Question

圆 $A$ 、 $B$ 和 $C$ 的半径都为 $1$ 。圆 $A$ 和圆 $B$ 共有一个切点。圆 $C$ 与 $\overline{AB}$ 的中点相切。位于圆 $C$ 内部、但在圆 $A$ 和圆 $B$ 外部的面积是多少？

Circles $A,$ $B,$ and $C$ each have radius $1.$ Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$

$3 - \dfrac{\pi}{2}$

$\dfrac{\pi}{2}$

$2$

$\dfrac{3\pi}{4}$

$1 + \dfrac{\pi}{2}$

答案：C

知识点：圆面积相切圆面积分割

难度评级：1540

小提示：

设圆心为 $A(-1,0)$ ， $B(1,0)$ ，和 $C(0,1)$ ；从 $C$ 到 $A, B$ 的距离都是 $\sqrt2$

Place the centers at $A(-1,0),$ $B(1,0),$ and $C(0,1);$ the distance from $C$ to each of $A, B$ is $\sqrt2$

大提示：

所求面积是圆 $C$ 的面积减去圆 $C$ 分别与圆 $A$ 和圆 $B$ 的两个透镜形重叠区域

The wanted area is the area of $C$ minus the two lens-shaped overlaps of $C$ with $A$ and with $B$

解答：

设 $A = (-1, 0)$ ， $B = (1, 0)$ ，则它们的切点是原点，也就是 $\overline{AB}$ 的中点。因此 $C = (0, 1)$ ，因为圆 $C$ 经过原点。

从 $C$ 到 $A$ （以及到 $B$ ）的距离为 $\sqrt2$ 。两个圆心距离为 $\sqrt2$ 的单位圆重叠成一个透镜形，其面积为 $\begin{aligned} 2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) \\ {}- \tfrac{\sqrt2}{2}\sqrt{4 - 2} \\ = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1 \end{aligned}\text{。}$

圆 $A$ 和圆 $B$ 只在原点相交，所以两个透镜形区域不重叠。所求面积为 $\pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2\text{。}$

因此，正确答案是 C。

Place $A = (-1, 0),$ $B = (1, 0),$ so their tangency point is the origin, the midpoint of $\overline{AB}.$ Then $C = (0, 1),$ since $C$ passes through the origin.

The distance from $C$ to $A$ (and to $B$ ) is $\sqrt2.$ Two unit circles whose centers are $\sqrt2$ apart overlap in a lens of area $\begin{aligned} 2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) \\ {}- \tfrac{\sqrt2}{2}\sqrt{4 - 2} \\ = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1. \end{aligned}$

Circles $A$ and $B$ meet only at the origin, so the two lenses do not overlap. The wanted area is $\pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2.$

Thus, the correct answer is C.

2011 AMC 12A 第 11 题

答案：C

难度评级：1540

小提示：

大提示：

解答：

其他年份的第 11 题