AMC12 2006 B

AMC12 2006 B · Q15

AMC12 2006 B · Q15. It mainly tests Circle theorems, Area & perimeter.

Circles with centers $O$ and $P$ have radii 2 and 4, respectively, and are externally tangent. Points $A$ and $B$ are on the circle centered at $O$, and points $C$ and $D$ are on the circle centered at $P$, such that $\overline{AD}$ and $\overline{BC}$ are common external tangents to the circles. What is the area of hexagon $AOBCPD$?

圆心为 $O$ 和 $P$ 的圆半径分别为 2 和 4，且外切。点 $A$ 和 $B$ 在以 $O$ 为圆心的圆上，点 $C$ 和 $D$ 在以 $P$ 为圆心的圆上，使得 $\overline{AD}$ 和 $\overline{BC}$ 是两圆的公外切线。六边形 $AOBCPD$ 的面积是多少？

(A) $18\sqrt{3}$ $18\sqrt{3}$

(B) $24\sqrt{2}$ $24\sqrt{2}$

(D) $24\sqrt{3}$ $24\sqrt{3}$

(E) $32\sqrt{2}$ $32\sqrt{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Draw the altitude from $O$ onto $DP$ and call the point $H$. Because $\angle OAD$ and $\angle ADP$ are right angles due to being tangent to the circles, and the altitude creates $\angle OHD$ as a right angle. $ADHO$ is a rectangle with $OH$ bisecting $DP$. The length $OP$ is $4+2$ and $HP$ has a length of $2$, so by pythagorean's, $OH$ is $\sqrt{32}$. $2 \cdot \sqrt{32} + \frac{1}{2}\cdot2\cdot \sqrt{32} = 3\sqrt{32} = 12\sqrt{2}$, which is half the area of the hexagon, so the area of the entire hexagon is $2\cdot12\sqrt{2} = \boxed{(B)24\sqrt{2}}$

从 $O$ 向 $DP$ 作高，垂足为 $H$。由于切线性质，$\angle OAD$ 与 $\angle ADP$ 为直角，而高使得 $\angle OHD$ 也是直角。于是 $ADHO$ 为矩形，且 $OH$ 平分 $DP$。$OP=4+2$，且 $HP=2$，由勾股定理得 $OH=\sqrt{32}$。 $2\cdot\sqrt{32}+\frac{1}{2}\cdot2\cdot\sqrt{32}=3\sqrt{32}=12\sqrt{2}$，这是六边形面积的一半，所以六边形的面积为 $2\cdot12\sqrt{2}=\boxed{(B)24\sqrt{2}}$。

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