Difference between revisions of "2007 AMC 12B Problems/Problem 6"

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Triangle <math>ABC</math> has side lengths <math>AB = 5</math>, <math>BC = 6</math>, and <math>AC = 7</math>. Two bugs start simultaneously from <math>A</math> and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point <math>D</math>. What is <math>BD</math>?  
 
Triangle <math>ABC</math> has side lengths <math>AB = 5</math>, <math>BC = 6</math>, and <math>AC = 7</math>. Two bugs start simultaneously from <math>A</math> and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point <math>D</math>. What is <math>BD</math>?  
  
<math>\mathrm {(A)} 1</math>  <math>\mathrm {(B)} 2</math>  <math>\mathrm {(C)} 3</math>  <math>\mathrm {(D)} 4</math>  <math>\mathrm {(E)) 5</math>
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<math>\mathrm {(A)}\ 1 \qquad \mathrm {(B)}\ 2 \qquad \mathrm {(C)}\ 3 \qquad \mathrm {(D)}\ 4 \qquad \mathrm {(E)}\ 5</math>
  
 
==Solution==
 
==Solution==
 
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[[Image:2007_12B_AMC-6.png]]
{{image}}
 
  
 
One bug goes to <math>B</math>. The path that he takes is <math>\dfrac{5+6+7}{2}=9</math> units long. The length of <math>BD</math> is <math>9-AB=9-5=4 \Rightarrow \mathrm {(D)}</math>
 
One bug goes to <math>B</math>. The path that he takes is <math>\dfrac{5+6+7}{2}=9</math> units long. The length of <math>BD</math> is <math>9-AB=9-5=4 \Rightarrow \mathrm {(D)}</math>
  
==See Also==
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==See also==
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{{AMC12 box|year=2007|ab=B|num-b=5|num-a=7}}
  
{{AMC12 box|year=2007|ab=B|num-b=5|num-a=7}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 04:21, 15 December 2020

Problem

Triangle $ABC$ has side lengths $AB = 5$, $BC = 6$, and $AC = 7$. Two bugs start simultaneously from $A$ and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point $D$. What is $BD$?

$\mathrm {(A)}\ 1 \qquad \mathrm {(B)}\ 2 \qquad \mathrm {(C)}\ 3 \qquad \mathrm {(D)}\ 4 \qquad \mathrm {(E)}\ 5$

Solution

2007 12B AMC-6.png

One bug goes to $B$. The path that he takes is $\dfrac{5+6+7}{2}=9$ units long. The length of $BD$ is $9-AB=9-5=4 \Rightarrow \mathrm {(D)}$

See also

2007 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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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