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

(→‎See also)
(5^2+5^2+4=54)
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==Solution==
 
==Solution==
  
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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 <math>\sqrt{(2+3)^2+(2+3)^2+2^2}=\sqrt{52} < 10</math> We now move forward one step at a time until they are ten feet away:
+
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 <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: A moves north to (2,3,2), B moves south to (-3,-4,0), distance of <math>\sqrt{(2+3)^2+(3+4)^2+2^2}=\sqrt{78} < 10</math>
 
7 steps: A moves north to (2,3,2), B moves south to (-3,-4,0), distance of <math>\sqrt{(2+3)^2+(3+4)^2+2^2}=\sqrt{78} < 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>
 
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>

Revision as of 20:53, 24 December 2013

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?

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

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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