Question

偶数个圆相互嵌套，第一个半径为 $1$ ，之后每次半径增加 $1$ ，并且所有圆共享一个公共点。从半径为 $2$ 的圆内但半径为 $1$ 的圆外的区域开始，每隔一圈给区域涂色。下面显示的是 $8$ 个圆的例子。至少需要多少个圆，才能使总阴影面积至少为 $2023\pi$ ？

An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi?$

$46$

$48$

$56$

$60$

$64$

答案：E

知识点：圆面积等差数列求和

难度评级：1560

解答：

半径为 $r$ 的圆面积为 $\pi r^2$ 。因此半径 $2k$ 与 $2k-1$ 之间的阴影环形区域面积为 $\pi\big((2k)^2 - (2k-1)^2\big)$ $= (4k-1)\pi$ 。有 $2n$ 个圆时，阴影总面积为 $\pi\sum_{k=1}^{n}(4k-1) = \pi(2n^2 + n)$ 。我们需要 $2n^2 + n \ge 2023$ 。当 $n = 31$ 时为 $1953$ ，当 $n = 32$ 时为 $2080$ 。所以 $n = 32$ ，也就是 $2n = 64$ 个圆。因此，正确答案是 E。

A circle of radius $r$ has area $\pi r^2.$ So the shaded ring between radius $2k$ and $2k-1$ has area $\pi\big((2k)^2 - (2k-1)^2\big)$ $= (4k-1)\pi.$ With $2n$ circles the shaded total is $\pi\sum_{k=1}^{n}(4k-1) = \pi(2n^2 + n).$ We want $2n^2 + n \ge 2023.$ At $n = 31$ it's $1953,$ at $n = 32$ it's $2080.$ So $n = 32,$ which means $2n = 64$ circles. Thus, E is the correct answer.

2023 AMC 10A 第 15 题

答案：E

难度评级：1560

解答：

其他年份的第 15 题