I stumbled upon the compound above while playing with facetings, starting with the rhombic triacontahedron. Here’s the compound’s dual.

I made these rotating models using* Stella 4d*, a program you can try for free at this website.

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I stumbled upon the compound above while playing with facetings, starting with the rhombic triacontahedron. Here’s the compound’s dual.

I made these rotating models using* Stella 4d*, a program you can try for free at this website.

This is the compound of five octahedra, each a different color.

Since the cube is dual to the octahedron, the compound of five cubes, below, is dual to the compound above.

Here are five cubes and five octahedra, compounded together, and shown with the same five colors used above.

This is the same compound, except with all squares/cubes having one color, and triangles/octahedra having another, made by changing the color-scheme used by *Stella 4d* (the program I use to make these images; it’s available here).

After seeing the two-color version of this ten-part compound, I decided to start stellating it, looking for stellations with an interesting appearance. Here is the 23rd stellation of the ten-part compound, colored by face-type.

Next, the 27th stellation, which is chiral, unlike the stellation showed above.

The 33rd stellation also has an interesting appearance (using, I admit, completely subjective criteria for “interesting”), while still having easily-noticable differences to the stellations shown above.

At the 35th stellation, another interesting chiral polyhedron is found. Unexpectedly, its direction of “twist” appears opposite that seen in the 27th stellation. (It could well be that this “twist-reversal” is a common phenomenon in stellation-series — simply one I have never noticed before.)

Next, the ten-part compound’s 39th stellation.

After the 39th stellation, I entered a sort of “desert,” with many stellations in a row which did not strike me as interesting, often with only tiny differences between one and the next. The 194th stellation, though, I liked.

Although I liked the 194th stellation, I didn’t want to risk trudging through another “desert” like the one which preceded it, so I jumped ahead to the final valid stellation, after which the series “wraps around” to its beginning.

Next, I made another rotating image of this final valid stellation, this time with the color-scheme set to “rainbow color mode.”

I couldn’t resist taking this one stellation further, to see the beginning of the stellation-series, since I knew I might have entered it somewhere in the middle, rather than at the beginning.

What I found, I immediately recognized as the rhombic triacontahedron. In some ways, this was surprising, and in other ways, it was not. The compound of five cubes is, itself, a stellation of the rhombic triacontahedron — but what I started stellating also included the compound of five octahedra, which, so far as I know, is not part of the rhombic triacontahedron’s (very) long stellation-series. Also, I know what the rhombic triacontahedron’s final stellation looks like, and it isn’t the final stellation shown above, but is, instead, this:

To try to better-understand just what was going on here, I went back, and deliberately left out the five-cube part of the ten-part compound (which is a stellation of the rhombic triacontahedron), which left me just with the compound of five octahedra — and then I had *Stella* produce this compound’s final stellation.

This was another polyhedron I recognized: the final stellation of the icosahedron. To verify that my memory was correct, I stellated it one more time. Sure enough, this is what I got:

This reminded me that the compound of five octahedra *is* the second stellation of the icosahedron, helping to explain some of this. I also noticed that the five-octahedron compound can be seen as a faceting of the icosidodecahedron. (The icosidodecahedron is dual to the rhombic triacontahedron, and faceting is the reciprocal function of stellation.) However, I have no idea why the final stellation of the ten-part compound above appears as it does.

It is my opinion that a productive polyhedral investigation usually does more than answer questions; it also raises new ones. At least in my mind, that’s exactly what has happened. Therefore, I think this was a perfectly good way to begin the new year.

To make the compound of five cubes, begin with a dodecahedron, as seen above. Next, add segments as new edges, and let them be all of the diagonals of all the dodecahedron’s faces. Then, remove the pentagonal faces, as well as the original edges. What’s left is five cubes, in this arrangement.

Using polyhedral manipulation software called *Stella 4d *(available at www.software3d.com/Stella.php), these five cubes can be removed one at a time. The first removal has this result:

That left four cubes, so the next removal leaves three:

And then only two:

And, finally, only one remains:

Because their edges were pentagon-diagonals for the original dodecahedron, each of these cubes has an edge length equal to the Golden Ratio, (1 + √5)/2, times the edge length of that dodecahedron.

Cubes are more interesting with jumping cube-images on their faces.

Software credit: see http://www.software3d.com/stella.php

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