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Difference between revisions of "2013 AMC 12B Problems/Problem 11"
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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>
 
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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==Solution 2==
 
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Denote the origin of the two bees as A. Let the final positions of the bees be B and C respectively. Since they are moving an equal distance away from the origin with each of their respective steps, two right angled triangles can be constructed from their final positions. Since these triangles are congruent, they each have a hypotenuse of five units so they are both (3,4,5) triangles. 'Resolving' these triangles into the moves of each bee shows that they must have taken 7 steps to reach their final position at which point Bee B must move east next and Bee C is moving west. Hence  <math>\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 14:38, 27 November 2019

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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