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2008 AMC 10B Problems/Problem 6

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Problem

Points $B$ and $C$ lie on $\overline{AD}$ . The length of $\overline{AB}$ is $4$ times the length of $\overline{BD}$ , and the length of $\overline{AC}$ is $9$ times the length of $\overline{CD}$ . The length of $\overline{BC}$ is what fraction of the length of $\overline{AD}$ ?

$\textbf{(A)}\ \frac{1}{36}\qquad\textbf{(B)}\ \frac{1}{13}\qquad\textbf{(C)}\ \frac{1}{10}\qquad\textbf{(D)}\ \frac{5}{36}\qq...$

Solution

Let $CD = 1$ . Then $AB = 4(BC + 1)$ and $AB + BC = 9\cdot1$ . From this system of equations we obtain $BC = 1$ . Adding $CD$ to both sides of the second equation, we obtain $AD = AB + BC + CD = 9 + 1 = 10$ . Thus, $\frac{BC}{AD} = \frac{1}{10} \implies\text{(C)}$

See also

2008 AMC 10B (Problems • Resources)
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

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Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

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