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I've got 8 points - A, B, C, D, E, F, G, H - and I need three specific sets of four (ABCD, ABEF, and CEGH) to describe tetrahedra of equal edge length in some multidimensional space.

I can embed ABCD and ABEF in $\mathbb R^3$, but I can't also put CEGH in $\mathbb R^3$, because then the edge connecting C and E is longer than the others, and there's nowhere to put G and H.

Of course I can go crazy and use the 7-simplex in $\mathbb R^7$ - i.e., so that all possible pairs of four are tetrahedra - but I'd like to avoid using extra dimensions.

The question: What is the minimal dimensionality I need for these 3 tetrahedra? And, any pointers as to how to find the cartesian coordinates of the vertices?

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    Why doesn't R^3 work? Allow AB to be the Z axis - say A at (0,0,-1) and B at (0,0,1); then start with C=(-2, -1, 0) and D=(-2, 1, 0) - so the tetrahedron ABCD goes in the negative X direction - and E=(2, -1, 0), F=(2, 1, 0), so ABEF goes in the positive X direction. Now you should be able to rotate one of these about the Z axis to give CE whatever length you need.2012-07-25
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    Yeah, I'm not seeing why you can't force $CE$ to be the same distance as $AB$ in $\mathbb R^3$.2012-07-25
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    Wow, you guys are right. Thank you. I'm terrible at thinking in 3d - rotation was the trick.2012-07-25
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    @CHCH In fact, you can pack five tetrahedra around a common edge this way, but not quite perfectly - there's a little bit of space left over. Folding into the fourth dimension to close up that space gives you the 600 cell (http://en.wikipedia.org/wiki/600-cell ).2012-07-25
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    Mind=blown. But can you point me in the right direction for getting an intuition for what you mean by "folding into", or thinking more fluidly about this? I guess I should work on understanding R^3 first...2012-07-26
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    CHCH: To notify other users of your comments, please use the [@-reply system](http://meta.stackexchange.com/questions/43019/how-do-comment-replies-work). I am guessing @Steven didn't see your last comment above. (And now I just pinged him.)2012-07-26
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    @CHCH The best example I can offer you for this: in $\mathbb{R}^2$ you can fit six equilateral triangles around a single point (this is the classic tiling of the plane by equilateral triangles). If you try to just fit five - by identifying the outside edges of the two end triangles - you'll find that you need to bend your plane a little bit to do it, and so you wind up 'folding' your vertex up into 3-space. If you continue to build cycles of five triangles around each vertex this way, you'll find your shape gradually closing back on itself and forming an icosahedron.2012-07-26
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    @CHCH The construction of the 600-cell in $\mathbb{R}^4$ works very similarly, except that instead of closing the gap between two triangles around a point you're closing the gap between two tetrahedra around an edge. There's less folding to be done - five tetrahedra come closer to completing a circle around your edge than five triangles do around your point - so you need more tetrahedra to eventually 'close up' the shape, which is why it has 600 tetrahedral faces (as opposed to the icosahedron's 20 triangles).2012-07-26

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