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Difference between revisions of "2013 AMC 12B Problems/Problem 11"

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8 steps: A moves east to (3,3,2), B moves west to (-4,-4,0), distance of <math>\sqrt{(3+4)^2+(3+4)^2+2^2}=\sqrt{102}>10</math>
 
8 steps: A moves east to (3,3,2), B moves west to (-4,-4,0), distance of <math>\sqrt{(3+4)^2+(3+4)^2+2^2}=\sqrt{102}>10</math>
  
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Thus they reach 10 feet away when A is moving east and B is moving west, between moves 7 and 8. Thus the answer is  <math>\textbf{(A)}</math>
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Thus they reach 10 feet away when A is moving east and B is moving west, between moves 7 and 8. Thus the answer is  <math>\boxed{\textbf{(A)}}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2013|ab=B|num-b=10|num-a=12}}
 
{{AMC12 box|year=2013|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:24, 19 October 2020

Problem

Two bees start at the same spot and fly at the same rate in the following directions. Bee $A$ travels $1$ foot north, then $1$ foot east, then $1$ foot upwards, and then continues to repeat this pattern. Bee $B$ travels $1$ foot south, then $1$ foot west, and then continues to repeat this pattern. In what directions are the bees traveling when they are exactly $10$ feet away from each other?

$\textbf{(A) } A\ \text{east, } B\ \text{west} \qquad \textbf{(B) } A\ \text{north, } B\ \text{south} \qquad \textbf{(C) } A\ \text{north, } B\ \text{west} \qquad \textbf{(D) } A\ \text{up, } B\ \text{south} \qquad \textbf{(E) } A\ \text{up, } B\ \text{west}$

Solution

Let A and B begin at (0,0,0). In 6 steps, A will have done his route twice, ending up at (2,2,2), and B will have done his route three times, ending at (-3,-3,0). Their distance is $\sqrt{(2+3)^2+(2+3)^2+2^2}=\sqrt{54} < 10$ We now move forward one step at a time until they are ten feet away: 7 steps: A moves north to (2,3,2), B moves south to (-3,-4,0), distance of $\sqrt{(2+3)^2+(3+4)^2+2^2}=\sqrt{78} < 10$ 8 steps: A moves east to (3,3,2), B moves west to (-4,-4,0), distance of $\sqrt{(3+4)^2+(3+4)^2+2^2}=\sqrt{102}>10$

Thus they reach 10 feet away when A is moving east and B is moving west, between moves 7 and 8. Thus the answer is $\boxed{\textbf{(A)}}$.

See Also

2013 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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