AMC10 2007 A

AMC10 2007 A · Q24

AMC10 2007 A · Q24. It mainly tests Angle chasing, Circle theorems.

Circles centered at $A$ and $B$ each have radius $2$, as shown. Point $O$ is the midpoint of $AB$, and $OA = 2\sqrt{2}$. Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B$, respectively, and $EF$ is a common tangent. What is the area of the shaded region $ECODF$?

以$A$和$B$为圆心、半径均为$2$的圆，如图所示。点$O$是$AB$的中点，且$OA = 2\sqrt{2}$。线段$OC$和$OD$分别切于以$A$和$B$为圆心的圆，且$EF$是公共切线。阴影区域$ECODF$的面积是多少？

(A) $8\sqrt{2} / 3$ $8\sqrt{2} / 3$

(B) $8\sqrt{2} - 4 - \pi$ $8\sqrt{2} - 4 - \pi$

(D) $4\sqrt{2} + \pi/8$ $4\sqrt{2} + \pi/8$

(E) $8\sqrt{2} - 2 - \pi/2$ $8\sqrt{2} - 2 - \pi/2$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Rectangle $ABFE$ has area $AE\cdot AB=2\cdot 4\sqrt2=8\sqrt2$. Right triangles $ACO$ and $BDO$ each have hypotenuse $2\sqrt2$ and one leg of length $2$. Hence they are each isosceles, and each has area $(1/2)(2^2)=2$. Angles $CAE$ and $DBF$ are each $45^\circ$, so sectors $CAE$ and $DBF$ each have area $$ \frac18\pi\cdot 2^2=\frac{\pi}{2}. $$ Thus the area of the shaded region is $$ 8\sqrt2-2\cdot 2-2\cdot\frac{\pi}{2}=8\sqrt2-4-\pi. $$

答案（B）：矩形 $ABFE$ 的面积为 $AE\cdot AB=2\cdot 4\sqrt2=8\sqrt2$。直角三角形 $ACO$ 和 $BDO$ 的斜边均为 $2\sqrt2$，且有一条直角边长为 $2$。因此它们都是等腰三角形，每个的面积为 $(1/2)(2^2)=2$。$\angle CAE$ 和 $\angle DBF$ 均为 $45^\circ$，所以扇形 $CAE$ 和 $DBF$ 的面积各为 $$ \frac18\pi\cdot 2^2=\frac{\pi}{2}. $$ 因此阴影部分的面积为 $$ 8\sqrt2-2\cdot 2-2\cdot\frac{\pi}{2}=8\sqrt2-4-\pi. $$

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