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

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

Latest revision as of 14:57, 10 January 2021

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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