AMC12 2011 A

AMC12 2011 A · Q11

AMC12 2011 A · Q11. It mainly tests Circle theorems, Area & perimeter.

Circles $A, B,$ and $C$ each has 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 $C$ 的半径都为 $1$。圆 $A$ 与圆 $B$ 有一个公共切点。圆 $C$ 在 $\overline{AB}$ 的中点处与其相切。圆 $C$ 内但在圆 $A$ 和圆 $B$ 外的面积是多少？

(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

The requested area is the area of $C$ minus the area shared between circles $A$, $B$ and $C$. Let $M$ be the midpoint of $\overline{AB}$ and $D$ be the other intersection of circles $C$ and $B$. The area shared between $C$, $A$ and $B$ is $4$ of the regions between arc $\widehat {MD}$ and line $\overline{MD}$, which is (considering the arc on circle $B$) a quarter of the circle $B$ minus $\triangle MDB$: $\frac{\pi r^2}{4}-\frac{bh}{2}$ $b = h = r = 1$ (We can assume this because $\angle DBM$ is 90 degrees, since $CDBM$ is a square, due to the application of the tangent chord theorem at point $M$) So the area of the small region is $\frac{\pi}{4}-\frac{1}{2}$ The requested area is area of circle $C$ minus 4 of this area: $\pi 1^2 - 4\left(\frac{\pi}{4}-\frac{1}{2}\right) = \pi - \pi + 2 = 2$ $\boxed{\textbf{C}}$.

所求面积等于圆 $C$ 的面积减去圆 $A$、$B$ 与 $C$ 的公共部分面积。设 $M$ 为 $\overline{AB}$ 的中点，$D$ 为圆 $C$ 与圆 $B$ 的另一个交点。圆 $C$、$A$、$B$ 的公共部分由 $4$ 个全等小区域组成，每个小区域是弧 $\widehat{MD}$ 与弦 $\overline{MD}$ 之间的部分。以圆 $B$ 上的弧来看，该小区域面积等于圆 $B$ 的一个四分之一减去 $\triangle MDB$ 的面积： $\frac{\pi r^2}{4}-\frac{bh}{2}$ $b = h = r = 1$ （可如此设定是因为 $\angle DBM$ 为 $90$ 度；由在点 $M$ 处应用切弦定理可知 $CDBM$ 为正方形。）因此小区域面积为 $\frac{\pi}{4}-\frac{1}{2}$ 所求面积为圆 $C$ 的面积减去 $4$ 个小区域面积： $\pi 1^2 - 4\left(\frac{\pi}{4}-\frac{1}{2}\right) = \pi - \pi + 2 = 2$ $\boxed{\textbf{C}}$.

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