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

(Problem)
(Solution)
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==Solution==
 
==Solution==
  
(12-4-1)*12/2 = 42 (D)
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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.
  
(SuperWill)
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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.
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We compute: <math>\frac{7}{11} {12 \choose 2}=\boxed{42}</math>
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//Can somebody better with latex than me put a  picture in here?
  
 
==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 16:19, 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

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: $\frac{7}{11} {12 \choose 2}=\boxed{42}$

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

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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