AMC10 2008 10B, Problem 6

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 $\overline{AD}$?

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, $\frac{BC}{AD} = \frac{1}{10} \implies\text{(C)}$ $\fbox{1/10}$.

Answer: 1/10