# For me, geometry for breakfast is not unusual. This morning, though, I’m sprinkling calculus on top before eating it.

It’s important to explain, right up front, that Ronald Reagan was president when I last took calculus. However, I have a new determination to learn the subject. I have a hunch this may go better without the “help” of actually being enrolled in a calculus class, since the way I learn things, and the way most people learn things, aren’t much alike.

My current calculus puzzle started when I noticed that taking the derivative of the volume of a sphere, in terms of the radius, (4/3)πr³, yields the formula for the surface area of a sphere, 4πr². That was both unexpected and exciting, so I tried applying the same idea to another solid: the cube. With edge length e, the volume of a cube is e³, and the derivative of that is 3e² . . . but that’s only half of the surface area of a cube, which is 6e².

Half? What’s going on here? I mentioned this puzzle on Facebook, where I have many on my friends’-list whose mathematical knowledge exceeds my own. It was pointed out to me that I’d made an important and unhelpful change by going from using the radius, for the sphere, to the edge length, for the cube.

So I’ll try this again, but do it in terms of the radius of the cube, rather than the edge length. For a cube, the radius extends from the center to any of the cube’s eight vertices. Both the light and dark blue segments in the diagram below are cube radii.

This radius is sqrt(3)/2 times the cube’s edge length, as can be verified by applying the Pythagorean Theorem twice, first to triangle ABC (which shows that the green face-diagonal is sqrt(2) times the edge length), and then to triangle BCD (which yields sqrt(3) times the edge length for the interior diagonal DC, half of which is the radius).

It then follows that, if r = [sqrt(3)/2]e, that e = [2/sqrt(3)]r, which “cleans up” to e = (2/3)sqrt(3)r, when the denominator is rationalized.

If a cube’s volume is e³, and e = (2/3)sqrt(3)r, it then follows that V = [(2/3)sqrt(3)r]³ = (8/27)(3)sqrt(3)r³ = (24/27)sqrt(3)r³ = [8sqrt(3)/9]r³. If I take the derivative of the last expression, I get [8sqrt(3)/3]r² for the derivative of the volume, which I now need to compare to the surface area of a cube, in terms of its radius, rather than edge length.

So here goes . . . SA = 6e² = 6[(2/3)sqrt(3)r]² = [48(3)/9]r² = 16r², which isn’t what I got for the derivative of the volume, above.

Well, I was using, as the radius, the radius of the cube’s circumscribed sphere. Perhaps I should have used the inscribed sphere, instead? The radius of the cube’s inscribed sphere is the “invisible” segment FM in the diagram above, which I’m going to call “a” (for “apothem,” because this looks like the 3-d version of the apothem of a regular polygon). The length of a is exactly one-half that of e, the cube’s edge length, which means that e = 2a. Therefore, V = e³ = (2a)³ = 8a³, the derivative of which is 24a².

Now to check the surface area, in terms of a: SA = 6e² = 6(2a)² = 24a², and that’s what I got when I took the derivative of the volume, in terms of a.

So this trick works for the cube if you use the radius of the inscribed sphere, but not the circumscribed sphere. This leaves me with three questions to address later:

1. Will this also work for other polyhedra? This is something I intend to explore in future blog-posts, starting with the tetrahedron and the octahedron.
2. Why did this work at all?
3. Why was it necessary to use the radius of the cube’s inscribed sphere, rather than its circumscribed sphere?

If any reader of this post knows the answer(s) to #2 and/or #3, sharing your knowledge in a comment would be very much appreciated.

# Open Octahedral Lattice of Cubes and Rhombicosidodecahedra

This pattern could be continued, indefinitely, into space.

Here is a second view, in rainbow color mode, and with all the squares hidden.

[These images were created with Stella 4d, software you may buy — or try for free — right here.]

# Six “Cubish” Polyhedra

I’m using the term “cubish polyhedra” here to refer to polyhedra which resemble a cube, if one looks only at the faces they have which feature the largest number of sides, always six in number, and with positions corresponding to the faces of a cube. In the first two examples shown, these faces are 36-sided polygons, also known as triacontakaihexagons. (Any of the images in this post may be enlarged with a click.)

Polyhedra fitting this description have appeared on this blog before, but it had not occurred to me to name them “cubish polyhedra” until today. The next two shown have icosakaioctagons, or 28-sided polygons, as their six faces which correspond to those of a cube. Also, and unlike the triacontakaihexagons in the first two cubish polyhedra above, these icosakaioctagons are regular.

The next two cubish polyhedra shown feature, on the left, six hexadecagons (16 sides per polygon) for “cubish faces,” which are shown in yellow — and on the right, six dodecagons (12 sides each), shown in orange. This last one, with the dodecagons, is unusual among cubish polyhedra in that all of its other faces are pentagons.

All six of these cubish polyhedra were made using Stella 4d: Polyhedron Navigator, a program you can find right here.

