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Difference between revisions of "2012 AMC 12A Problems/Problem 22"
(→‎Video Solution by Richard Rusczyk)
 
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<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 23\qquad\textbf{(E)}\ 24 </math>
 
<math> \textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 23\qquad\textbf{(E)}\ 24 </math>
  
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==Solution==
 +
 +
<center><asy>
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pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(-1,3), F=(1,3), G=(1,1), H=(-1,1), I=(1,0), J=(2,1), K=(1,2), L=(0,1), M=(-0.5,0.5), N=(-1,2), O=(-0.5,2.5), P=(0,3), Q=(1.5,2.5), R=(1,2), S=(1.5,0.5), T=(0,1);
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draw(A--B--C--D--E--F);
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draw(H--A);
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draw(A--D);
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draw(H--E);
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draw(F--C);
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draw(H--G--F);
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draw(G--B);
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label("\(A\)",A,SW);
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label("\(B\)",B,SE);
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label("\(C\)",C,NE);
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label("\(D\)",D,SW);
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label("\(E\)",E,NW);
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label("\(F\)",F,NE);
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label("\(G\)",G,NE);
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label("\(H\)",H,SW);
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</asy></center>
 +
 +
<center><asy>
 +
pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(-1,3), F=(1,3), G=(1,1), H=(-1,1), I=(1,0), J=(2,1), K=(1,2), L=(0,1), M=(-0.5,0.5), N=(-1,2), O=(-0.5,2.5), P=(0,3), Q=(1.5,2.5), R=(1,2), S=(1.5,0.5), T=(0,1);
 +
draw(A--B--C--D--E--F);
 +
draw(H--A);
 +
draw(A--D);
 +
draw(H--E);
 +
draw(F--C);
 +
draw(I--J--K--L--I);
 +
draw(I--K);
 +
draw(J--L);
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draw(K--Q--P--O--K);
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draw(K--P);
 +
draw(Q--O);
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draw(O--L--M--N--O);
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draw(O--M);
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draw(L--N);
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label("\(A\)",A,SW);
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label("\(B\)",B,SE);
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label("\(C\)",C,NE);
 +
label("\(D\)",D,SW);
 +
label("\(E\)",E,NW);
 +
label("\(F\)",F,NE);
 +
label("\(H\)",H,SW);
 +
</asy></center>
 +
 +
 +
We need two 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.
 +
 +
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>.
 +
 +
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>.
 +
 +
'''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.
 +
 +
'''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.
 +
 +
'''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.
 +
 +
'''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.
 +
 +
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>.
 +
 +
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.
 +
 +
'''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.
 +
 +
==Video Solution by Richard Rusczyk==
 +
https://artofproblemsolving.com/videos/amc/2012amc12a/251
 +
 +
~dolphin7
 +
 +
==See Also==
 
{{AMC12 box|year=2012|ab=A|num-b=21|num-a=23}}
 
{{AMC12 box|year=2012|ab=A|num-b=21|num-a=23}}
 +
{{MAA Notice}}

Latest revision as of 16:13, 4 April 2020

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

[asy] pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(-1,3), F=(1,3), G=(1,1), H=(-1,1), I=(1,0), J=(2,1), K=(1,2), L=(0,1), M=(-0.5,0.5), N=(-1,2), O=(-0.5,2.5), P=(0,3), Q=(1.5,2.5), R=(1,2), S=(1.5,0.5), T=(0,1); draw(A--B--C--D--E--F); draw(H--A); draw(A--D); draw(H--E); draw(F--C); draw(H--G--F); draw(G--B); label("\(A\)",A,SW); label("\(B\)",B,SE); label("\(C\)",C,NE); label("\(D\)",D,SW); label("\(E\)",E,NW); label("\(F\)",F,NE); label("\(G\)",G,NE); label("\(H\)",H,SW); [/asy]
[asy] pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(-1,3), F=(1,3), G=(1,1), H=(-1,1), I=(1,0), J=(2,1), K=(1,2), L=(0,1), M=(-0.5,0.5), N=(-1,2), O=(-0.5,2.5), P=(0,3), Q=(1.5,2.5), R=(1,2), S=(1.5,0.5), T=(0,1); draw(A--B--C--D--E--F); draw(H--A); draw(A--D); draw(H--E); draw(F--C); draw(I--J--K--L--I); draw(I--K); draw(J--L); draw(K--Q--P--O--K); draw(K--P); draw(Q--O); draw(O--L--M--N--O); draw(O--M); draw(L--N); label("\(A\)",A,SW); label("\(B\)",B,SE); label("\(C\)",C,NE); label("\(D\)",D,SW); label("\(E\)",E,NW); label("\(F\)",F,NE); label("\(H\)",H,SW); [/asy]


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.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012amc12a/251

~dolphin7

See Also

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

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