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.

Very nice design work there, Robert. Happy New Year to you and your family.

Leslie

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The same to you and yours! Thanks for following. =)

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Haha, I sort of jumped all the way toward the end while staring at the first shape. I was counting how many of those sort of rhombic pyramids were covering the surface of the compound octahedra, and realized, hey, there are 6 of those rhombic pyramids for each vertex of each octahedron. 6×5 is 30. So I thought hey, I bet if you shave off all those to create a convex figure, it’ll be a rhombic triacontahedron!

And I was right! What’s it called when you “stellate” a polyhedron like that? like, take a rhombic triacontahedron, and slap pyramids on each of its rhombic faces?

What do you get if you do that to a deltoidal icositetrahedron’s faces? If you do it to a cube, you get an octahedron, but bigger. What do you get if you do it to a Rhombic Dodecahedron? A Blinski’s Dodecahedron?

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But, you know, As I look closer, it’s that’s not really what it is. Why is the compound of 5 octahedra so similar to what happens to a rhombic triacontahedron when you put pyramids on all the faces?

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I’m not sure, to be honest!

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Do you know the name of that kind of “stellation”? I know it’s not technically a stellation…. Well, before I posted my reply here, I think I found the answer .It might be Kleetope, or the Kis operator. But that’s usually a much shallower pyramid than would be here.

Do you think you could get Stella4D to make a rhombicuboctahedron with pyramids on each face that have the same outer vertex configuration as the compound of 5 octahedra?

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I’m not sure about the stellation-types; I simply count mouse-clicks with

Stellato keep track of them. As for your question about the rhombcuboctahedron modification, I’m reasonably sure that is possible inStella.LikeLike

What if you do the exact same thing, except using dodecahedra and icosahedra?

What will the final stellation of the compound look like?

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I don’t know. I think it’s possible, but I don’t know how to do it (yet).

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