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.]
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.
[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.]
I’m going to start this experiment with a single octahedron, with faces in two colors, placed so that two faces which share an edge are always of different colors.
Next, I will augment the red faces — and only the red faces — with identical octahedra.
The regions with four blue, adjacent faces look as though they might hold icosahedra — but I checked, and they don’t quite fit. I will therefore continue the same process — augmenting only the red faces with more octahedra of the original type.
I’ve now decided that I definitely like this game, so I’ll keep playing it.
Immediately above, at the fourth of these images, some of the octahedra have started to overlap slightly, but I’m choosing to not be bothered by that — I’m continuing the now-established pattern, just in order to see where it takes me.
The regions of overlap are now far more obvious, but I’m continuing, anyway. Why? Because this is fun, that’s why! Right now, Stella 4d, the program I use to do these polyhedral manipulations, is chugging away on the next one. (This program is avilable at http://www.software3.com/Stella.php.) Ah, it’s ready — here it is!
Rather than repeat this process again, I now have another question: what would the convex hull of this figure look like? (A convex hull of a non-convex polyhedron is the smallest convex polyhedron which can contain a given non-convex polyhedron.) With Stella 4d, that’s easily answered.
I must admit this: that was nothing like what I expected — but such unexpected discoveries are a large part of what makes these polyhedral investigations with Stella 4d so much fun. And now, to close this particular polyhedral journey, I will have Stella 4d produce, for me, the dual of the convex hull shown above. (In case you aren’t familiar with duality regarding polyhedra, it describes the relationship between the octahedron, with which this post began, and the familar cube. Basically, with duals, faces and verticies are “flipped” over edges, although that is an extremely informal and imprecise way to describe the at the process.)
And with that, my friends, I bid you good night!
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.
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.]
The images on the faces of this polyhedron were created with Geometer’s Sketchpad and MS-Paint. Projecting these images onto these faces, and then creating this rotating image, was accomplished using Stella 4d: Polyhedron Navigator— a program you can try for yourself, for free, at http://www.software3d.com/Stella.php.
This Jack-o-Lantern picture was found with a Google image-search, and then I projected it onto a rhombicosidodecahedron, and created this rotating .gif file, using Stella 4d — a program available at http://www.software3d.com/Stella.php. Happy Halloween!
The faces of this polyhedron are decorated with the same type of curvy tessellation seen in the last post here, and it was created using Geometer’s Sketchpad and MS-Paint. Projecting these images onto the faces of this rhombic dodecahedron, in different colors, and then creating this rotating images of it, required a third program, Stella 4d: Polyhedron Navigator. This latter program, an indispensable tool for polyhedral investiagations, may be tried for free, as a trial version, at http://www.software3d.com/Stella.php.
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.
When Norman Johnson first found, and named, all the Johnson solids in the latter 1960s, he came across a number of “near-misses” — polyhedra which are almost Johnson solids. If you aren’t familiar with the Johnson solids, you can find a definition of them here. The “near-miss” which is most well-known features regular enneagons (nine-sided polygons):
This is the dual of the above polyhedron:
As with all polyhedra and their duals, a compound can be made of these two polyhedra, and here it is:
Finding this polyhedron interesting, I proceeded to use Stella 4d (polyhedron-manipulation software, available at http://www.software3d.com/Stella.php) to make its convex hull.
Here, then, is the dual of this convex hull:
Stella 4d has a “try to make faces regular” function, and I next used it on the polyhedron immediately above. If this function cannot work, though — because making the faces regular is mathematically impossible — one sometimes gets completely unexpected, and interesting, results. Such was the case here.
Next, I found the dual of this latest polyhedron.
The above polyhedron’s “wrinkled” appearance completely surprised me. The next thing I did to change it, once more, was to create this wrinkled polyhedron’s convex hull. A convex hull of a non-convex polyhedron is simply the smallest convex polyhedron which can contain the non-convex polyhedron, and this process often has interesting results.
Next, I created this latest polyhedron’s dual:
I then attempted “try to make faces regular” again, and, once more, had unexpected and interesting results:
The next step was to take the convex hull of this latest polyhedron. In the result, below, all of the faces are kites — two sets of twenty-four each.
I next stellated this kite-faced polyhedron 33 times, looking for an interesting result, and found this:
This looked like a compound to me, so I told Stella 4d to color it as a compound, if possible, and, sure enough, it worked.
The components of this compound looked like triakis tetrahedra to me. The triakis tetrahedron, shown below, is the dual of the truncated tetrahedron. However, I checked the angle measurement of a face, and the components of the above compound-dual are only close, but not quite, to being the same as the true triakis tetrahedron, which is shown below.
This seemed like a logical place to end my latest journey through the world of polyhedra, so I did.
Software credit: I made this using Stella 4d, which is available at http://www.software3d.com/Stella.php.
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