AMC10 2011 B

AMC10 2011 B · Q20

AMC10 2011 B · Q20. It mainly tests Triangles (properties), Area & perimeter.

Rhombus ABCD has side length 2 and ∠B = 120°. Region R consists of all points inside the rhombus that are closer to vertex B than any of the other three vertices. What is the area of R ?

菱形 ABCD 边长为 2，∠B = 120°。区域 R 由菱形内所有比其他三个顶点更靠近顶点 B 的点组成。R 的面积是多少？

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

(B) $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{3}}{2}$

(D) $1 + \frac{\sqrt{3}}{3}$ $1 + \frac{\sqrt{3}}{3}$

(E) 2 2

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $E$ and $H$ be the midpoints of $\overline{AB}$ and $\overline{BC}$, respectively. The line drawn perpendicular to $AB$ through $E$ divides the rhombus into two regions: points that are closer to vertex $A$ than $B$, and points that are closer to vertex $B$ than $A$. Let $F$ be the intersection of this line with diagonal $AC$. Similarly, let point $G$ be the intersection of the diagonal $AC$ with the perpendicular to $BC$ drawn from the midpoint of $BC$. Then the desired region $R$ is the pentagon $BEFGH$. Note that $\triangle AFE$ is a $30-60-90^\circ$ triangle with $AE=1$. Hence the area of $\triangle AFE$ is $\frac12\cdot 1\cdot \frac{1}{\sqrt3}=\frac{\sqrt3}{6}$. Both $\triangle BFE$ and $\triangle BGH$ are congruent to $\triangle AFE$, so they have the same areas. Also $\angle FBG=120^\circ-\angle FBE-\angle GBH=60^\circ$, so $\triangle FBG$ is an equilateral triangle. In fact, the altitude from $B$ to $FG$ divides $\triangle FBG$ into two triangles, each congruent to $\triangle AFE$. Hence the area of $BEFGH$ is $4\cdot \frac{\sqrt3}{6}=\frac{2\sqrt3}{3}$.

答案（C）：设$E$与$H$分别为$\overline{AB}$与$\overline{BC}$的中点。过$E$作垂直于$AB$的直线把菱形分成两部分：一部分是到顶点$A$比到$B$更近的点，另一部分是到顶点$B$比到$A$更近的点。设$F$为此直线与对角线$AC$的交点。类似地，从$BC$的中点作垂直于$BC$的直线，并令点$G$为该直线与对角线$AC$的交点。则所求区域$R$为五边形$BEFGH$。注意$\triangle AFE$是一个$30-60-90^\circ$三角形，且$AE=1$。因此$\triangle AFE$的面积为$\frac12\cdot 1\cdot \frac{1}{\sqrt3}=\frac{\sqrt3}{6}$。$\triangle BFE$与$\triangle BGH$都与$\triangle AFE$全等，所以它们的面积相同。又有$\angle FBG=120^\circ-\angle FBE-\angle GBH=60^\circ$，因此$\triangle FBG$是等边三角形。事实上，从$B$向$FG$作高会把$\triangle FBG$分成两个三角形，每个都与$\triangle AFE$全等。故$BEFGH$的面积为$4\cdot \frac{\sqrt3}{6}=\frac{2\sqrt3}{3}$。

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