Difference between revisions of "2019 AMC 12B Problems/Problem 11"
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Adding the two cases, the answer is hence <math>36 + 6 = \boxed{\textbf{(D) }42}</math>. | Adding the two cases, the answer is hence <math>36 + 6 = \boxed{\textbf{(D) }42}</math>. | ||
+ | |||
+ | ==Solution 3 (Quick Solution)== | ||
+ | As in solution 1, we see that for an arbitrary edge of the cube, only <math>7</math> of the other edges can be chosen to form a plane. | ||
+ | |||
+ | For each of the <math>12</math> edges, this gives <math>(7)(12) = 84</math> total possible pairings. However each pair has been counted twice (e.g edge <math>A</math> with <math>B</math>, and edge <math>B</math> with <math>A</math>) so we divide by 2 to get <math>\boxed{\textbf{(D) }42}</math>. | ||
==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}} |
Latest revision as of 20:51, 6 October 2020
Problem
How many unordered pairs of edges of a given cube determine a plane?
Solution 1
Without loss of generality, choose one of the edges of the cube to be among the two selected. We now calculate the probability that a randomly-selected second edge makes the pair satisfy the condition in the problem statement.
For two lines in space to determine a common plane, they must either intersect or be parallel (in other words, they cannot be skew lines). If all line segments are extended to lines, the first (arbitrarily chosen) edge's line intersects
lines and is parallel to another
. Thus
of the
remaining line segments (which could be chosen for the second edge) give a pair of lines determining a common plane. To see this, observe that in the diagram below, the red edge is parallel to the
green edges and intersects with the
blue edges.
This means that the probability that a randomly-selected pair of edges determine a plane is , and we calculate that there are
total pairs of edges that could be chosen (without the restriction). Thus the answer is
.
Solution 2
As in Solution 1, we observe that the two edges must either be parallel or intersect. Clearly the edges will intersect if and only if they are part of the same face. We can thus divide into two cases:
Case 1: The two edges are part of the same face. There are faces, and
ways to choose
of the
edges of the square, giving a total of
possibilities.
Case 2: The two edges are parallel and not part of the same face. Observe that each of the edges is parallel to exactly
edge that is not part of its face. The edges can thus be paired up, giving
possibilities for this case.
Adding the two cases, the answer is hence .
Solution 3 (Quick Solution)
As in solution 1, we see that for an arbitrary edge of the cube, only of the other edges can be chosen to form a plane.
For each of the edges, this gives
total possible pairings. However each pair has been counted twice (e.g edge
with
, and edge
with
) so we divide by 2 to get
.
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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