Difference between revisions of "2013 AMC 12B Problems/Problem 11"
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== Solution == | == 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 <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: | + | 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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| − | 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>. | + | 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> |
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| + | 8 steps: <math>A</math> moves east to <math>(3,3,2)</math>, <math>B</math> moves west to <math>(-4,-4,0)</math>, distance of <math>\sqrt{(3+4)^2+(3+4)^2+2^2}=\sqrt{102}>10</math> | ||
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| + | Thus they reach <math>10</math> feet away when <math>A</math> 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 14:56, 10 January 2021
Problem
Two bees start at the same spot and fly at the same rate in the following directions. Bee 
travels 
foot north, then foot east, then foot upwards, and then continues to repeat this pattern. Bee 
travels foot south, then foot west, and then continues to repeat this pattern. In what directions are the bees traveling when they are exactly 
feet away from each other?
Solution
Let A and B begin at 
. In 
steps, will have done his route twice, ending up at 
, and will have done his route three times, ending at 
. Their distance is 
We now move forward one step at a time until they are ten feet away:
7 steps: moves north to 
, moves south to 
, distance of 
8 steps: moves east to 
, moves west to 
, distance of 
Thus they reach feet away when is moving east and B is moving west, between moves 7 and 8. Thus the answer is 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
