Difference between revisions of "2019 AMC 12B Problems/Problem 11"
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==Solution== | ==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: <math>\frac{7}{11} {12 \choose 2}=\boxed{42}</math> | ||
+ | |||
+ | //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 15:19, 14 February 2019
Problem
How many unordered pairs of edges of a given cube determine a plane?
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:
//Can somebody better with latex than me put a picture in here?
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 |
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