2019 AMC 12B Problems/Problem 11

Revision as of 23:12, 16 February 2019 by Vedadehhc (talk | contribs) (Solution 1)

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. 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,-8/5,3/4); 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]

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

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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