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Difference between revisions of "2019 AMC 12B Problems/Problem 11"

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==Solution 1==
 
==Solution 1==
  
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WLOG, pick one of the 12 edges of the cube to be among the two selected. We seek the answer by computing the probability that a random choice of second edge satisfies the problem statement.
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Without loss of generality, choose one of the <math>12</math> edges of the cube to be among the two selected. We now calculate the probability that a randomly-selected second edge makes the pair satisfy the condition in the problem statement.
  
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For two line segments in space to correspond to a common plane, they must correspond to lines that either intersect or are parallel. If all 12 line segments are extended to lines, our first edge's line intersects 4 lines and is parallel to another 3. Thus 7 of the 11 line segments satisfy the problem statement. In the diagram below, the red edge is parallel to the 3 green edges and intersects with the 4 blue edges.
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For two lines in space to determine a common plane, they must either intersect or be parallel (in other words, they cannot be skew lines). If all <math>12</math> line segments are extended to lines, the first (arbitrarily chosen) edge's line intersects <math>4</math> lines and is parallel to another <math>3</math>. Thus <math>4+3=7</math> of the <math>12-1=11</math> remaining line segments (which could be chosen for the second edge) give a pair of lines determining a common plane. To see this, observe that in the diagram below, the red edge is parallel to the <math>3</math> green edges and intersects with the <math>4</math> blue edges.
  
 
<asy>
 
<asy>
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</asy>
 
</asy>
  
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We compute: <math>\frac{7}{11} {12 \choose 2}=\boxed{\textbf{(D) }42}</math>
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This means that the probability that a randomly-selected pair of edges determine a plane is <math>\frac{7}{11}</math>, and we calculate that there are <math>{12 \choose 2} = 66</math> total pairs of edges that could be chosen (without the restriction). Thus the answer is <math>\frac{7}{11} \cdot 66 =\boxed{\textbf{(D) }42}</math>.
  
 
==Solution 2==
 
==Solution 2==
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Case 1: The two edges are on the same face. There are <math>6 \cdot {4 \choose 2}=36</math> possibilities.
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As in Solution 1, we observe that the two edges must either be parallel or intersect. Clearly the edges will intersect if and only if they are part of the same face. We can thus divide into two cases:
  
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Case 2: The two edges are parallel but not on the same face. There are <math>6</math> possibilities.
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'''Case 1''': The two edges are part of the same face. There are <math>6</math> faces, and <math>{4 \choose 2}=6</math> ways to choose <math>2</math> of the <math>4</math> edges of the square, giving a total of <math>6 \cdot 6 = 36</math> possibilities.
  
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<math>36 + 6 = \boxed{\textbf{(D) }42}</math>
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'''Case 2''': The two edges are parallel and not part of the same face. Observe that each of the <math>12</math> edges is parallel to exactly <math>1</math> edge that is not part of its face. The edges can thus be paired up, giving <math>\frac{12}{2} = 6</math> possibilities for this case.
 +
 
 +
Adding the two cases, the answer is hence <math>36 + 6 = \boxed{\textbf{(D) }42}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2019|ab=B|num-b=10|num-a=12}}
 
{{AMC12 box|year=2019|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:32, 18 February 2019

Problem

How many unordered pairs of edges of a given cube determine a plane?

$\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66$

Solution 1

Without loss of generality, choose one of the $12$ edges of the cube to be among the two selected. We now calculate the probability that a randomly-selected second edge makes the pair satisfy the condition in the problem statement.

For two lines in space to determine a common plane, they must either intersect or be parallel (in other words, they cannot be skew lines). If all $12$ line segments are extended to lines, the first (arbitrarily chosen) edge's line intersects $4$ lines and is parallel to another $3$. Thus $4+3=7$ of the $12-1=11$ remaining line segments (which could be chosen for the second edge) give a pair of lines determining a common plane. To see this, observe that in the diagram below, the red edge is parallel to the $3$ green edges and intersects with the $4$ blue edges.

[asy] import three; import three; unitsize(1cm); size(200); currentprojection=perspective(-6/5,-8/5,7/8); draw((0,1,0)--(0,0,0)--(1,0,0), blue); draw((1,0,0)--(1,1,0)--(0,1,0)); draw((0,0,0)--(0,0,1), red); draw((0,1,0)--(0,1,1), green); draw((1,1,0)--(1,1,1), green); draw((1,0,0)--(1,0,1), green); draw((0,1,1)--(0,0,1)--(1,0,1), blue); draw((1,0,1)--(1,1,1)--(0,1,1)); [/asy]

This means that the probability that a randomly-selected pair of edges determine a plane is $\frac{7}{11}$, and we calculate that there are ${12 \choose 2} = 66$ total pairs of edges that could be chosen (without the restriction). Thus the answer is $\frac{7}{11} \cdot 66 =\boxed{\textbf{(D) }42}$.

Solution 2

As in Solution 1, we observe that the two edges must either be parallel or intersect. Clearly the edges will intersect if and only if they are part of the same face. We can thus divide into two cases:

Case 1: The two edges are part of the same face. There are $6$ faces, and ${4 \choose 2}=6$ ways to choose $2$ of the $4$ edges of the square, giving a total of $6 \cdot 6 = 36$ possibilities.

Case 2: The two edges are parallel and not part of the same face. Observe that each of the $12$ edges is parallel to exactly $1$ edge that is not part of its face. The edges can thus be paired up, giving $\frac{12}{2} = 6$ possibilities for this case.

Adding the two cases, the answer is hence $36 + 6 = \boxed{\textbf{(D) }42}$.

See Also

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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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