# Two Compounds with Pyritohedral Symmetry: the Icosidodecahedron / Truncated Octahedron Compound, and the Rhombic Triacontahedron / Tetrakis Cube Compound

Stella 4d, a program you can try here, was used to create these two compounds. Both have pyritohedral symmetry: the symmetry of a standard volleyball. The two compounds are also duals.

# At 47, My Age Is a Prime Number Again =D

For some reason, I like having my age be a prime number of years. Today, I turn 47, so I get to have a prime-number-age for a whole year now. This hasn’t happened since I was 43, so I made this 47-pointed star to celebrate:

I also make birthday-stars for composite-number ages as well, just because it’s fun, and you can find at least two others on this blog, on January 12, in past years. Also, I wouldn’t want to have to wait until I’m 53 (my next prime age) to make another one of these.

At the moment, I certainly don’t feel 47. There are times when I feel twenty-two . . .

There are also times when I feel six.

At the moment, however, I feel about thirty. For that reason, I put the 47-pointed stars on the thirty faces of a rotating rhombic triacontahedron, because (a) it’s my birthday, (b) I want to, and (c) I can.

Image/music credits:

1. I created this using Geometer’s Sketchpad and MS-Paint.
2. “When Yer Twenty-Two,” by The Flaming Lips, via a YouTube posting.
3. Two panels from a Calvin and Hobbes cartoon, by Bill Watterson. (Calvin is perpetually six years old.)
4. Created using the image at the top of this post, and the program Stella 4d: Polyhedron Navigator, which is available here.

# My First Solution to the Zome Cryptocube Puzzle, with Special Guest Appearances by Jynx the Kitten

Last month, in a special Christmas promotion, the Zometool company (www.zometool.com) briefly sold a new kit (which will return later) — a fascinating game, or puzzle, called the “Cryptocube.” Zome usually comes in a variety of colors, with each color having mathematical significance, but the Cryptocube is produced in black and white, which actually (in my opinion) makes it a better puzzle. Here’s how the Crypocube challenge works:  you use the black parts to make a simple cube, and then use the smaller white parts to invent a structure which incorporates the cube, is symmetrical, is attractive, and can survive having the twelve black cube-edges removed, leaving only the cube’s eight black vertices in place. I had a lot of fun making my first Cryptocube, and photographed it from several angles.

If this was built using standard Zome colors, the round white figure inside the cube, a rhombic triacontahedron, would be red, and the pieces outside the cube, as well as those joining the rhombic triacontahedron to the cube (from inside the cube), would be yellow.

It isn’t only humans who like Zome, by the way. Jynx the Kitten had to get in on this!

Jynx quickly became distracted from the Cryptocube by another puzzle, though: he wanted to figure out how to pull down the red sheet I had attached to the wall, as a photographic backdrop for the Cryptocube. Jynx takes his feline duties as an agent of entropy quite seriously.

As usually happens, Jynx won (in his never-ending struggle to interfere with whatever I’m doing, in this case by pulling the sheet down) and it took me quite a while to get the red sheet back up, in order to take kitten-free pictures of my Cryptocube solution, after removal of the black cube’s edges.

Here’s the view from another angle.

The Cryptocube will be back, available on the Zometool website, later in 2015. In the meantime, I have advice for anyone not yet familiar with Zome, but who wants to try the Cryptocube when it returns: go ahead and get some Zome now, at the link above, in the standard colors (red, blue, and yellow, plus green in advanced kits), and have fun building things with it over the next few months. The reason to do this, before attempting to solve the Crypocube, is simple: the colors help you learn how the Zome system works, which is important before trying to solve a Zome puzzle without these colors visible. After gaining some familiarity with the differing shapes of the red, blue, yellow, and green pieces, working with them in white becomes much easier.

On a related note, Zome was recommended by Time magazine, using the words “Zometool will make your kids smarter,” as one of the 14 best toys of 2013. I give Zome my own strong, personal recommendation as well, and, as a teacher who uses my own Zome collection in class, for instructional purposes, I can attest that Time‘s 2013 statement about Zome is absolutely correct. Zome is definitely a winner!

# If You Have Enough Platonic Dodecahedra Around, and Glue Them Together Just Right, You Can Make a Rhombic Triacontahedron.

Aren’t you glad to know that? As soon as I found out isocahedra can form a rhombic dodecahedron (see last post), I knew this would be true as well. Why? Zome explains why, actually. It’s at http://www.zometool.com. Anything buildable with yellow Zome can be built out of icosahedra. Dodecahedra con form anything buildable with red Zome. Finally, if you can make it with blue Zome, it can be built out of rhombic triacontahedra. It follows that rhombicosidoecahedra can build anything Zome-constructable — but one look at a Zomeball makes that easy to believe, since Zomeballs are modified rhombicosidodecahedra.

Anyway, here’s the rhombic triacontahedron, made of dodecahedra:

[Image created with Stella 4d; see http://www.software3d.com/Stella.php for more info re: this program.]

# Polygons Related to the Golden Ratio, and Associated Figures in Geometry, Part 2: Quadrilaterals

The golden ratio, also known as φ, has a value of [1 + sqrt(5)]/2, or ~1.61803. It is associated with a great many figures in geometry, and also appears in numerous other contexts. The most well-known relationship between a geometric figure and the golden ratio is the golden rectangle, which has a length:width ratio equal to the golden ratio. An interesting property of the golden rectangle is that, if a square is removed from it, the remaining portion is simply a smaller golden rectangle — and this process can be continued without limit.

