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A Compound of Four Octahedra

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12 pieces
92 solves

Thanks for sharing. Here is your html-code:

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This is a compound of four octahedra. It is actually a uniform compound. This means that you could map any vertex onto any other vertex by using one if The symmetries. Since it is a symmetry operation, it would then mean that you map all the vertices on a vertex.

The whole model is clearly surrounded by a cube, though it isn't so clear from this picture, which was on purpose. The compound looks a bit like a crystal, I think. That is why a thought I might use only one colour. Instead I decided to give each octahedron its own colour.

The model is quite easy to build, you will only need to cut and paste 144 pieces.

1. Syke0:09
2. marunka2770:10
3. Dilubreuer0:11
4. Wurm0:12
5. smart_ginger0:13
6. Beekay0:14
7. Bubble0:15
8. davitai0:15
9. JennyG660:15
10. Data0:15

You've amazed me again!

After all those hours of staring at it, I thought I had winkled all its secrets out of it. The pyramids on the sides of a cube took me *completely* by surprise. But yes, it is so. And I love the idea of dividing the cube internally into eight smaller, regular pyramids. That explanation also makes the ratio obvious; I just didn't see it. Now I really am wondering by what route I arrived at it.

Thanks for the pointers to the javascript code. I'm not familiar with .off format, so there's another learning gradient for me.

:D Beb.

@blueeyedblond
The rhombic dodecahedron consist of a cube with pyramids on each square. Those pyramids can be obtained from one cube by taking the cubes centre as the top of the pyramid and one face as its base.

In fact the classical way of explaining someone that the rhombic dodecahedron is a space filler is by imagining filling space with cubes an colouring in a checkered way, so that each white cube is surrounded by black cubes and vice versa. Now the rhombic dodecahedra are formed by taking all whites and diving all blacks into these pyramids. Each pyramid is added to its neighbouring cube.

Anyway, I think this is the easiest way to understand that the ratio becomes 1: V2.

About the JavaScript code. It only loads and shows a model in a file at the server side. The file is in the .off format. The JavaScript code can be found on github @marcelteun. The url to the .off files can be found in the HTML files.

You're way ahead of me.. in several directions.

I meant, top layer consisting of one ball and second top layer consisting of two balls per side (= total three balls)

I won't come into the 4D discussion of the pyramid (well, not yet, anyway) because I'm still having trouble "seeing" a mental model of the assertion on your website that in a 4D space, the intersection of two polyhedrons is a plane. EVERY way I look at it, I end up with another 3D intersection (actually a venn diagram). So, if you're right, I'm missing something. :(

Why is the short:long ratio in that case 1:1.4142? I knew it was close to that... but why?
Having built it, I CAN'T REMEMBER how I arrived at the relative dimensions :(
I might have used trigonometry to calculate angles and then just drawn it out. ???

The "playground" structure - an organised "stack" of rhombic dodecahedra with selected internal panels missing so as to create internal climb-through and crawl-through chambers - a sort of artificial cave system.

Assuming that the javascript code will show at my end, I might have a look at it and see if I can adapt it to another shape - interesting exercise!

Yes, I agree with you about either regular three colour or four colour versions. I didn't go as far as colouring mine but in the years since, I have probably spent hours looking at it from both cubic and hexagonal cross sectional points of view, and taking note of the various ways of slicing the planes through it that you referred to in your first reply. The more I look at it the more I am impressed by the elegance and complexity of what it represents. I could so easily have passed through life and missed it completely - as MOST other people sadly do.

One question: what is the noun from which "rhombic" is derived? Each side of the shape is a... what?

Thanks, Beb.

"one and two balls each"
The top layers consist of three and one sphere.

"3D and 4D objects"
I actually thought about it to continue this to the fourth dimension. The first pyramid would then consist of 5 hyperspheres, then 15. I guess you will have to continue to a 4D pyramid consisting of 126 hyperspheres, until you get one that is surrounded by hyperspheres.

"5:7"
The real ratio is 1:V2, where 'V' is the square root. This is more or less 0.7071, while 5:7 ~= 0.7143. That is only one decimal.

The big structure you're thinking about. Are you still thinking about spheres? Or are you thinking about rhombic dodecahedra?

