[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 154
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Это может сказать только Санёк. Я в поисковике такой задачи не нашел.
I tell you this problem was posed in an issue of Science and Life (and it came from some international convention of puzzlers or something similar), the solution is unambiguous
если можно организовать такую структуру из тетраэдра, то почему нельзя из куба
The problem seems to be unsolvable. I'll try to prove it tomorrow. Odds and evens are more likely to work.
// That's it for today.
если можно организовать такую структуру из тетраэдра, то почему нельзя из куба
No and that's it!
If it were possible, the forex distribution would be normal or at least strictly Cauchy. But it's a hybrid with bifurcation manners.
// Mathemat, pardon for the offtop... ;-)
про муравьев:
A=3+6-9
B=8+4-12
C=10+1-11
D=5+2-7
Бинго! :)))
You owe me another one of those plates. Just a different one. You guessed it yourself... ;)
We construct a cube in Cartesian space so that its side is 1, one of its vertices is (0,0,0) and the other vertex on the main diagonal is (1,1,1).
There are three square faces adjacent to each end of the main diagonal, forming two trihedrons A and B (they are not closed in volume, they have none). The whole surface of the cube is a union of these trihedrons. Now consider one such tetrahedron (A).
Draw its graph (by its edges). The graph consists of a hexagon and one point in its centre connected to three vertices of the hexagon through one. The point in the centre is the vertex of the main diagonal.
We apply weights a, b, c, d, e, f clockwise to the sides of the hexagon in the trihedron A. The same sides (and their weights with orientations) will be in the second hexagon, since the trihedrons are aligned exactly by it. The graphs will differ only in their insides, i.e. in the contacts of the main diagonal. And also by the vertices of the hexagon to which they are connected. If A has vertices 1, 3, 5 (conditionally), B has vertices 2, 4, 6.
It is important now to understand that a, b, c, d, e, f together with the orientations of the edges completely determine the whole layout. No, it does not mean that a, b, c, d, e, f together with orientations can be assigned at random.
OK, now nothing prevents us from superimposing these graphs on each other along the hexagon. We will get a single, "unbroken" graph of the cube. Now we have to think about proper partitioning of the hexagon.
By the way, the unified graph can be seen at once if we draw the cube on the plane the way we are used to drawing it. The outer contour on the plane is our hexagon, common to the trihedrons.
All that's left is to apply the hexagon markings. Well here I'm getting into a stupor ("stupor + corkscrew")... Better, of course, not just to find a solution, but to find a general principle of marking. Or at least find many solutions - if there really are many of them.
Maybe that'll help. With the odds odds the following has been figured out so far. I was wrong - the parity-oddness decomposition by edges exists, but it's the ONLY one.
// with the accuracy of cube rotations.
See figure. In the picture even edges are marked in black, odd edges in red.
Try it your way (referring to the drawing as well). I'm going to bed.
// There's something about the pictures that doesn't load well. It took me ten times to load them.
I've got a hunch that modulo 13 deductions might help... It's not just to make a hexagon from scratch.
Может кто знает ?
I remember, in principle... But badly - I was just a kid then... )))
Well, in general, BV didn't just happen. "It was for a reason" (C). Perhaps, pratsivilization unsuccessfully launched a collider. And then, boom!
Is the problem still open?
It's clear: it's either 24 or 25.