Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
m (minor edit)
Line 1: Line 1:
−
==Problem==
+
== Problem ==
 
Two bees start at the same spot and fly at the same rate in the following directions. Bee <math>A</math> travels <math>1</math> foot north, then <math>1</math> foot east, then <math>1</math> foot upwards, and then continues to repeat this pattern. Bee <math>B</math> travels <math>1</math> foot south, then <math>1</math> foot west, and then continues to repeat this pattern. In what directions are the bees traveling when they are exactly <math>10</math> feet away from each other?
 
Two bees start at the same spot and fly at the same rate in the following directions. Bee <math>A</math> travels <math>1</math> foot north, then <math>1</math> foot east, then <math>1</math> foot upwards, and then continues to repeat this pattern. Bee <math>B</math> travels <math>1</math> foot south, then <math>1</math> foot west, and then continues to repeat this pattern. In what directions are the bees traveling when they are exactly <math>10</math> feet away from each other?
  
−
<math>\textbf{(A)}\ A</math> east, <math>B</math> west<br \><math>\qquad \textbf{(B)}\ A</math> north, <math>B</math> south<br \><math> \qquad \textbf{(C)}\ A</math> north, <math>B</math> west<br \><math> \qquad \textbf{(D)}\ A</math> up, <math>B</math> south<br \><math> \qquad \textbf{(E)}\ A</math> up, <math>B</math> west<br \>
+
<math>\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}</math>
−
 
 
−
==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 (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>
Line 12: Line 11:
 
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>
  
−
== 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 00:21, 19 October 2020

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 $\textbf{(A)}$

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_11&oldid=135326"