"Did you write the code yourself?"
Yes I did

Btw I found the model that I built so long ago. Actually I built two models, one with three and one with four colours. In both cases the colours can be divided in a regular way, meaning that you can map all faces with colour 'x' on faces with colour 'y' by one symmetry of the whole model.

@marcelteun

HEY! This is the most exciting conversation I've yet had on jigidi!

You are absolutely right in everything you have said in your reply post. :D
And I bow to your picking me up about my lack of definition. Well done! :)

In fact, the base layer would have to be of minimum side 5; the "middle" ball would be in the layer with side 4 (second from the bottom) and the top two layers (with respectively one and two balls each) would be irrelevant to the problem.

I made mine more than fifteen years ago, but you are the first person I have met since then who would even enter into a conversation with me about it. Mentally manipulating 3D and 4D objects just seems to go right over most people's heads.

One of several surprises was the ratio of the long and short dimensions of the rhombic planes - 5:7 apparently accurate to at least three decimal places. Instead of cutting out twelve separate pieces, I worked out how to draw one piece so it represented several sides, using exactly these dimensions, then wherever possible, I folded instead of cutting and joining. Thus I was able to assemble the actual object with a minimum of pieces, without any distortion or stress.

Something I would like to do is to actually build the pyramid, to about five levels high, in something durable like stainless steel or a high density plastic, in a suitable scale for a kids' playground, with enough internal panels missing that it became an "indoor" climbing maze NOT based on the right angles and horizontal floors that we are all used to in most of our man-made constructions. With a bit of thought, the missing panels could be selected so as to produce two or three different routes through the maze with possibly a hole the kids could stick their heads out of somewhere near the top. Again, I have lain in bed at night, visualising various combinations of missing panels to see where the maze would lead.

I've copy-and-pasted your link below and manipulated the 3D object. Fascinating! Apparently my graphics hardware is up to the job. Did you write the code yourself?

And since writing the above line, I have returned to the link, surfed around a bit and eventually landed on your home page, which I have bookmarked. I am sure we will carry this conversation further! :D Beb.

@blueeyedblond on this page I added a 3D model that can be investigated interactively (if you have JavaScript enabled and reasonable new HW):
http://tunnissen.eu/polyh/uniform_compounds/4xW2-UC12.html
Then you will be able to see the cube shape.

@blueeyedblond I haven't given this a lot of thought, but there are many ways of making tetrahedron pyramids with spheres. One way is to put three spheres in a triangle and then put a fourth one in the middle on top of the others. Now you have four spheres. Then there is no answer on your first question, since there is no sphere that is in the middle. So I think that your first questing should be: what is the least amount of spheres that you should use to get one sphere that is in the middle. I think that that should be 35 spheres making 5 layers, where the middle is in the second layer from the bottom. I that case the answer to your first question would be 12.

For your second question I think you mean three layers, if you only count the horizontal ones as layers. If you count all the planes parallel to the sides as layers, then it would be four times as much.

Third questing: that would be a rhombic dodecahedron, which has both a cube and an octahedron inside (take either the long or short diagonals). And yes I can and I did built that one, though I'm not sure where it is.

PS: Your colouration of the polyhedron in your puzzle is a nice touch. It does make it easy to distinguish each octahedron from the others. I'm not so sure about the cube yet, so if it truly IS surrounded by a cube, then I agree that you did a good job of obscuring this. ;) Beb.

Nice one @marcelteun !

I lay in bed one night thinking about a triangular pyramid of cannon balls.
Obviously, they touch the other cannon balls around them in a regular pattern.

So:
FIRST QUESTION: for any cannon ball in the middle of the pyramid, how many other cannon balls does it touch?
SECOND QUESTION: How many layers of the pyramid are represented by the other cannon balls it touches?
THIRD QUESTION (this is the curly one!): If you take the plane of each of the tangential points of contact identified in Question One and extend it until it is intersected by one or more of the other planes, what will be the resulting shape of each plane and how many other planes will intersect it?
FOURTH (and final) QUESTION: Can you build the polyhedron represented by the sum of these intersecting planes?

I'll let you mull on these questions for a while to see if you can work it out for yourself, as I did. I DID eventually build the polyhedron. It is a rather fascinating little structure and its shape and relative dimensions surprised me. It also follows that if you had the patience to build one for each cannon ball in the original pyramid, with sufficient precision, they would all stack as neatly together as the cannon balls - in a solid stack with NO air gaps between the faces.