1 + 1) + (8 + 4 + 4 + 1 + 1) = 1 + 6 + 18 = 25 cells. This also equals 5(1) + 3(4) + 1(8).
At n = 4, this number increases to (1) + (4 + 1 + 1) + (8 + 4 + 4 + 1 + 1) + (12 + 8 + 8 + 4 + 4 + 1 + 1) = 1 + 6 + 18 + 38 = 63. This also equals 7(1) + 5(4) + 3(8) + 1(12).
At n = 5, this sum is now (1) + (4 + 1 + 1) + (8 + 4 + 4 + 1 + 1) + (12 + 8 + 8 + 4 + 4 + 1 + 1) + (16 + 12 + 12 + 8 + 8 + 4 + 4 + 1 + 1) = 1 + 6 + 18 + 38 + 66 = 129. This also equals 9(1) + 7(4) + 5(8) + 3(12) + 1(16), which can also be written as 9 + 28 + 40 + 36 + 16.

Here is the next octahedral lattice of rhombic triacontahedra, with n = 6.

Now, at n = 6, this sum of the number of cells is (1) + (4 + 1 + 1) + (8 + 4 + 4 + 1 + 1) + (12 + 8 + 8 + 4 + 4 + 1 + 1) + (16 + 12 + 12 + 8 + 8 + 4 + 4 + 1 + 1) + (20 + 16 + 16 + 12 + 12 + 8 + 8 + 4 + 4 + 1 + 1) = 1 + 6 + 18 + 38 + 66 + 102 = 129 interior cells + 102 exterior cells = 231 cells, total. This also equals 11(1) + 9(4) + 7(8) + 5(12) + 3(16) + 1(20), which can also be written as 11 + 36 + 56 + 60 + 48 + 20. In terms of n, when n = 6, 11 + 36 + 56 + 60 + 48 + 20 may also be written as (2n – 1)(1) + (2n – 3)[4(n – 5)] + (2n – 5)[4(n – 4)] + (2n – 7)[4(n – 3)] + (2n – 9)[4(n – 2)] + (2n – 11)[4(n – 1)].

Obviously, I am trying to find a way to express the number of cells in the nth figure, in terms of n, but with only limited success, so far. The first six terms are 1, 7, 25, 63, 129, 231, and patterns of numbers above could easily be used to predict the seventh term, but that’s not my goal. What I want is a simple formula which will give me the total number of cells, in terms of n, for the nth of these octahedral lattices of rhombic triacontahedra. I’ll go ahead and find the seventh term, though, in the hope that it will help me figure out the pattern. When n = 7, the total number of cells will equal 13(1) + 11(4) + 9(8) + 7(12) + 5(16) + 3(20) + 1(24) = 13 + 44 + 72 + 84 + 80 + 60 + 24 = 377. Having found that number, I might as well throw in another picture, also. Stella 4d: Polyhedron Navigator, makes these rotating images easy to make, and you can try that program’s trial version for free, or purchase the fully-functioning version I use, at http://www.software3d.com/Stella.php.

This did help: it helped me avoid a dead end (later edit: or so I thought). I had noticed that the first six terms (1, 7, 25, 63, 129, 231) stayed close to, or matched exactly, the cubes of the first six counting numbers (1, 8, 27, 64, 125, 216), with only the sixth term deviating far from the sixth perfect cube. With the seventh term, 377, the deviation from the seventh perfect cube, 343, grows even wider, so, at this point, I don’t think the solution is related to perfect cubes in any way. (However, please keep reading; sometimes things which appear to be mathematical dead ends are actually only illusions of dead ends.)

I’m not yet willing to give up, though. I will next analyze the differences in successive terms, the differences between those differences, and so on. For this, the eighth term might be helpful, so I’ll go ahead and find it. From one of the patterns above, I can see that the 8th term will equal 15(1) + 13(4) + 11(8) + 9(12) + 7(16) + 5(20) + 3(24) + 1(28) = 15 + 52 + 88 + 108 + 112 + 100 + 72 + 28 = 575. The cube of 8 is 512, so, as I expected, the deviation from the sequence of perfect cubes continues to widen. Here, also, is a picture for the next of these figures, when n = 8.

Now it is time for the analysis of differences in this sequence, the differences in those differences, and so on. Here goes….

Now this is helpful! It tells me that the solution to this problem will take the form a third-order polynomial, also known as a cubic equation — so my earlier idea that this sequence was not related to the perfect cubes was, I now know, completely false.

I next struggled, for several hours, to find the cubic equation for the solution to this problem, without success. After finally giving up on finding the solution myself, I asked my wife to assist me (she’s also a math teacher, and her knowledge of algebra exceeds my own, for I specialize in geometry). She performed a cubic regression, and this is how we now know that the solution to this problem is that the number of cells for the nth figure in this sequence equals (4/3)n³ – 2n² + (8/3)n – 1. I spent many hours on this problem, but my wife finished solving it in mere minutes!

It’s no wonder I couldn’t find this solution, for I was only considering integers as coefficients, since only whole numbers for answers make sense — which I thought, incorrectly, would require the coefficients to be integers. However, for every value of n I have tested, the number of thirds in the answer always ends up being a multiple of three, cancelling threes in all denominators, and yielding whole numbers for answers. In retrospect, this makes sense, considering that the octahedron is a dipyramid, and that there is a 1/3 in the pyramid’s volume formula.

I was originally seeking to make an informative and interesting blog-post when I started this. However, I didn’t anticipate that I would learn as much as I did from the experience. I’m giving my wife joint credit for this solution, for I would not have been able to solve this problem without her help.

[Later edit: more information about this sequence can be found on these two websites: http://en.wikipedia.org/wiki/Centered_octahedral_number and https://oeis.org/A001845. I did not know about these sites until after I had finished this post.]

Cuboctahedral Cluster of Rhombic Triacontahedra

Posted on 27 September 2014 by RobertLovesPi

Due to their high number of planes of symmetry, rhombic triacontahedra make excellent building blocks to build other polyhedra. To make this, I used a program called Stella 4d, which you can try right here.

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.

A Cluster-Polyhedron Formed By 15 Truncated Octahedra, Plus Variations

Posted on 21 September 2014 by RobertLovesPi

To form the cluster-polyhedron above, I started with one truncated octahedron in the center, and then augmented each of its fourteen faces with another truncated octahedron. Since the truncated octahedron is a space-filling polyhedron, this cluster-polyhedron has no gaps, nor overlaps. The same cluster-polyhedron is below, but colored differently: each set of parallel faces gets a color of its own.

This is the cluster-polyhedron’s sixth stellation, using the same coloring-scheme as in the last image:

Here’s the sixth stellation again, but with the coloring scheme that Stella 4d: Polyhedron Navigator (the program I use to make these images) calls “color by face type.” If you’d like to try Stella for yourself, you can do so here.

Also colored by face-type, here are the 12th, 19th, and 86th stellations.

Leaving stellations now, and returning to the original cluster-polyhedron, here is its dual.

This image reveals little about this dual, however, for much of its structure is internal. So that this internal structure may be seen, here is the same polyhedron, but with only its edges visible.