Question

如图，线段 $\overline{AD}$ 被点 $B$ 和 $C$ 三等分，使得 $AB=BC=CD=2$ 。三个半径为 $1$ 的半圆 $AEB$ ， $BFC$ ，和 $CGD$ 的直径都在 $\overline{AD}$ 上，并分别在 $E$ ， $F$ ，和 $G$ 处与直线 $EG$ 相切。一个半径为 $2$ 的圆以 $F$ 为圆心。图中阴影区域，即在该圆内但在三个半圆外的区域，其面积可表示为

$\dfrac{a}{b}\cdot\pi-\sqrt{c}+d,$

其中 $a,b,c$ ，和 $d$ 为正整数，且 $a$ 与 $b$ 互质。 $a+b+c+d$ 是多少？

As shown in the figure, line segment $\overline{AD}$ is trisected by points $B$ and $C$ so that $AB=BC=CD=2.$ Three semicircles of radius $1,$ $AEB,$ $BFC,$ and $CGD,$ have their diameters on $\overline{AD},$ and are tangent to line $EG$ at $E,$ $F,$ and $G,$ respectively. A circle of radius $2$ has its center on $F.$ The area of the region inside the circle but outside the three semicircles, shaded in the figure, can be expressed in the form

$\dfrac{a}{b}\cdot\pi-\sqrt{c}+d,$

where $a,b,c,$ and $d$ are positive integers and $a$ and $b$ are relatively prime. What is $a+b+c+d?$

$13$

$14$

$15$

$16$

$17$

答案：E

知识点：圆面积面积分割坐标几何

难度评级：1830

解答：

设 $A=(0,0)$ ， $\ B=(2,0)$ ， $\ C=(4,0)$ ， $\ D=(6,0)$ ，则三个半圆的圆心为 $(1,0),(3,0),(5,0)$ ，顶点为 $E=(1,1)$ ， $\ F=(3,1)$ ， $\ G=(5,1)$ 。该圆的圆心为 $F=(3,1)$ ，半径为 $2$ ，所以它经过 $E$ 和 $G$ ，面积为 $4\pi$ 。

中间半圆 $BFC$ 完全在该圆内，去掉面积 $\dfrac{\pi}{2}$ 。由对称性，两个外侧半圆各自落在圆内的重叠面积相同，为 $E$ 。 $x$ $x=3-\sqrt3$ 。 $\begin{aligned} I &=\int_1^{3-\sqrt3}1\,dx\\ &\quad-\int_1^{3-\sqrt3} \sqrt{4-(x-3)^2}\,dx\\ &=2-\dfrac{\sqrt3}{2}-\dfrac{\pi}{3}. \end{aligned}$ $\begin{aligned} R&=\dfrac{\pi}{4}-I\\ &=\dfrac{7\pi}{12}-2+\dfrac{\sqrt3}{2}. \end{aligned}$

阴影面积为因此 $a=7,\ b=3,\ c=3,\ d=4$ ，所以 $a+b+c+d=17$ 。 $4\pi-\dfrac{\pi}{2}-2R=\dfrac{7}{3}\pi-\sqrt3+4.$

所以正确答案是 E。

Put $A=(0,0),$ $\ B=(2,0),$ $\ C=(4,0),$ $\ D=(6,0),$ so the semicircles are centered at $(1,0),(3,0),(5,0)$ and their tops are $E=(1,1),$ $\ F=(3,1),$ $\ G=(5,1).$ The circle has center $F=(3,1)$ and radius $2,$ so it passes through $E$ and $G,$ and has area $4\pi.$

The middle semicircle $BFC$ lies entirely inside the circle, removing area $\dfrac{\pi}{2}.$ For the left semicircle, the overlap starts at $E$ and includes its right-hand quarter-circle, except for the region below the large circle. The large circle meets the $x$ -axis at $x=3-\sqrt3.$ The excluded area is $\begin{aligned} I &=\int_1^{3-\sqrt3}1\,dx\\ &\quad-\int_1^{3-\sqrt3} \sqrt{4-(x-3)^2}\,dx\\ &=2-\dfrac{\sqrt3}{2}-\dfrac{\pi}{3}. \end{aligned}$ Therefore the overlap has area $\begin{aligned} R&=\dfrac{\pi}{4}-I\\ &=\dfrac{7\pi}{12}-2+\dfrac{\sqrt3}{2}. \end{aligned}$ By symmetry, the right semicircle contributes the same overlap.

The shaded area is $4\pi-\dfrac{\pi}{2}-2R=\dfrac{7}{3}\pi-\sqrt3+4.$ Hence $a=7,\ b=3,\ c=3,\ d=4,$ so $a+b+c+d=17.$

Thus, E is the correct answer.

2019 AMC 12B 第 15 题

答案：E

难度评级：1830

解答：

其他年份的第 15 题