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.
The hemisphere problem referred to here was described in the previous post. To reword it somewhat, consider this hemisphere, half of a sphere of radius r. The orange cross-section is a circle parallel to the hemisphere’s yellow, circular base.
We are to find the height of the yellow section with the orange circular top (which I shall call x), as a fraction of r, such that the yellow and red sections above have equal volumes.
Since the volume of a hemisphere is (2/3)πr³ (that’s half a sphere’s volume), each of these two sections must have half the hemisphere’s volume, or (1/3)πr³.
Moreover, the top (red) portion is a “spherical cap,” described here on Wikipedia, as was pointed out to me, by a friend, on Facebook. On that Wikipedia page, you can find this diagram, as well as the formula shown below for the volume of the purple spherical cap in the diagram.
Now, as our goal is to find x, as described at the top of this post, it important to remember that r = x + h, where r is the radius of the original sphere (and height of the hemisphere), h is the height of the spherical cap, and x is the height of the hemisphere, after the spherical cap is removed. We now have two expressions for the volume of the spherical cap: (1/3)πr³ (because it is a fourth of the volume of the original sphere), and (1/6)πh(3a² + h²) from the Wikipedia article on the spherical cap (so all of this assumes, then, accuracy in that Wikipedia article). Setting them equal to each other,
(1/3)πr³ = (1/6)πh(3a² + h²)
Next, I’ll clean this up by multiplying left and right by 6/π, to cancel fractions and π from both sides.
2r³ = h(3a² + h²)
A right triangle exists in the blue-and-purple figure above, and the unlabeled leg is x, the problem’s original goal. I’ll add an “x” to this diagram.
Using this right triangle, and the Pythagorean Theorem, it can be seen that a² = r² – x². Also, since r = x + h, it follows that h = r – x. By substitution for a and h, then,
2r³ = h(3a² + h²)
becomes
2r³ = (r – x)[3(r² – x²) + (r – x)²], which then distributes to
2r³ = (r – x)(3r² – 3x² + r² – 2rx + x²), which expands as
2r³ = 3r³ – 3rx² + r³ – 2r²x + rx² -3r²x + 3x³ – r²x + 2rx² – x³, which simplifies to
0 = 2r³ – 6r²x + 2x³, which becomes, by division:
0 = r³ – 3r²x + x³.
I’m trying to find x, as a fraction of r, meaning that x = kr, and I want k. On that basis, I’ll now substitute kr for each x in the last equation above.
0 = r³ – 3r²(kr) + (kr)³, which then becomes
0 = r³ – 3kr³ + k³r³, or
3kr³ = r³ + k³r³, and then dividing by r³ yields
3k = 1 + k³, which can be rearranged to
3k – k³ = 1, which factors,on the left side, to yield
(k)(3 – k²) = 1.
For k and (3 – k²) to have a product of one, they must be reciprocals. Therefore, 1/k = 3 – k². I can then graph y = 1/k, as well as y = 3 – k², and find the solution by seeing where the graphed functions intersect above the k-axis, with a “k” value between zero and one, since no value of k less than zero or greater than one would make sense, as a solution to the original problem. (I’ll be using k coordinates, and a k-axis, in place of the usual x coordinates and x-axis.) Here’s the initial graph:
The only intersection in the specified range of zero to one is between point A and point B, so I brought them together, as closely as I could get Geometer’s Sketchpad to let me.
With points A and B almost on top of each other, k = 0.34958 by one equation, and k = 0.34820 by the other. To two significant figures, then, I can conclude that the horizontal cut in the original problem should be made 35% of the way from the base of the hemisphere to the hemisphere’s top.
Using a graphing calculator, a more precise answer of 0.34729636 was obtained. I’d still like to have an exact answer, but this will do for now.
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Later addition: a helpful reader led me to a Wolfram Alpha site where I could get an exact answer, as well as a decimal approximation with a greater degree of precision. In the pic below, I have omitted the two solutions of the third-order polynomial which are not the one solution of interest. Here’s the one which is:
Now, however, I have another mystery: how can an exact answer, with all those imaginary units in it, have a real-only approximation? To this question, at least for now, I don’t even have the beginnings of an answer.
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Even later post-script: I have been assured by friends on Facebook that the imaginary units in the above exact solution somehow cancel, although I must concede that I still do not see how, myself. I’ve also been shown another way to express the solution, for 2sin(π/18) is also ≈ 0.3472963553338606977034333. This surprised me, due the the lack of any explicit appearance by a π/18 (= 10°) angle in the original problem, and the fact that no trigonometric functions were used to solve it.
A hemisphere rests with its circular base on a horizontal, level surface, and is to be cut into two pieces of equal volume. If the hemisphere’s radius is r, at what fraction of r above the floor should the horizontal cut be made?
[Solution in next post: https://robertlovespi.wordpress.com/2015/04/06/working-towards-a-solution-of-the-hemisphere-problem/]
This is my second solution to the Zome Cryptocube puzzle. In this puzzle, you start with a black cube, build a white, symmetrical, aethetically-pleasing geometrical structure which incorporates it, and then, finally, remove the cube. In addition, I added a rule of my own, this time around: I wanted a solution which is chiral; that is, it exists in left- and right-handed forms.
It took a long time, but I finally found such a chiral solution, one with tetrahedral symmetry. Above, it appears with the original black cube; below, you can see what it looks like without the black cube’s edges.
Two polyhedra are shown in this post — one which is chiral, and a similar one which is not. The non-chiral polyhedron in this pair is above. Its mirror-image is not any different from itself, except if you consider the direction of rotation.
The similar polyhedron below, however, features an overall “twist,” causing it to qualify as a chiral polyhedron. In its mirror-image (not shown, unless you use a mirror to make it visible), the “twisting” goes in the opposite direction. The direction of rotation would be reversed as well, of course, in a reflected image.
Multiple terms exist for mirror-image pairs of chiral polyhedra, the most well-known of which are the snub cube ansd snub dodecahedron, two of the thirteen Archimedean Solids. Some prefer to call them “enantiomers,” but many others prefer the more familiar term “reflections,” which I often use. I’ve also seen such polyhedra referred to as “left-handed” and “right-handed” forms, but I avoid these anthropomorphic terms related to handedness, simply because, if there is an established rule which would let me know whether any given chiral polyhedron is left- or right-handed, I’m not familiar with it. (Also, polyhedra do not have hands.) I could not, therefore, tell you if the example shown above would be correctly described as left- or right-handed — either because no such rule exists, or there is such a rule, but it is unknown to me. If the latter, I would appreciate it if someone would provide the details in a comment.
Both images above were created with Stella 4d, software you can try, for free, right here.
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