# The Compound of Five Cubes, Augmented with Thirty Snub Cubes: Three Versions

This cluster-polyhedron was made with Stella 4d, software you can try at this website. Above, it is colored by face-type, referring to each face’s position within the overall cluster. In the image below, the original compound of five cubes contained one cube each, of five colors, and then each snub cube “inherited” its color from the cube to which it was attached.

In the next version, the colors are chosen by the number of sides of each face.

# One Dozen Precious Metal Cubes: A Problem Involving Geometry, Chemistry, and Finance (Solution Provided, with Pictures)

The troy ounce is a unit of mass, not weight, and is used exclusively for four precious metals. At this time, the prices per troy ounce, according to this source for current precious metal prices, for these four elements, are:

• Gold, \$1,094
• Platinum, \$965
• Silver, \$14.82

(As a side note, it is rare for platinum to have a lower price per troy ounce than gold, as is now the case. I would explain the reasons this is happening, except for one problem: I don’t understand the reasons, myself, well enough to do so. Yet.)

A troy ounce equals 31.1034768 grams, but, for most purposes, 31.103 g, or even 31.1 g, works just fine.

Also, as you can see here, these “troy elements” are all in one part of the periodic table. This is related to the numerous similarities in these elements’ physical and chemical properties, which is itself related, of course, to the suitability of these four elements for such things as jewelry, coinage, and bullion.

To determine the volume of a given mass of one of these metals, it is also necessary to know their densities, so I looked them up, using Google (they are not listed on the periodic table above):

• Gold, 19.3 g/cm³
• Platinum, 21.46 g/cm³
• Silver, 10.49 g/cm³

In chemistry, of course, one must often deal with elements (as well as other chemicals) in terms of the numbers of units (such as atoms or molecules), except for one problem: this is absurdly impractical, due to the outrageously small size of atoms. Despite this, though, it is necessary to count such things as atoms in order to do much chemistry at all, so chemists have devised a “workaround” for this problem: when counting units of pure chemicals, they don’t count such things as atoms or molecules directly, but count them a mole at a time. A mole is defined as a number of things equal to the number of atoms in exactly 12 grams of pure carbon-12. To three significant figures, this number is 6.02 x 10²³. To deal with moles, since atoms have differing masses, we need to know the molar mass (mass of one mole) of whatever we are dealing with to convert, both directions, between moles and grams. Here are the molar masses of the four troy-measured elements, as seen on the periodic table above, below each element’s symbol.

• Gold, 196.97 g
• Platinum, 195.08 g
• Silver, 107.87 g

I’ve given these numbers  as the information needed to solve the following problem: rank one dozen precious metal cubes (descriptions follow) by ascending order of volume. There are three cubes each of gold, palladium, platinum, and silver. Four of the twelve (one of each element) have a mass of one troy ounce each. Another four each have a value, at the time of this writing, of \$1,000. The last set of four each contain one mole of the element which composes the cube, and, again, there is one of each of these same four elements in the set.

If you would like to do this problem for yourself, the time to stop reading is now. Otherwise (or to check your answers against mine), just scroll down.

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In the solutions which follow, a rearrangement of the formula for density (d=m/v) is used; solved for v, this equation becomes v = m/d. In order, then, by both volume and edge length, from smallest to largest, here are the twelve cubes:

Smallest cube: one troy ounce of platinum

One tr oz, or 31.103 g, of platinum would have a volume of v = m/d = 31.103 g / (21.46 g/cm³) = 1.449 cm³. A cube with this volume would have an edge length equal to the its volume’s cube root, or 1.132 cm. (This explanation for the calculation of the edge length, given the cube’s volume, is omitted in the items below, since the mathematical procedure is the same each time.)

Second-smallest cube: \$1000 worth of gold

Gold worth \$1000, at the time of this posting, would have a troy mass, and then a mass in grams, of \$1000.00/(\$1,094.00/tr oz) = (0.914077 tr oz)(31.103 g/tr oz) = 28.431 g. This mass of gold would have a volume of v = m/d = 28.431 g / (19.3 g/cm³) = 1.47 cm³. A cube with this volume would have an edge length of  1.14 cm.

Third-smallest cube: \$1000 worth of platinum

Platinum worth \$1000, at the time of this posting, would have a troy mass, and then a mass in grams, of \$1000.00/(\$965.00/tr oz) = (1.0363 tr oz)(31.103 g/tr oz) = 32.231 g. This mass of platinum would have a volume of v = m/d = 32.231 g / (21.46 g/cm³) = 1.502 cm³. A cube with this volume would have an edge length of  1.145 cm.

Fourth-smallest cube: one troy ounce of gold

One tr oz, or 31.1 g, of gold would have a volume of v = m/d = 31.1 g / (19.3 g/cm³) = 1.61 cm³. A cube with this volume would have an edge length of 1.17 cm.

Fifth-smallest cube: one troy ounce of palladium

One tr oz, or 31.1 g, of palladium would have a volume of v = m/d = 31.1 g / (11.9 g/cm³) = 2.61 cm³. A cube with this volume would have an edge length of 1.38 cm.

