red-blue cubes

[radosr]

You are given 8 unit cubes such that 24 of the faces are painted blue and 24 of the faces are painted red. Is it always possible to use these cubes to form a 2x2x2 cube that has the same number of blue and red unit squares on its surface?

 

[Penglu]

so what's the answer? always possible?

 

[pruse]

the answer is it's always possible.

Number the 8 cubes 1-8 and number each of the cube's vertices 1-8 as well. Let (r_{ij}) denote the number of red faces meeting at vertex (j) of cube (i). Assume WLOG that (r_{i1}\leq r_{i2}\leq \cdots r_{i8}) for all (i). Note that

((r_{11}+\cdots+r_{81})+(r_{12}+\cdots+r_{82})+\cdots+(r_{18}+\cdots+r_{88})=4\cdot 24=96)

since each of the 24 red faces is counted 4 times -- once for each of its 4 vertices. By our assumption above,

((r_{11}+\cdots+r_{81})\leq(r_{12}+\cdots+r_{82})\leq\cdots\leq(r_{18}+\cdots+r_{88}))

from which, together with the above equality, we get that (r_{11}+\cdots+r_{81}\leq \frac{96}{8}=12) and (r_{18}+\cdots+r_{88}\geq 12).

Now in the following sequence of steps
(r_{11}+r_{21}+\cdots+r_{81}\rightarrow)
(r_{12}+r_{21}+\cdots+r_{81}\rightarrow)
(\cdots)
(r_{18}+r_{21}+\cdots+r_{81}\rightarrow)
(r_{18}+r_{22}+\cdots+r_{81}\rightarrow)
(\cdots)
(r_{18}+r_{28}+\cdots+r_{81}\rightarrow)
(\cdots)
(\cdots)
(r_{18}+r_{28}+\cdots+r_{88})

the first sum is (\leq 12), the last is (\geq 12), and each increment is at most 1. This means we must pass through the value 12 at some point. When we do, take the vertices in that sum and make them the vertices of the large cube: this cube then has exactly 12 red faces on its surface.

 

[radosr]

Peter, amazing solution (as always)!

 

[pruse]

Peter, amazing solution (as always)!

Thanks Rados

 

[ThisGuy]

Peter's solution inspired me to a (to my mind, at least) simpler solution. Now suppose it were not possible to arrange the cubes to produce an equal number of red and blue faces on the large cube and assume (this probably doesn't need assuming but we can assume it wlog so let's just do that) that there is a configuration of the small cubes which has fewer than 12 red faces on the exterior of the large cube. (As in Peter's solution, a configuration of the smaller cubes consists of a choice of a vertex from each of the cubes.) Now consider a configuration of the small cubes which has as many red faces on the exterior of the large cube as possible without having more than 12 (exactly 12 is, by assumption, impossible). If any of the vertices in the configuration were not a vertex on that cube with the maximum number of incident red faces then by following a path from the chosen vertex to a maximum vertex (each step only changing the number of red faces on the large cube by 1), we would at some point increase the number of red faces on the large cube by 1 which is, by assumption, impossible. But if our configuration consists only of vertices with maximum number of incident red faces then the exterior of the large cube must have at least half the the total number of red faces, contradicting the fact that it has less than 12.

 

[pruse]

Peter's solution inspired me to a (to my mind, at least) simpler solution. Now suppose it were not possible to arrange the cubes to produce an equal number of red and blue faces on the large cube and assume (this probably doesn't need assuming but we can assume it wlog so let's just do that) that there is a configuration of the small cubes which has fewer than 12 red faces on the exterior of the large cube. (As in Peter's solution, a configuration of the smaller cubes consists of a choice of a vertex from each of the cubes.) Now consider a configuration of the small cubes which has as many red faces on the exterior of the large cube as possible without having more than 12 (exactly 12 is, by assumption, impossible). If any of the vertices in the configuration were not a vertex on that cube with the maximum number of incident red faces then by following a path from the chosen vertex to a maximum vertex (each step only changing the number of red faces on the large cube by 1), we would at some point increase the number of red faces on the large cube by 1 which is, by assumption, impossible. But if our configuration consists only of vertices with maximum number of incident red faces then the exterior of the large cube must have at least half the the total number of red faces, contradicting the fact that it has less than 12.

it's basically the same as my solution, only recast as an argument by contradiction. nice though