2019 AMC 12B Problems/Problem 11: Difference between revisions
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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. | 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. | 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: <math>\frac{7}{11} {12 \choose 2}=\boxed{\textbf{(D) }42}</math> | We compute: <math>\frac{7}{11} {12 \choose 2}=\boxed{\textbf{(D) }42}</math> | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 22:12, 16 February 2019
Problem
How many unordered pairs of edges of a given cube determine a plane?
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]](http://latex.artofproblemsolving.com/3/3/5/33509ceac791f33f271b3b901edd45096393d5f3.png)
We compute: 
Solution 2
Case 1: The two edges are on the same face. There are 
possibilities.
Case 2: The two edges are parallel but not on the same face. There are 
possibilities.
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
