AMC12 2008 B

AMC12 2008 B · Q8

AMC12 2008 B · Q8. It mainly tests Ratios & proportions, Ratios in geometry.

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 $\overline{AD}$?

点 $B$ 和 $C$ 在 $\overline{AD}$ 上。$\overline{AB}$ 的长度是 $\overline{BD}$ 长度的 $4$ 倍，且 $\overline{AC}$ 的长度是 $\overline{CD}$ 长度的 $9$ 倍。$\overline{BC}$ 的长度是 $\overline{AD}$ 长度的几分之几？

(A) $\frac{1}{36}$ $\frac{1}{36}$

(B) $\frac{1}{13}$ $\frac{1}{13}$

(D) $\frac{5}{36}$ $\frac{5}{36}$

(E) $\frac{1}{5}$ $\frac{1}{5}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Since $\overline{AB}=4\overline{BD}$ and $\overline{AB}+\overline{BD}=\overline{AD}$, $\overline{AB}=\frac{4}{5}\overline{AD}$. Since $\overline{AC}=9\overline{CD}$ and $\overline{AC}+\overline{CD}=\overline{AD}$, $\overline{AC}=\frac{9}{10}\overline{AD}$. Thus, $\overline{BC}=\overline{AC}-\overline{AB}=\left(\frac{9}{10}-\frac{4}{5}\right)\overline{AD} = \frac {1}{10}\overline{AD} \Rightarrow C$.

由于 $\overline{AB}=4\overline{BD}$ 且 $\overline{AB}+\overline{BD}=\overline{AD}$，所以 $\overline{AB}=\frac{4}{5}\overline{AD}$。由于 $\overline{AC}=9\overline{CD}$ 且 $\overline{AC}+\overline{CD}=\overline{AD}$，所以 $\overline{AC}=\frac{9}{10}\overline{AD}$。因此 $\overline{BC}=\overline{AC}-\overline{AB}=\left(\frac{9}{10}-\frac{4}{5}\right)\overline{AD} = \frac {1}{10}\overline{AD} \Rightarrow C$。

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