Sixth-smallest cube: one troy ounce of silver

One tr oz, or 31.103 g, of silver would have a volume of v = m/d = 31.103 g / (10.49 g/cm³) = 2.965 cm³. A cube with this volume would have an edge length of 1.437 cm.

Sixth-largest cube: \$1000 worth of palladium

Palladium worth \$1000, at the time of this posting, would have a troy mass, and then a mass in grams, of \$1000.00/(\$600.00/tr oz) = (1.6667 tr oz)(31.103 g/tr oz) = 51.838 g. This mass of palladium would have a  volume of v = m/d = 51.838 g / (11.9 g/cm³) = 4.36 cm³. A cube with this volume would have an edge length of  1.63 cm.

Fifth-largest cube: one mole of palladium

A mole of palladium, or 106.42 g of it, would have a volume of v = m/d = 106.42 g / (11.9 g/cm³) = 8.94 cm³. A cube with this volume would have an edge length of 2.07 cm.

Fourth-largest cube: one mole of platinum

A mole of platinum, or 195.08 g of it, would have a volume of v = m/d = 195.08 g / (21.46 g/cm³) = 9.090 cm³. A cube with this volume would have an edge length of 2.087 cm.

Third-largest cube: one mole of gold

A mole of gold,  or 196.97 g of it, would have a volume of v = m/d = 196.97 g / (19.3 g/cm³) = 10.2 cm³. A cube with this volume would have an edge length of  2.17 cm.

Second-largest cube: one mole of silver

A mole of silver, or 107.87 g of it, would have a volume of v = m/d = 107.87 g / (10.49 g/cm³) = 10.28 cm³. A cube with this volume would have an edge length of 2.175 cm.

Largest cube: \$1000 worth of silver

Silver worth \$1000, at the time of this posting, would have a troy mass, and then a mass in grams, of \$1000.00/(\$14.82/tr oz) = (67.48 tr oz)(31.103 g/tr oz) = 2099 g. This mass of gold would have a volume of v = m/d = 2099 g / (10.49 g/cm³) = 200.1 cm³. A cube with this volume would have an edge length of  5.849 cm.

Finally, here are pictures of all 12 cubes, with 1 cm³ reference cubes for comparison, all shown to scale, relative to one another.

A third of these cubes change size from day-to-day, and sometimes even moment-to-moment during the trading day, if their value is held constant at \$1000 — which reveals, of course, which four cubes they are. The other eight cubes, by contrast, do not change size — no precious metal prices were used in the calculation of those cubes’ volumes and edge lengths, precisely because the size of those cubes is independent of such prices, due to the way those cubes were defined in the wording of the original problem.

# Two Different Forty-Part Polyhedral Compounds

The polyhedron above is a compound of twenty cubes and twenty octahedra, colored by symmetry-based face-type. If the same compound is viewed in “rainbow color mode,” it looks like this:

With this particular compound, though, there are two versions — without taking coloring into consideration at all. The other version simply has the twenty cubes and twenty octahedron in a different, but still symmetrical, arrangement:

The compound above uses this second arrangement, colored by face type, and the next image is the same (second) compound, but in “rainbow color mode.”

These rotating polyhedral images were made with Stella 4d, software you can try for yourself, right here.

# Two Polyhedral Meta-Compounds

The polyhedral compound above is actually a compound of two compounds: the compound of three cubes (red, yellow, and blue), as well as the cube/octahedron base/dual compound (green and purple). The dual of this five-part compound is shown below, still with the cube/octahedron compound in green and purple (it is its own dual), and with the three parts of the compound of three octahedra in red, yellow, and blue.

I created these using the “add/blend from memory” function of Stella 4d: Polyhedron Navigator, one of this program’s capabilities which I have only recently begun to explore. You may try this software for yourself, for free, right here.

# A Chiral Solution to the Zome Cryptocube Puzzle

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.

# A Kite-Faced Polyhedron Based on the Cube, Octahedron, and Rhombic Dodecahedron

Above is the entire figure, showing all three set of kites. The yellow set below, though, lie along the edges of a rhombic dodecahedron.

The next set, the blue kites, lie along the edges of an octahedron.

Finally, the red set of kites lies along the edges of a cube — the dual to the octahedron delineated by the blue kites.

These images were made using Stella 4d, which is available here.

# Three Polyhedral Clusters of Icosahedra

In the last post on this blog, there were three images, and the first of these was a rotating icosahedron, rendered in three face-colors. After making it, I decided to see what I could build, using these tri-colored icosahedra as building blocks. Augmenting the central icosahedron’s red and blue faces with identical icosahedra creates this cubic cluster of nine icosahedra:

If, on the other hand, this augmentation is performed only on the blue faces of the central icosahedron, the result is a tetrahedral cluster of five icosahedra:

The next augmentation I performed started with this tetrahedral cluster of five icosahedra, and added twelve more of these icosahedra, one on each of the blue faces of the four outer icosahedra. The result is a cluster of 17 icosahedra, with an overall icosahedral shape.

All of these images were made using Stella 4d, which is available at http://www.software3d.com/Stella.php.