This is the familiar dodecahedron/icosahedron compound, but with each face of both components of the compound altered by the excavation of an equal-edge-length pyramid. To make it, as well as the rotating image below, I used Stella 4d, which you can find here.
Also, here is the dual of the compound above:
In this video, the great dodecahedron is stellated, by twentieths, into the great stellated dodecahedron, while a selection from Ludwig van Beethoven’s Ninth Symphony plays. The images for this video were created using Stella 4d, a program you can try for yourself (free trial download available), right here: http://www.software3d.com/Stella.php.
I made this with Stella 4d, software you can try here.
This compound was created using Stella 4d, software you can try for yourself here.
The sine and cosine curves themselves are shown for reference, and all four possible combinations of sine and/or cosine which nest one function inside another (with values for the outer function to evaluate ranging from -2π to 2π) are also shown.
Shown above: the compound of the icosahedron and the rhombic dodecahedron. Below is its dual, the compound of the dodecahedron and the cuboctahedron.
Both these compounds were created using the “add/blend polyhedron from memory” function in Stella 4d: Polyhedron Navigator. To check out this program for yourself, just follow this link.
I’m in my twentieth year of teaching mostly science and mathematics, so it is understandable that most people are surprised to learn that I majored in, of all things, history.
It’s true. I focused on Western Europe, especially modern France, for my B.A., and post-WWII Greater China for my M.A. My pre-certification education classes, including student teaching, were taken between these two degree programs.
Student teaching in social studies did not go well, for the simple reason that I explain things by reducing them to equations. For some reason, this didn’t work so well in the humanities, so I took lots of science and math classes, and worked in a university physics department, while working on my history M.A. degree, so that I could job-hunt in earnest, a year later, able to teach physics and chemistry. As it ended up, I taught both my first year, along with geometry, physical science, and both 9th and 12th grade religion. Yes, six preps: for an annual salary of US$16,074.
History to mathematics? How does one make that leap? In my mind, this explains how:
…And that at least starts to explain how someone with two history degrees ended up with both a career, and an obsession, way over on the mathematical side of academia.
This compound was created using Stella 4d, software you can try here.
Pyritohedral symmetry, seen by example both above and below, is most often described at the symmetry of a volleyball:
[Image of volleyball found here.]
To make the rotating polyhedral compound at the top, from an octahedron and an icosahedron, I simply combined these two polyhedra, using Stella 4d, which may be purchased (or tried for free) here.
In the process, I demonstrated that it is possible to combine a figure with octahedral (sometimes called cuboctahedral) symmetry, with a figure with icosahedral (sometimes called icosidodecahedral) symmetry, to produce a figure with pyritohedral symmetry.
Now I can continue with the rest of my day. No matter what happens, I’ll at least know I accomplished something.
That hemisphere problem (see previous two posts) was quite difficult. I’m going to unwind a bit with the much easier cone version of the same problem: at what height x above the ground, expressed as a fraction of h, must a cone of height h and radius r be cut, in order for the two pieces produced by the cut to have equal volume? The fact that a path down the lateral surface of a cone is a straight line, not a curve, should make this much easier than the hemisphere problem.
Since the volume of a cone is (1/3)πr²h, and the smaller cone created above the cut would be half that volume, it follows that
(1/3)πr²h = (2/3)π(r′)²h′ [equation 1]
By cancellation of (1/3)π, this equation becomes
r²h = 2(r’)²h’ [equation 2]
Also, based on divisions of the cone’s altitude, we know that
h = h′ + x [equation 3]
Furthermore, since the problem asks that the height x be expressed as a fraction of h, we can let that fraction (a decimal between zero and one) be represented by f, so that
x = fh [equation 4]
Also, by using similar right triangles’ corresponding legs, we know that
r/h = r′/h′ [equation 5]
which rearranges to
rh′ = r′h [equation 6]
There is a proportionality constant in play here, p, defined as the fraction of the length of one part of the larger cone which equals the length of the corresponding part of the smaller cone. As equations, then,
r′ = pr and h′ = ph [equations 7a and 7b]
Also, because p is the fraction of h which is h′, and f is the fraction of h which is x, and h = h′ + x, it follows that
p + f = 1 [equation 8].
Next, by substituting equations 7a and 7b into equation 2 for r′ and h′, we know that
r²h = 2(pr)²ph [equation 9]
Which reduces to
1 = 2p³ [equation 10]
When equation 10 is solved for p, it becomes
p = (1/2)^(1/3) [equation 11]
And, since equation 8 states that p + f = 1, it follows that f = 1 – p, and f is the fraction we seek. By substituting equation 11 for p in f = 1 – p, the following value for f can be determined:
f = 1 – (1/2)^(1/3) [equation 12]
This leads to the following cleaned-up solution to the problem, shown in standard exact form, and with a decimal approximation as well.
The cut, therefore, should be made approximately 20.6% of the way from the bottom to the top of the full cone.
To check this answer, I need only find the volume of the smaller cone, times two, and show that it equals the value of the larger cone.
2(volume of smaller cone) = (2/3)π(r′)²h′ = (2/3)π(pr)²ph =
(2/3)πp³r²h = (2/3)π(cube root of ½)³r²h = (2/3)π(1/2)r²h = (1/3)πr²h,
which is the volume of the full cone, as it should be. The problem has now been solved, and the solution f (by way of p, which equals 1 – f, by a rearrangement of equation 8) has been checked.
RobertLovesPi.net 