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

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[[Image:2007_12B_AMC-6.png]]
 
[[Image:2007_12B_AMC-6.png]]
  
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>92-AB=9-5=4 \Rightarrow \mathrm {(D)}</math>
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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==
 
==See also==

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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