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Difference between revisions of "2012 AMC 12A Problems/Problem 22"
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==Solution==
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We need to handle different kinds of planes that only intersect <math>Q</math> at the mentioned segments (we call them ''traces'' in this solution). These will be all the possible <math>p_j</math>'s.
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First, there are two kinds of segments joining the midpoints of every pair of edges belonging to the same face of <math>Q</math>: ''long traces'' are those connecting the midpoint of opposite sides of the same face of <math>Q</math>, and ''short traces'' are those connecting the midpoint of adjacent sides of the same face of <math>Q</math>.
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Suppose <math>p_j</math> contains a short trace <math>t_1</math> of a face of <math>Q</math>. Then it must also contain some trace <math>t_2</math> of an adjacent face of <math>Q</math>, where <math>t_2</math> share a common endpoint with <math>t_1</math>. So, there are three possibilities for <math>t_2</math>, each of which determines a plane <math>p_j</math> containing both <math>t_1</math> and <math>t_2</math>.
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'''Case 1:''' <math>t_2</math> makes an acute angle with <math>t_1</math>. In this case, <math>p_j \cap Q</math> is an equilateral triangle made by three short traces. There are <math>8</math> of them, corresponding to the <math>8</math> vertices.
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'''Case 2:''' <math>t_2</math> is a long trace. <math>p \cap Q</math> is a rectangle. Each pair of parallel faces of <math>Q</math> contributes <math>4</math> of these rectangles so there are <math>12</math> such rectangles.
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'''Case 3:''' <math>t_2</math> is the short trace other than the one described in case 1, i.e. <math>t_2</math> makes an obtuse angle with <math>t_1</math>. It is possible to prove that <math>p \cap Q</math> is a regular hexagon (See note #1 for a proof) and there are <math>4</math> of them.
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'''Case 4:''' <math>p_j</math> contains no short traces. This can only make <math>p_j \cap Q</math> be a square enclosed by long traces.  There are <math>3</math> such squares.
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In total, there are <math>8+12+4+3=27</math> possible planes in <math>P</math>. So the maximum of <math>k</math> is <math>27</math>.
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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 <math>6\times 4 + 4\times 3 = 36</math> traces. So the minimum of <math>k</math> is <math>7</math>. The answer to this problem is then <math>27-7=20</math> ... <math>\framebox{C}</math>.
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'''Note 1''': Indeed, let <math>t_1=AB</math> where <math>B=t_1\cap t_2</math>, and <math>C</math> be the other endpoint of <math>t_2</math> that is not <math>B</math>. Draw a line through <math>A</math> parallel to <math>t_1</math>. This line passes through the center <math>O</math> of the cube and therefore we see that the reflection of <math>A,B,C</math>, denoted by <math>A', B', C'</math>, respectively, lie on the same plane containing <math>A,B,C</math>. Thus <math>p_j \cap Q</math> is the regular hexagon <math>ABCA'B'C'</math>. 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 <math>4</math> short traces. Therefore, there are <math>4</math> such hexagons.
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'''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.

Revision as of 04:19, 7 December 2012

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 $k$?

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

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Solution

We need to handle 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.

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