AMC12 2003 B

AMC12 2003 B · Q16

AMC12 2003 B · Q16. It mainly tests Circle theorems, Area & perimeter.

Three semicircles of radius $1$ are constructed on diameter $\overline{AB}$ of a semicircle of radius $2$. The centers of the small semicircles divide $\overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?

在半径为 $2$ 的半圆的直径 $\overline{AB}$ 上作三个半径为 $1$ 的半圆。小半圆的圆心将 $\overline{AB}$ 分成四段等长线段，如图所示。求阴影区域的面积：该区域位于大半圆内但在小半圆外。

(A) $\pi - \sqrt{3}$ $\pi - \sqrt{3}$

(B) $\pi - \sqrt{2}$ $\pi - \sqrt{2}$

(D) $\pi + \frac{\sqrt{3}}{2}$ $\pi + \frac{\sqrt{3}}{2}$

(E) $\frac{7}{6}\pi - \frac{\sqrt{3}}{2}$ $\frac{7}{6}\pi - \frac{\sqrt{3}}{2}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into $\frac{5}{6}$ of a circle with radius $1$ and two equilateral triangles with side length $1$. This gives the area of the white region as $\frac{5}{6}\pi+\frac{2\cdot\sqrt3}{4}=\frac{5}{6}\pi+\frac{\sqrt3}{2}$. The area of the shaded region is the area of the white region subtracted from the area of the large semicircle. This is equivalent to $2\pi-\left(\frac{5}{6}\pi+\frac{\sqrt3}{2}\right)=\frac{7}{6}\pi-\frac{\sqrt3}{2}$. Thus the answer is $\boxed{\textbf{(E)}\ \frac{7}{6}\pi-\frac{\sqrt3}{2}}$. The reason why it is $\frac{5}{6}$ of a circle and why the triangles are equilateral are because, first, the radii are the same and they make up the equilateral triangles. Secondly, the reason it is $\frac{5}{6}$ of a circle is because the middle sector has a degree of $180-2 \cdot 60 = 60$ and thus $\frac{60}{360}=\frac{1}{6}$ of a circle. The other two have areas of $\frac{180-60}{360}=\frac{1}{3}$ of a triangle each. Therefore, the total fraction of the circle(since they have the same radii) is $\frac{1}{6} + 2 \cdot \frac{1}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}.$

通过从半圆交点向各自圆心画四条线，我们把白色区域分割成一个半径为 $1$ 的圆的 $\frac{5}{6}$，以及两个边长为 $1$ 的等边三角形。因此白色区域面积为 $\frac{5}{6}\pi+\frac{2\cdot\sqrt3}{4}=\frac{5}{6}\pi+\frac{\sqrt3}{2}$。阴影区域面积等于大半圆面积减去白色区域面积，即 \[2\pi-\left(\frac{5}{6}\pi+\frac{\sqrt3}{2}\right)=\frac{7}{6}\pi-\frac{\sqrt3}{2}.\] 因此答案是 $\boxed{\textbf{(E)}\ \frac{7}{6}\pi-\frac{\sqrt3}{2}}$。之所以是圆的 $\frac{5}{6}$，以及三角形是等边的原因是：首先，半径相同，它们构成等边三角形。其次，之所以是圆的 $\frac{5}{6}$，是因为中间的扇形圆心角为 $180-2 \cdot 60 = 60$，因此它占整圆的 $\frac{60}{360}=\frac{1}{6}$。另外两个扇形各自的面积占比为 $\frac{180-60}{360}=\frac{1}{3}$。因此总的圆面积占比（因为它们半径相同）为 $\frac{1}{6} + 2 \cdot \frac{1}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}.$

Topics