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.]

A Great Rhombcuboctahedron, Decorated with Circles and Hexagons

Posted on 5 November 2014 by RobertLovesPi

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.

Hexagons and Circles

Posted on 5 November 2014 by RobertLovesPi

R. Buckminster Fuller, On Beauty

Posted on 3 November 2014 by RobertLovesPi

A Star with 49 Points, to Celebrate the 49-Hour Weekend Caused by the End of Daylight Saving Time, Tomorrow, in Most of the USA

Posted on 1 November 2014 by RobertLovesPi

This is the weekend that Daylight Saving Time (or DST) ends in most parts of the USA, which means that this is the only weekend of the year, here, which lasts 49 hours, rather than the usual 48.

To celebrate this once-a-year event, I created the design above, based on the number 49. I started by making one heptagram, inscribed in a circle. The heptagram I used is one of two which exist, and is also called the {7/3} star heptagon. It looks like this:

After making one of these, I then rotated it 1/49th of a full rotation, repeatedly, until I had seven of them inscribed in the circle. Seven times seven, of course, is 49, so this created one type (many are possible) of 49-pointed star. Also, I had already extended the line segments to form lines, so that this geometrical design would extend outside the circle. Next came thickening and blackening these lines, as well as the circle, and re-coloring the red points to be black, as well.

All of this work was performed using Geometer’s Sketchpad. I then took a screenshot, moved the design to MS-Paint, and used that program to add the colors seen in the image at the top of this post.

I don’t like Daylight Saving Time, and never have, but I do enjoy the end of it, when it arrives once each year, and we get our “missing” hour returned to us — the one which was stolen from one of our weekends in the Spring.

To those who live in areas which do not observe DST, such as most of Arizona, you are fortunate — at least in this one respect. Heart attacks actually increase when DST starts each year — a fact which can be easily verified with Google. There are other problems with DST, as well. Daylight Saving Time (one of the worst ideas Benjamin Franklin ever had) should be abolished. Everywhere.

A Halloween Rhombicosidodecahedron

Posted on 31 October 2014 by RobertLovesPi

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!

A Rhombic Dodecahedron, Decorated with Curvy Tessellations

Posted on 30 October 2014 by RobertLovesPi

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.

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.

“How Tall Are You?”

Posted on 12 October 2014 by RobertLovesPi

When I am asked for my height, anywhere — especially at school — I answer the question honestly. I am 1.80 meters tall.

I also live in the USA, one of only three remaining countries (the other two holdouts are Liberia and Myanmar) which have stubbornly refused to adopt the metric system. However, I am every bit as stubborn as other Americans, but, on this issue, I choose to be stubborn in the opposite direction.

It should surprise no one who knows me well that my classroom, whether I am teaching science or mathematics, is, by design, an all-metric zone. After all, like >99% of people, I have ten fingers (assuming thumbs are counted as fingers), ten toes, and almost always use the familiar base-ten number system when counting, measuring, doing arithmetic, or doing actual mathematics. (Doing arithmetic is not the same thing as doing real mathematics, any more than spelling is equivalent to writing.) Using the metric system is consistent with these facts, and using other units is not.

Admittedly, I do sometimes carry this to an extreme, but I do so to make a point. Metric units are simply better than non-metric units. Why should anyone need to memorize the fact that there are 5,280 feet in one mile? It actually embarrasses me that I have that particular conversion-factor memorized. By “extreme,” I mean that I have been known to paint the non-metric side of meter sticks black, simply to make it impossible for students in my classes to confuse inches and centimeters, and prevent them from measuring anything with the incorrect units.

To those who object that American students need to understand non-metric units, I simply point out that there are plenty of other teachers who take care of that. This is, after all, the truth.

Often, after giving my height as 1.80 meters, I am asked to give it in other units. Unless the person asking is a police officer (in, say, a traffic-stop situation), however, I simply refuse to answer with non-metric units. What do I say, instead? “I’m also 180 centimeters tall. Would you like to know my height in kilometers?”

If pressed on this subject in class — and it comes up, because we do lab exercises where the height of people must be measured — I will go exactly this far: I am willing to tell a curious student that there are 2.54 centimeters in an inch, 12 inches in a foot, and 3.28 feet in a meter. Also, I’m willing to loan calculators to students. Beyond that, if a student of mine really wants to know my height in non-metric units, he or she simply has to solve the problem for themselves — something which has not yet happened. I do not wish to tell anyone my height in feet and inches, for I do not enjoy headaches, and uttering my height, in those units I despise, would certainly give me one. Also, obviously, you won’t find my height, expressed in non-metric units, on my blog, unless someone else leaves it here, in a comment — and I am definitely not asking anyone to do that.

I might, just for fun, at some point, determine my height in cubits. For all I know, a person’s height, measured with their own cubits, might be a near-constant. That would be an interesting thing to investigate, and my students, now that I’ve thought about the question, might find themselves investigating this very issue, next week. The variability of cubits, from one person to another, makes them at least somewhat interesting. It also makes cubits almost completely useless, which explains why they haven’t been used since biblical times, but that’s not the point. One can still learn things while investigating something which is useless, if one is sufficiently clever about it.

Feet and inches, however, are not interesting — at all. They are obsolete, just as cubits are, and they are also . . . offensive. It is not a good thing to insult one’s own brain.