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

Revision as of 22:13, 16 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. In the diagram below, the red edge is parallel to the 3 green edges and intersects with the 4 blue edges.

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

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.

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

2019 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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

Revision as of 22:13, 16 February 2019

Contents

Problem

Solution 1

Solution 2

See Also

@@ Line 15: / Line 15: @@
 unitsize(1cm);
 size(200);
-currentprojection=perspective(-6/5,-8/5,3/4);
+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));