While the golden ratio is related to many polyhedra, this relationship does not always involve golden rectangles, but sometimes it does. For example, it is possible to modify a rhombicosidodecahedron, by replacing that figure’s squares with golden rectangles (with the longest side adjacent to the triangles, not the pentagons), to obtain a “Zomeball” — the node which is at the heart of the Zometool ball-and-stick modeling system for polyhedra, and other phenomena. The entire Zome system is based on the golden ratio. Zome kits are available for purchase at http://www.zometool.com, and this image of a Zomeball was found at http://www.graphics.rwth-aachen.de/media/resource_images/zomeball.png.

In some cases, the relationship between a golden rectangle, and a polyhedron, is more subtle. For example, consider three mutually-perpendicular golden rectangles, each with the same center:

While this is not, itself, a polyhedron, it is possible to create a polyhedron from it, by creating its convex hull. A convex hull is simply the smallest convex polyhedron which can contain a given figure in space. For the three golden rectangles above, the convex hull is the icosahedron, one of the Platonic solids:

In addition to the golden rectangle, there are also other quadrilaterals related to the golden ratio. For example, a figure known as a golden rhombus is formed by simply connecting the midpoints of the sides of a golden rectangle. The resulting rhombus has diagonals which are in the golden ratio.

One of the Archimedean solids, the icosidodecahedron, has a dual called the rhombic triacontahedron. The rhombic triacontahedron has thirty faces, and all of them are golden rhombi.

There are also other polyhedra which have golden rhombi for faces. One of them, called the rhombic hexacontahedron (or “hexecontahedron,” in some sources), is actually the 26th stellation of the rhombic triacontahedron, itself. It has sixty faces, all of which are golden rhombi.

Other quadrilaterals related to the golden ratio can be formed by reflecting the golden triangle and golden gmonon (described in the post right before this one) across each of their bases, to form two other types of rhombus.

In these two rhombi, the golden ratio shows up as the side-to-short-diagonal ratio (in the case of the 36-144-36-144 rhombus), and the long-diagonal-to-side ratio (in the case of the 72-108-72-108 rhombus). These two rhombi have a special property:  together, they can tile a plane in a pattern which never repeats itself, but, despite this, can be continued indefinitely. This “aperiodic tiling” was discovered by Roger Penrose, a physicist and mathematician. The image below, showing part of such an aperiodic tiling, was found at https://en.wikipedia.org/wiki/Penrose_tiling.

There are also at least two other quadrilaterals related to the golden ratio, and they are also formed from the golden triangle and the golden gnomon. The procedure for making these figures, which could be called the “golden kite” and the “golden dart,” is similar to the one for making the rhombi for the Penrose tiling above, but has one difference: the two triangles are each reflected over a leg, rather than a base.

In the case of this kite and dart, it is the longer and shorter edges, in each case, which are in the golden ratio — just as is the case with the golden rectangle. Another discovery of Roger Penrose is that these two figures, also, can be used to form aperiodic tilings of the plane, as seen in this image from http://www.math.uni-bielefeld.de/~gaehler/tilings/kitedart.html.

There is yet another quadrilateral which has strong connections to the golden ratio. I call it the golden trapezoid, and this shows how it can be made from a golden rectangle, and how it can be broken down into golden triangles and golden gnomons. However, I have not yet found an interesting polyhedron, not tiling pattern, based on golden trapezoids — but I have not finished my search, either.

[Image credits:  see above for the sources of the pictures of the two Penrose tilings, as well as the Zomeball, shown in this post. Other “flat,” nonmoving pictures I created myself, using Geometer’s Sketchpad and MS-Paint. The rotating images, however, were created using a program called Stella 4d, which is available at http://www.software3d.com/Stella.php.]

# A Rhombic Triacontahedron, Made of Icosidodecahedra

It turns out that it is possible to use multiple icosidodecahedra as building blocks to build that polyhedron’s dual, the rhombic triacontahedron. This construction appears below, in four different coloring-schemes.

These rotating virtual models were constructed using Stella 4d, a program available at http://www.software3d.com/Stella.php.

# An Alteration of the Icosahedron/Dodecahedron Compound

The dual of the icosahedron is the dodecahedron, and a compound can be made of those two solids. If one then takes the convex hull of this solid, the result is a rhombic triacontahedron. One can then made a compound of the rhombic triacontahedron and its dual, the icosidodecahedron — and then take the convex hull of that compound. If one then makes another compound of that convex hull and its dual, and then makes a convex hull of that compound, the dual of this latest convex hull is the polyhedron you see above.

I did try to make the faces of this solid regular, but that attempt did not succeed.

All of these polyhedral manipulations were were performed with Stella 4d:  Polyhedron Navigator, available at http://www.software3d.com/Stella.php.

# A Polyhedral Demonstration of the Fact That Nine Times Thirty Equals 270, Along with Its Interesting Dual

It would really be a pain to count the faces of this polyhedron, in order to verify that there are 270 of them. Fortunately, it isn’t necessary to do so. The polyhedron above is made of rhombus-shaped panels which correspond to the thirty faces of the rhombic triacontahedron. Each of these panels contains nine faces: one square, surrounded by eight triangles. Since (9)(30) = 270, it is therefore possible to see that this polyehdron has 270 faces, without actually going to the trouble to count them, one at a time.

The software I used to make this polyhedron may be found at http://www.software3d.com/Stella.php, and is called Stella 4d. With Stella 4d, a single mouse-click will let you see the dual of a polyhedron. Here’s the dual of the one above.

This polyhedron is unusual, in that it has faces with nine sides (enneagons, or nonagons), as well as fifteen sides (pentadecagons). However, these enneagons and pentadecagons aren’t regular — yet — but they will be in the next post.

# Building, and Analyzing, Octahedral Lattices of Rhombic Triacontahedra

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.