2012 AMC 12A Problems/Problem 22

Problem

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 $k$ ?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 23\qquad\textbf{(E)}\ 24$

Solution

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.

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2012 AMC 12A Problems/Problem 22

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