rolling cube $a$,$b$ and $c$ are natural numbers. we have a $(2a+1)\times (2b+1)\times (2c+1)$ cube. this cube is on an infinite plane with unit squares. you call roll the cube to every side you want. faces of the cube are divided to unit squares and the square in the middle of each face is coloured (it means that if this square goes on a square of the plane, then that square will be coloured.) prove that if any two of lengths of sides of the cube are relatively prime, then we can colour every square in plane. time allowed for this question was 1 hour.
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Tags: geometry, 3D geometry, number theory, relatively prime, number theory proposed
ashkan31194
22.08.2011 22:46
goodar2006 wrote: rolling cube $a$,$b$ and $c$ are natural numbers. we have a $(2a+1)\times (2b+1)\times (2c+1)$ cube. this cube is on an infinite plane with unit squares. you call roll the cube to every side you want. faces of the cube are divided to unit squares and the square in the middle of each face is coloured (it means that if this square goes on a square of the plane, then that square will be coloured.) prove that if any two of lengths of sides of the cube are relatively prime, then we can colour every square in plane. time allowed for this question was 1 hour.
MAN shenide boodam in soalo yeki az talahaye chand sal ghabl ba ghazie feshordegi hal karde!
goodar2006
23.08.2011 06:24
chi?? ghaziyeye feshordegi?? man ta unjayi ke shenide budam ba bezu hal mishe, ye tamim ham dare ke ine:
there is a generalization for this question, prove that if only the middle square of only one face is colored, it's still possible to color all the plane.
Pathological
02.11.2019 18:52
Call a point tasty if we can land on it with the $(2a+1) \times (2b+1)$ side of the cube, with the $2a+1$ side being vertical. Assign each square of the plane an ordered pair in $\mathbb{Z}^2$ in the obvious manner. Claim. If $(x, y)$ is tasty, then so are $(x, y \pm 2), (x \pm 2, y).$ Proof. We will show that $(x, y)$ tasty implies that $(x, y \pm 2(a+b+1)), (x, y \pm 2(a+c+1)), (x, y \pm 2(b+c+1))$ are all tasty. Together with a similar result ($(x \pm 2(a+b+1), y)$, etc.), this would imply the claim because $\gcd(2(a+b+1), 2(a+c+1), 2(b+c+1)) = 2.$ Let's show that $(x, y + 2(a+b+1))$ is tasty; the other ones follow similarly. First, make an arbitrary sequence of moves so that we reach a position where the cube is on the $(2a+1) \times (2c+1)$ side, with the $2c+1$ side horizontal. Now, turn the cube over the top $2c+1$ side two times in a row. This will have moved the cube up by $2(a+b+1)$, while not changing the orientation of the cube. After this, simply pretend that we did not do those two moves, and revert the moves made before it. With eleven similar results, the claim follows. $\blacksquare$ From here, the finish is straightforward by conducting casework on the parities of $a+b, b+c, c+a$. If necessary, a redefinition of tastiness will be useful. $\square$