AMC8 2019

AMC8 2019 · Q24

AMC8 2019 · Q24. It mainly tests Area & perimeter, Ratios in geometry.

In triangle $\triangle ABC$, point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$. Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$. Given that the area of $\triangle ABC$ is $360$, what is the area of $\triangle EBF$?

在三角形 $\triangle ABC$ 中，点 $D$ 将边 $\overline{AC}$ 分成 $AD:DC=1:2$。$E$ 是 $\overline{BD}$ 的中点，$F$ 是直线 $\overline{BC}$ 和直线 $\overline{AE}$ 的交点。已知 $\triangle ABC$ 的面积为 $360$，$\triangle EBF$ 的面积是多少？

(A) 24 24

(B) 30 30

(D) 36 36

(E) 40 40

Answer

Correct choice: (B)

正确答案：（B）

Solution

We use the line-segment ratios to infer area ratios and height ratios. Areas: $AD:DC = 1:2 \implies AD:AC = 1:3 \implies [ABD] =\frac{[ABC]}{3} = 120$. $BE:BD = 1:2 \text{ (midpoint)} \implies [ABE] = \frac{[ABD]}{2} = \frac{120}{2} = 60$. Heights: Let $h_A$ = height (of altitude) from $\overline{BC}$ to $A$. $AD:DC = 1:2 \implies CD:CA = 2:3 \implies \text{height } h_D$ from $\overline{BC}$ to $D$ is $\frac{2}{3}h_A$. $BE:BD = 1:2 \text{ (midpoint)} \implies \text{height } h_E$ from $\overline{BC}$ to $E$ is $\frac{1}{2} h_D = \frac{1}{2}(\frac{2}{3} h_A) = \frac{1}{3} h_A$. Conclusion: $\frac{[EBF]} {[ABF]} = \frac{[EBF]} {[EBF] + [ABE]} = \frac{[EBF]} {[EBF]+60}$, and also $\frac{[EBF]} {[ABF]} = \frac{h_E}{h_A} = \frac{1}{3}$. So, $\frac{[EBF]} {[EBF] + 60} = \frac{1}{3}$, and thus, $[EBF] = \boxed{\textbf{(B) }30}$

我们使用线段比率推断面积比率和高度比率。面积： $AD:DC = 1:2 \implies AD:AC = 1:3 \implies [ABD] =\frac{[ABC]}{3} = 120$。 $BE:BD = 1:2 \text{ (中点)} \implies [ABE] = \frac{[ABD]}{2} = \frac{120}{2} = 60$。高度：令 $h_A$ = 从 $\overline{BC}$ 到 $A$ 的高度（垂线）。 $AD:DC = 1:2 \implies CD:CA = 2:3 \implies \text{高度 } h_D$ 从 $\overline{BC}$ 到 $D$ 是 $\frac{2}{3}h_A$。 $BE:BD = 1:2 \text{ (中点)} \implies \text{高度 } h_E$ 从 $\overline{BC}$ 到 $E$ 是 $\frac{1}{2} h_D = \frac{1}{2}(\frac{2}{3} h_A) = \frac{1}{3} h_A$。结论： $\frac{[EBF]} {[ABF]} = \frac{[EBF]} {[EBF] + [ABE]} = \frac{[EBF]} {[EBF]+60}$，并且也 $\frac{[EBF]} {[ABF]} = \frac{h_E}{h_A} = \frac{1}{3}$。所以，$\frac{[EBF]} {[EBF] + 60} = \frac{1}{3}$，因此，$[EBF] = \boxed{\textbf{(B) }30}$

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