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If we cut along the plane orthogonal to the largest diagonal of a Rubik's cube, what is the maximum number of small cubes can we cut?

I thought this should be $9$, but apparently this is not the right answer. Any ideas?

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Could we cut the Rubik's cube in such a way that the cross-section is in the shape of a hexagon? – Alexander L Apr 30 '12 at 4:37
up vote 4 down vote accepted

The Rubik's cube consists of 27 small cubes if you count the invisible cube at its center. Counting this center cube, I can see how to intersect 19 small cubes by having the orthogonal plane pass through the center of the Rubik's cube. Just as the orthogonal plane intersects all six faces of the Rubik's cube, it also intersects all six faces of this small interior cube. That means it also intersects each of the six small cubes that shares a face with the small interior cube. These six cubes are the small cubes containing the center squares on the outer faces of the Rubik's cube. The intersection of the orthogonal plane with these six small cubes happens inside the Rubik's cube; the intersection is not visible from the outside.

On the other hand, there are 12 small cubes for which the intersection with the orthogonal is visible from the outside. Visualize the longest diagonal as oriented in the vertical direction. Two opposite vertices lie on this vertical axis, with three edges emanating from each of them. The cube has six additional edges that form a zigzag pattern circling the vertical axis. The orthogonal plane is then horizontal, and intersects each edge of the zigzag at its center, making a hexagonal cross section with the outer faces of the Rubik's cube. The intersection of the orthogonal plane with each of the six faces of the cube forms a line joining the midpoints of adjacent edges. This line crosses three small squares of the face, one of which lies on a small corner cube, and two of which lie on a small edge cube. One face of each of six small corner cubes is crossed, as are two faces of each of six small edge cubes.

So there are 12 visible crossings and seven invisible ones (counting the small interior cube), for a total of 19.

Addendum: I had added the following remark, which does not seem to be correct after all: I believe this plane actually contains the vertices of six more small cubes, although it doesn't intersect their interiors. By making the plane slightly off-center, you could intersect the interiors of three of these six small cubes, for a total of 22 intersections.

Addendum II: The correct statement is that if we insist that the plane intersect the interior of the small cubes, then 19 is the maximum; if we allow intersection in a vertex only, then 22 is possible. Here is an arithmetical argument:

Let the Rubik's cube be a 6-by-6 cube made of 27 2-by-2 cubes. The 2-by-2 cubes are $C(v)=\{v+u\mid u\in[0,2]^3\}$ where $v\in\{-3,-1,1\}^3$. Let the long diagonal be along the $(1,1,1)$ direction, so that orthogonal planes take the form $x+y+z=c$. In order for such a plane to intersect $C((x,y,z))$ it is necessary that $x+y+z\le c\le x+y+z+6$. There is

  1. one cube with $x+y+z=-9$, namely $C(-3,-3,-3)$
  2. three cubes with $x+y+z=-7$, namely $C(-3,-3,-1)$ and permutations
  3. six cubes with $x+y+z=-5$, namely $C(-3,-1,-1)$, $C(-3,-3,1)$ and permutations
  4. seven cubes with $x+y+z=-3$, namely $C(-1,-1,-1)$, $C(-3,-1,1)$ and permutations
  5. six cubes with $x+y+z=-1$, namely $C(-3,1,1)$, $C(-1,-1,1)$ and permutations
  6. three cubes with $x+y+z=1$, namely $C(-1,1,1)$ and permutations
  7. one cube with $x+y+z=3$, namely $C(1,1,1)$

For the plane $x+y+z=0$, the 19 cubes in classes 3, 4, and 5 satisfy the intersection criterion $x+y+z\le 0\le x+y+z+6$. In general, the plane $x+y+z=c$ intersects at most three of the seven classes, unless $c$ is an odd integer, in which case it can intersect four of them. If we let $c=-1$, then the plane intersects the 22 cubes in classes 2, 3, 4, and 5, but for the nine cubes in classes 2 and 5 this is a vertex-only intersection.

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I get 13 non-trivial cuts including the hidden cube. By non-trivial, I mean the plane intersects the interior of the cube. – copper.hat Apr 30 '12 at 5:55
If I include any intersection, I get a maximum of 22. – copper.hat Apr 30 '12 at 6:00
By interior of the cube above, I meant the interior of a small cube. – copper.hat Apr 30 '12 at 6:05
I'm curious which ones we disagree on. Since the orthogonal plane is not parallel to any faces or edges, if it intersects a small cube at all, it must intersect either in a vertex only, or must have intersection with the interior. If we include the hidden cube, I think we certainly need to include the six face-center cubes. Is it the exterior faces we disagree on? – Will Orrick Apr 30 '12 at 6:13
Will: I did a mostly brute-force count using a small program. I am checking my code and answer and trying to see if I can create an answer more appropriate than code. – copper.hat Apr 30 '12 at 6:18

Here is a more brute-force approach:

I label each small cube by $(i,j,k) \in \{0,1,2\}^3$. A cube $(i,j,k)$ occupies the space $[i,i+1]\times[j,j+1]\times[k,k+1] \subset \mathbb{R}^3$.

Let $\sigma$ denote the long diagonal hyperplane, ie, $\sigma((x,y,z)) = x+y+z$. The values that $\sigma$ takes on a given cube $(i,j,k)$ are given by the interval $[i+j+k, i+j+k+3]$. There are 7 such intervals on the Rubik's cube. The following table shows these intervals, along with the small cubes corresponding to these intervals.

To determine which cubes intersect the hyperplane $\sigma((x,y,z)) = \bar \sigma$, we just find the intervals containing $\bar \sigma$ and count the total. To find the number of non-trivial cuts, we find the intervals containing $\bar \sigma$ in their interior and count the total.

$$\begin{array}{ccc} [\min \sigma, \max \sigma] & \#\mathbb{small \;cubes} & \mathbb{small \;cubes} \\\hline [0, 3] & 1 & (0, 0, 0) \\\ [1, 4] & 3 & (0, 0, 1) (0, 1, 0) (1, 0, 0) \\\ [2, 5] & 6 & (0, 0, 2) (0, 1, 1) (0, 2, 0) (1, 0, 1) (1, 1, 0) (2, 0, 0) \\\ [3, 6] & 7 & (0, 1, 2) (0, 2, 1) (1, 0, 2) (1, 1, 1) (1, 2, 0) (2, 0, 1) (2, 1, 0) \\\ [4, 7] & 6 & (0, 2, 2) (1, 1, 2) (1, 2, 1) (2, 0, 2) (2, 1, 1) (2, 2, 0) \\\ [5, 8] & 3 & (1, 2, 2) (2, 1, 2) (2, 2, 1) \\\ [6, 9] & 1 & (2, 2, 2) \end{array}$$

It is only necessary to consider integral values of $\bar\sigma$ when searching for intersections, since if $\bar\sigma \in [a,b]$ (with $a,b$ integers), then $\lfloor \bar\sigma \rfloor \in [a,b]$. Similarly for the interval $(a,b)$, if $\bar\sigma \in (a,b)$ (with $a,b$ integers), then $\frac{1}{2} (\lfloor \bar\sigma \rfloor +\lceil \bar\sigma \rceil) \in (a,b)$. Consequently, when searching for non-trivial cuts, it is only necessary to consider multiples of $\frac{1}{2}$ when searching.

A slightly tedious calculation shows that the number of intersections is maximized with $\bar\sigma = 4$ giving $3+6+7+6=22$ intersections, and the number of non-trivial cuts is maximized with $\bar\sigma = 4.5$ giving $6+7+6=19$ non-trivial cuts.

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