AMC10 2011 A

AMC10 2011 A · Q18

AMC10 2011 A · Q18. It mainly tests Circle theorems, Area & perimeter.

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 $B$?

圆 $A$、$B$ 和 $C$ 半径均为 1。圆 $A$ 和 $B$ 有一个公共切点。圆 $C$ 与 $\overline{AB}$ 的中点有切点。求圆 $C$ 内但在圆 $A$ 和圆 $B$ 外的面积？

(A) 3 $- \frac{\pi}{2}$ 3 $- \frac{\pi}{2}$

(B) $\frac{\pi}{2}$ $\frac{\pi}{2}$

(D) $\frac{3\pi}{4}$ $\frac{3\pi}{4}$

(E) 1 + $\frac{\pi}{2}$ 1 + $\frac{\pi}{2}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $D$ be the midpoint of $\overline{AB}$, and let circle $C$ intersect circles $A$ and $B$ at $E$ and $F$, respectively, distinct from $D$. The shaded portion of unit square $ADCE$ has area $1-\frac{\pi}{4}$, as does the shaded portion of unit square $BDCF$. The portion of the shaded region which is outside these squares is a semicircle of radius $1$ and has area $\frac{\pi}{2}$. The total shaded area is $2\left(1-\frac{\pi}{4}\right)+\frac{\pi}{2}=2$. OR Let $D, E,$ and $F$ be defined as in the first solution, and let $G$ be diametrically opposite $D$ on circle $C$. The shaded area is equal to the area of square $DFGE$, which has diagonal length $2$. Its side length is $\sqrt{2}$, and its area is $(\sqrt{2})^2=2$.

答案（C）：设$D$为$\overline{AB}$的中点，并令圆$C$分别与圆$A$和圆$B$相交于$E$与$F$，且$E,F$均不同于$D$。单位正方形$ADCE$的阴影部分面积为$1-\frac{\pi}{4}$，单位正方形$BDCF$的阴影部分面积也为$1-\frac{\pi}{4}$。阴影区域中位于这两个正方形之外的部分是半径为$1$的半圆，面积为$\frac{\pi}{2}$。因此阴影总面积为$2\left(1-\frac{\pi}{4}\right)+\frac{\pi}{2}=2$。或者令$D,E,F$与第一种解法中的定义相同，并令$G$为圆$C$上与$D$直径相对的点。阴影面积等于正方形$DFGE$的面积，该正方形的对角线长为$2$，边长为$\sqrt{2}$，其面积为$(\sqrt{2})^2=2$。

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