AMC12 2004 A

AMC12 2004 A · Q24

AMC12 2004 A · Q24. It mainly tests Circle theorems, Area & perimeter.

A plane contains points $A$ and $B$ with $AB = 1$. Let $S$ be the union of all disks of radius $1$ in the plane that cover $\overline{AB}$. What is the area of $S$?

平面上有点 $A$ 和 $B$，且 $AB = 1$。设 $S$ 为平面中所有半径为 $1$ 且覆盖 $\overline{AB}$ 的圆盘的并集。求 $S$ 的面积。

(A) $2\pi + \sqrt{3}$ $2\pi + \sqrt{3}$

(B) $\dfrac{8\pi}{3}$ $\dfrac{8\pi}{3}$

(D) $\dfrac{10\pi}{3} - \sqrt{3}$ $\dfrac{10\pi}{3} - \sqrt{3}$

(E) $4\pi - 2\sqrt{3}$ $4\pi - 2\sqrt{3}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

As the red circles move about segment $AB$, they cover the area we are looking for. On the left side, the circle must move around pivoted on $B$. On the right side, the circle must move pivoted on $A$ However, at the top and bottom, the circle must lie on both A and B, giving us our upper and lower bounds. This egg-like shape is $S$. The area of the region can be found by dividing it into several sectors, namely \begin{align*} A &= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\ A &= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi (1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\ A &= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\ A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}

当红色圆在 $\overline{AB}$ 周围移动时，它们覆盖了我们要找的区域。左侧，圆必须以 $B$ 为支点绕动。右侧，圆必须以 $A$ 为支点绕动。然而在上方和下方，圆必须同时经过 $A$ 和 $B$，从而给出上、下边界。这个蛋形区域就是 $S$。将该区域分割成若干扇形，可得面积为 \begin{align*} A &= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\ A &= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi (1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\ A &= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\ A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*}

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