*Stella 4d: Polyhedron Navigator* has a “put models on vertices” function which I used to build this cluster of cubes. If you’d like to try this software for yourself, there is a free trial download available at http://www.software3d.com/Stella.php.

# Tag Archives: cube

# The Eleventh Stellation of the Truncated Octahedron Is an Interesting Polyhedral Compound

This compound has three parts: two tetrahedra, plus one smaller cube. I made it using Stella 4d: Polyhedron Navigator, which you can try for free at http://www.software3d.com/Stella.php.

# Some Ten-Part Polyhedral Compounds

While examining different facetings of the dodecahedron, I stumbled across one which is also a compound of ten elongated octahedra.

Here’s what this compound looks like with the edges and vertices hidden:

Next, I’ll put the edges and vertices back, but hide nine of the ten components of the compound. This makes it easier to see the single elongated octahedron which is still shown.

Here’s what this elongated octahedron looks like with all those vertices and edges hidden from view.

I made all these polyhedral transformations using *Stella 4d*, a program you can try for yourself at this website. *Stella *includes a “measurement mode,” and, using that, I was able to determine that the short edge to long edge ratio in these elongated octahedra is 1:sqrt(2).

The next thing I wanted to try was to make the octahedra regular. *Stella* has a function for that, too, and here’s the result: a compound of ten regular octahedra.

My last step in this polyhedral exploration was to form the dual of this solid. Since the octahedron’s dual is the cube, this dual is a compound of ten cubes.

# The Cubic Rhombicosidodecahedroid

I call the polyhedron above the cubic rhombicosidodecahedroid because it combines a cube’s six squares (shown in green) with the overall appearance of a rhombicosidodecahedron. For comparison, the latter two polyhedra are shown below.

I made these rotating images using *Stella 4d: Polyhedron Navigator*. This program may be tried for free at http://www.software3d.com/Stella.php.

# Some Tetrahedral Stellations of the Truncated Cube

I created these with *Stella 4d*, which you may try for free at this website. To make a given polyhedral stellation appear larger, simply click on it.

# The Truncated Cube, with Two Variations Featuring Regular Dodecagons

This is the truncated cube, one of the thirteen Archimedean solids.

If the truncation-planes are shifted, and increased in number, in just the right way, this variation is produced. Its purple faces are regular dodecagons, and the orange faces are kites — two dozen, in eight sets of three.

Applying yet another truncation, of a specific type, produces the next polyhedron. Here, the regular dodecagons are blue, and the red triangles are equilateral. The yellow triangles are isosceles, with a vertex angle of ~41.4 degrees.

All three of these images were produced using *Stella 4d*, available at this website.

# 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:

- 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.
- Why did this work at all?
- 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.