AMC12 2008 B

AMC12 2008 B · Q13

AMC12 2008 B · Q13. It mainly tests Triangles (properties), Area & perimeter.

Vertex $E$ of equilateral $\triangle{ABE}$ is in the interior of unit square $ABCD$. Let $R$ be the region consisting of all points inside $ABCD$ and outside $\triangle{ABE}$ whose distance from $AD$ is between $\frac{1}{3}$ and $\frac{2}{3}$. What is the area of $R$?

正方形 $ABCD$（边长为1）的内部有正三角形 $\triangle{ABE}$ 的顶点 $E$。令 $R$ 为方形 $ABCD$ 内且 $\triangle{ABE}$ 外、距边 $AD$ 的距离在 $\frac{1}{3}$ 与 $\frac{2}{3}$ 之间的所有点的区域。$R$ 的面积是多少？

(A) $\frac{12 - 5\sqrt{3}}{72}$ $\frac{12 - 5\sqrt{3}}{72}$

(B) $\frac{12 - 5\sqrt{3}}{36}$ $\frac{12 - 5\sqrt{3}}{36}$

(D) $\frac{3 - \sqrt{3}}{9}$ $\frac{3 - \sqrt{3}}{9}$

(E) $\frac{\sqrt{3}}{12}$ $\frac{\sqrt{3}}{12}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

The region is the shaded area: We can find the area of the shaded region by subtracting the pentagon from the middle third of the square. The area of the middle third of the square is $\left(\frac13\right)(1)=\frac13$. The pentagon can be split into a rectangle and an equilateral triangle. The base of the equilateral triangle is $\frac13$ and the height is $\left(\frac13\right)\left(\frac12\right)(\sqrt{3})=\frac{\sqrt{3}}{6}$. Thus, the area is $\left(\frac{\sqrt3}{6}\right)\left(\frac13\right)\left(\frac12\right)=\frac{\sqrt3}{36}$. The base of the rectangle is $\frac13$ and the height is the height of the equilateral triangle minus the height of the smaller equilateral triangle. This is: $\frac{\sqrt3}{2}-\frac{\sqrt3}{6}=\frac{\sqrt3}{3}$ Therefore, the area of the shaded region is $\frac13-\frac{\sqrt3}{9}-\frac{\sqrt3}{36}=\boxed{\text{(B) }\frac{12-5\sqrt3}{36}}.$

该区域为阴影部分：我们可以用正方形中间三分之一的面积减去其中的五边形来求阴影面积。正方形中间三分之一的面积为 $\left(\frac13\right)(1)=\frac13$。该五边形可分成一个矩形和一个正三角形。正三角形的底为 $\frac13$，高为 $\left(\frac13\right)\left(\frac12\right)(\sqrt{3})=\frac{\sqrt{3}}{6}$。因此其面积为 $\left(\frac{\sqrt3}{6}\right)\left(\frac13\right)\left(\frac12\right)=\frac{\sqrt3}{36}$。矩形的底为 $\frac13$，高为大正三角形的高减去小正三角形的高，即： $\frac{\sqrt3}{2}-\frac{\sqrt3}{6}=\frac{\sqrt3}{3}$ 因此阴影区域的面积为 $\frac13-\frac{\sqrt3}{9}-\frac{\sqrt3}{36}=\boxed{\text{(B) }\frac{12-5\sqrt3}{36}}.$

Topics