One brain teaser

ONE, of course.

(What a load "big word" nonsense).

1 cubic unit of length

Actually not quite! The pedantic answer is ....one unit of volume equal to the cube of one unit of length...but since the OP omitted to specify units the answer can equally do the same.

Yeah, I thumbed up lood's answer, because I just figured he was messing with us, just like he does 90% of 110% of the time.

...

The problem isn't as trivial as it looks, however...

Are the circles supposed to be co-planar? Or are we supposed to arrange them in 3-dimensional space so that every circle touches every other circle? I need clarification before I can tell you whether or not I agree that the answer is actually ONE.

I suggest that the word "regular" in the OP rules out non-coplanar arrangements of the circles. For example, if the circles were circumcircles of the faces of a tetrahedron, it would be impossible for all four centres to be vertices of a regular hexahedron.
Conversely, if we interpret the OP to mean that all the circles must touch each other, then there is no answer.

If on the other hand we allow the circles to be inscribed circles of the faces of a cube, then their centres could be vertices of a cube volume [sqrt(8)] cubed
[change "circumcircles" to "inscribed circles" for tetrahedron above]

Aaaaaah!........that volume should have read "[sqrt(1/8)] cubed"

Yes, that's exactly what I was picturing. I was picturing our four "centre points" as the centres of the inscribed circles of a regular tetrahedron.



So you're saying the cube couldn't even incorporate our "centre points" as one of its two 4-element independent sets? I had a hunch that perhaps the cube could incorporate them as one of its independent sets, but I had not yet run the numbers to check whether that would work. Perhaps I was mistaken.

(For anyone following along, an "independent set" in graph theory is a set of vertices, no two of which are adjacent.)

If , instead of a cube, "a regular hexahedron" can be formed by two tetrahedons back-to-back, then there is a solution and I must retract by "big words" comment.

The volume of such a hexahedron would be
2 times (sqrt2)/12 times ((sqrt3)/2 )^3

http://whistleralley.com/polyhedra/tetrahedron.htm

BUT A google search on "regular hexahedron" only gives "cube".

Well, fresco, I've thought it over some more, and I still think that there is a solution where:

(1) "regular hexahedron" DOES mean "cube", just like I thought it did;
(2) each pair of circles touch at exactly one point;
(3) the circles are all inscribed on the sides of a single regular tetrahedron;
(4) the regular hexahedron (or "cube") incorporates the four circle centres as the vertices of one of its 4-element stable sets.

I plan to prove it, and then to post the proof as soon as I can figure out the best way to upload my sketches to the Internet.

Cheers,
Oylok

But that might not be for some time, because I don't like the graphics tools currently at my disposal....

For now, I will just give an overview, with none of the coordinate systems or graphics that might help people see that I'm right.

Before I start, I need to introduce a new term, which will hopefully render the explanation clearer, and which will describe four points in space. If you have four points in space, such that the distance between any two is always the same, then we'll say those points form a "tetrahedral pattern." They have the same spatial relationship to each other as the hydrogen atoms in a methane molecule, I think. They form the corners of their own little regular tetrahedron.

Now, we want to line up four circles of equal size, so that they "just touch" in six places--each pair of circles touching exactly once. You can do that by inscribing them on the sides of a regular tetrahedron, which (to avoid confusing with other tetrahedra) I'm going to call "the BIG tetrahedron." (The other requirement we make of the circles is that we have to be able to superimpose 4 vertices of a cube on their centres, but we'll see in time that this can be done.)

The first claim I'll make is that the four circles' centres are in a "tetrahedral pattern." You can see that this must be the case. Let's say that the distance between two of the centres ('a' and 'b', say) is X. Well, we can rotate the big tetrahedron around so that the two other centres ('c' and 'd', say) take the place of where 'a' and 'b' were, so the distance between 'c' and 'd' has to be X as well. In other words, all circle centres are the same distance apart, so their pattern is tetrahedral.

The second claim I'll make is that the stable set on a cube also forms a "tetrahedral pattern" in space. Let's use the unit cube whose vertices are {(0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), (1,1,1)}. One of the stable sets of this cube is {(0,0,0), (0,1,1), (1,0,1), (1,1,0)}. Clearly, any two of these points are a distance sqrt(2) apart. So the stable set forms a tetrahedral pattern in space.

The great thing about that is that by magnifying, rotating and translating one tetrahedral pattern of points, we can move them around so they cover up another tetrahedral pattern. So by magnifying, rotating and translating our unit cube we can cover up the 4 circle centres with the 4 vertices in cube's stable set, without our cube losing that lovely "cube shape." Therefore it is possible for a cube to incorporate the centres of 4 circles which all "just touch each other."

(I may have botched the calculation, but I believe the volume of the cube we want is sqrt(6) / 36. I will have to double check.)

Nice job!

Well done ! I believe your proof boils down to saying that a tetrahedron can be formed by joining the ends of diagonals perpendicular to each other from two opposite faces of a cube.