AMC10 2007 A

AMC10 2007 A · Q18

AMC10 2007 A · Q18. It mainly tests Similarity, Area & perimeter.

Consider the $12$-sided polygon $ABCDEFHIJKL$, as shown. Each of its sides has length $4$, and each two consecutive sides form a right angle. Suppose that $AG$ and $CH$ meet at $M$. What is the area of quadrilateral $ABCM$?

考虑如图所示的 12 边形 $ABCDEFHIJKL$。它的每条边长均为 $4$，且每两条连续边形成直角。假设 $AG$ 和 $CH$ 相交于 $M$。四边形 $ABCM$ 的面积是多少？

(A) $\frac{44}{3}$ $\frac{44}{3}$

(B) 16 16

(D) 20 20

(E) $\frac{62}{3}$ $\frac{62}{3}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Extend $\overline{CD}$ past $C$ to meet $\overline{AG}$ at $N$. Since $\triangle ABG$ is similar to $\triangle NCG$, $$ NC = AB\cdot \frac{CG}{BG} = 4\cdot \frac{8}{12}=\frac{8}{3}. $$ This implies that trapezoid $ABCN$ has area $$ \frac12\cdot\left(\frac{8}{3}+4\right)\cdot 4=\frac{40}{3}. $$ Let $v$ denote the length of the perpendicular from $M$ to $NC$. Since $\triangle CMN$ is similar to $\triangle HMG$, and $$ \frac{GH}{NC}=\frac{4}{8/3}=\frac{3}{2}, $$ the length of the perpendicular from $M$ to $HG$ is $\frac{3}{2}v$. Because $$ v+\frac{3}{2}v=8, $$ we have $$ v=\frac{16}{5}. $$ Hence the area of $\triangle CMN$ is $$ \frac12\cdot \frac{8}{3}\cdot \frac{16}{5}=\frac{64}{15}. $$ So $$ \text{Area}(ABCM)=\text{Area}(ABCN)+\text{Area}(\triangle CMN)=\frac{40}{3}+\frac{64}{15}=\frac{88}{5}. $$

答案（C）：将 $\overline{CD}$ 延长过点 $C$，与 $\overline{AG}$ 相交于 $N$。由于 $\triangle ABG$ 与 $\triangle NCG$ 相似， $$ NC = AB\cdot \frac{CG}{BG} = 4\cdot \frac{8}{12}=\frac{8}{3}. $$ 因此，梯形 $ABCN$ 的面积为 $$ \frac12\cdot\left(\frac{8}{3}+4\right)\cdot 4=\frac{40}{3}. $$ 设 $v$ 为从 $M$ 到 $NC$ 的垂线长度。由于 $\triangle CMN$ 与 $\triangle HMG$ 相似，且 $$ \frac{GH}{NC}=\frac{4}{8/3}=\frac{3}{2}, $$ 所以从 $M$ 到 $HG$ 的垂线长度为 $\frac{3}{2}v$。因为 $$ v+\frac{3}{2}v=8, $$ 可得 $$ v=\frac{16}{5}. $$ 因此，$\triangle CMN$ 的面积为 $$ \frac12\cdot \frac{8}{3}\cdot \frac{16}{5}=\frac{64}{15}. $$ 所以 $$ \text{Area}(ABCM)=\text{Area}(ABCN)+\text{Area}(\triangle CMN)=\frac{40}{3}+\frac{64}{15}=\frac{88}{5}. $$

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