AMC10 2021 B

AMC10 2021 B · Q7

AMC10 2021 B · Q7. It mainly tests Area & perimeter, Geometry misc.

In a plane, four circles with radii $1,3,5,$ and $7$ are tangent to line $\ell$ at the same point $A,$ but they may be on either side of $\ell$. Region $S$ consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region $S$?

在平面上一条直线$\ell$上有四个半径分别为$1,3,5,$和$7$的圆，它们在同一点$A$处与直线$\ell$相切，但可以位于$\ell$的两侧。区域$S$由位于恰好一个圆内部的所有点组成。$S$区域的最大可能面积是多少？

(A) 24\pi 24\pi

(B) 32\pi 32\pi

(D) 65\pi 65\pi

(E) 84\pi 84\pi

Answer

Correct choice: (D)

正确答案：（D）

Solution

Suppose that line $\ell$ is horizontal, and each circle lies either north or south to $\ell.$ We construct the circles one by one: 1. Without the loss of generality, we draw the circle with radius $7$ north to $\ell.$ 2. To maximize the area of region $S,$ we draw the circle with radius $5$ south to $\ell.$ 3. Now, we need to subtract the circle with radius $3$ at least. The optimal situation is that the circle with radius $3$ encompasses the circle with radius $1,$ in which we do not need to subtract more. That is, the two smallest circles are on the same side of $\ell,$ but can be on either side. The diagram below shows one possible configuration of the four circles: Together, the answer is $\pi\cdot7^2+\pi\cdot5^2-\pi\cdot3^2=\boxed{\textbf{(D) }65\pi}.$

假设直线$\ell$是水平的，每个圆位于$\ell$的北侧或南侧。我们逐一构造圆： 1. 不失一般性，将半径为$7$的圆画在$\ell$的北侧。 2. 为了最大化区域$S$的面积，将半径为$5$的圆画在$\ell$的南侧。 3. 现在，至少需要减去半径为$3$的圆。最优情况是半径为$3$的圆包含半径为$1$的圆，这样无需额外减去。这意味着两个最小圆位于$\ell$的同一侧，但可以是任意侧。下图展示了一个可能的四个圆的配置：总面积为$\pi\cdot7^2+\pi\cdot5^2-\pi\cdot3^2=\boxed{\textbf{(D) }65\pi}$。

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