Distinct planes $p_1,p_2,....,p_k$ intersect the interior of a cube $Q$. Let $S$ be the union of the faces of $Q$ and let $P =\bigcup_{j=1}^{k}p_{j}$. The intersection of $P$ and $S$ consists of the union of all segments joining the midpoints of every pair of edges belonging to the same face of $Q$. What is the difference between the maximum and minimum possible values of $k$?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 23\qquad\textbf{(E)}\ 24$
We need two different kinds of planes that only intersect $Q$ at the mentioned segments (we call them traces in this solution). These will be all the possible $p_j$'s.
First, there are two kinds of segments joining the midpoints of every pair of edges belonging to the same face of $Q$: long traces are those connecting the midpoint of opposite sides of the same face of $Q$, and short traces are those connecting the midpoint of adjacent sides of the same face of $Q$.
Suppose $p_j$ contains a short trace $t_1$ of a face of $Q$. Then it must also contain some trace $t_2$ of an adjacent face of $Q$, where $t_2$ share a common endpoint with $t_1$. So, there are three possibilities for $t_2$, each of which determines a plane $p_j$ containing both $t_1$ and $t_2$.
Case 1: $t_2$ makes an acute angle with $t_1$. In this case, $p_j \cap Q$ is an equilateral triangle made by three short traces. There are $8$ of them, corresponding to the $8$ vertices.
Case 2: $t_2$ is a long trace. $p \cap Q$ is a rectangle. Each pair of parallel faces of $Q$ contributes $4$ of these rectangles so there are $12$ such rectangles.
Case 3: $t_2$ is the short trace other than the one described in case 1, i.e. $t_2$ makes an obtuse angle with $t_1$. It is possible to prove that $p \cap Q$ is a regular hexagon (See note #1 for a proof) and there are $4$ of them.
Case 4: $p_j$ contains no short traces. This can only make $p_j \cap Q$ be a square enclosed by long traces.  There are $3$ such squares.
In total, there are $8+12+4+3=27$ possible planes in $P$. So the maximum of $k$ is $27$.
On the other hand, the most economic way to generate these long and short traces is to take all the planes in case 3 and case 4. Overall, they intersect at each trace exactly once (there is a quick way to prove this. See note #2 below.) and also covered all the $6\times 4 + 4\times 3 = 36$ traces. So the minimum of $k$ is $7$. The answer to this problem is then $27-7=20$ ... $\framebox{C}$.


Note 1: Indeed, let $t_1=AB$ where $B=t_1\cap t_2$, and $C$ be the other endpoint of $t_2$ that is not $B$. Draw a line through $A$ parallel to $t_1$. This line passes through the center $O$ of the cube and therefore we see that the reflection of $A,B,C$, denoted by $A', B', C'$, respectively, lie on the same plane containing $A,B,C$. Thus $p_j \cap Q$ is the regular hexagon $ABCA'B'C'$. To count the number of these hexagons, just notice that each short trace uniquely determine a hexagon (by drawing the plane through this trace and the center), and that each face has $4$ short traces. Therefore, there are $4$ such hexagons.
Note 2: The quick way to prove the fact that none of the planes described in case 3 and case 4 share the same trace is as follows: each of these plane contains the center and therefore the intersection of each pair of them is a line through the center, which obviously does not contain any traces.