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

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<math>\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66</math>
 
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66</math>
  
==Solution==
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==Solution 1==
  
 
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.
 
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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//Can somebody better with latex than me put a  picture in here?
 
//Can somebody better with latex than me put a  picture in here?
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==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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Case 2: The two edges are parallel but not on the same face. There are <math>6</math> possibilities.
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<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 20:27, 14 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

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.

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.

We compute: $\frac{7}{11} {12 \choose 2}=\boxed{\textbf{(D) }42}$

//Can somebody better with latex than me put a picture in here?

Solution 2

Case 1: The two edges are on the same face. There are $6 \cdot {4 \choose 2}=36$ possibilities.

Case 2: The two edges are parallel but not on the same face. There are $6$ possibilities.

$36 + 6 = \boxed{\textbf{(D) }42}$

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
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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