# 2008 AMC 10B Problems/Problem 6

## 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 $\overline{AD}$?

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

## Solution 1

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

## Solution 2

Let $x = BD$ and $y = CD$. Therefore, $AB = 4x$ and $AC = 9y$, as shown in the diagram(the labels on the bottom are for that line segment while the labels on the top are from one point to the left to one point to the right).

$[asy] dot((0,0)); label("A", (0,0), S); dot((5,0)); label("B", (5,0), S); dot((10,0)); label("C", (10,0), S); dot((15,0)); label("D", (15,0), S); draw((0,0)--(5,0)); draw((5,0)--(10,0)); draw((10,0)--(15,0)); draw((0,0)--(10,0)); draw((10,0)--(15,0)); label("4x", (0,0)--(5,0), S); label("9y", (0,0)--(10,0), N); label("y", (10,0)--(15,0), S); label("x", (5,0)--(15,0), N); [/asy]$

From this, we can see that $AD = 10y = 5x$, and since $BC = BD - CD = x-y$. Now, our ratio is $\frac{x-y}{AD}$. We can split this into 2 fractions: $\frac{x}{AD} - \frac{y}{AD} = \frac{x}{5x} - \frac{y}{10y} = \frac{1}{5} - \frac{1}{10} = \boxed{\textbf{(C)}\ \frac{1}{10}}$

~idk12345678