2013 AMC 12B Problems/Problem 11

Revision as of 14:57, 10 January 2021 by Angelalz (talk | contribs) (Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS