Show how to cover a size 4 cube with sixteen 1x6 strips. The strips are allowed to bend over an edge to cover multiple faces.
Is it possible to cover a size 5 cube with 25 of the 1x6 strips?
Imagine the cube as two U shaped pieces of three squares each. Each can be unfolded into a 4 by 12 rectangle, easily covered by the strips.
The 5x5 sides of the 5-cube each has an area of 25. This isn't divisible by 6. In fact only when you consider all the sides do you get an area divisible by 6. This means no subset of the faces can be covered as in the 4-cube.
After much playing around, I highly doubt this is possible. I don't have a proof. But here are some things I know:
- If you look at one corner of a cube, three strips cannot meet there.
- Every strip must fold over exactly one edge, so of the 60 units of edge length, 25 are crossed and 35 are not.
- If you color the entire cube like a checkerboard and looks at the 150 squares, 78 are black and 72 are white. Each strip covers 4 of one color and 2 of the other. This means 14 of the strips are of the 4 black type and 11 are of the 4 white type. This parity argument seems promising but I'm stuck.
Extensions:
If N is even, a N-cube can be covered with N^2 1x6 strips.
(Proof is simple)
If N is a multiple of 3, a N-cube can be covered with N^2 1x6 strips.
(Proof is simple)
N=1 is clearly not possible.
Is there any other N for which it is possible?
Edited on August 18, 2016, 6:32 pm
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Posted by Jer
on 2016-08-18 17:09:11 |
Solution to Part 1. Thoughts on part 2. Extensions.
