AMC10 2023 A

AMC10 2023 A · Q15

AMC10 2023 A · Q15. It mainly tests Arithmetic sequences basics, Area & perimeter.

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 $2023\pi$?

有偶数个圆嵌套，起始半径为$1$，每次增加$1$，所有圆共有一个公共点。每隔一个圆之间的区域被涂阴影，从半径$2$的圆内但半径$1$的圆外的区域开始。下方显示了$8$个圆的示例。需要最少多少个圆才能使总阴影面积至少为$2023\pi$？

(A) 46 46

(B) 48 48

(D) 60 60

(E) 64 64

Answer

Correct choice: (E)

正确答案：（E）

Solution

Notice that the area of the shaded region is $(2^2\pi-1^2\pi)+(4^2\pi-3^2\pi)+(6^2\pi-5^2\pi)+ \cdots + (n^2\pi-(n-1)^2 \pi)$ for any even number $n$. Using the difference of squares, this simplifies to $(1+2+3+4+\cdots+n) \pi$. So, we are basically finding the smallest $n$ such that $\frac{n(n+1)}{2}>2023 \Rightarrow n(n+1) > 4046$. Since $60(61) > 60^2=3600$, the only option higher than $60$ is $\boxed{\textbf{(E) } 64}$.

注意阴影区域的面积为$(2^2\pi-1^2\pi)+(4^2\pi-3^2\pi)+(6^2\pi-5^2\pi)+ \cdots + (n^2\pi-(n-1)^2 \pi)$，其中$n$为偶数。使用平方差公式，这简化为$(1+2+3+4+\cdots+n) \pi$。因此，我们要找最小的$n$使得$\frac{n(n+1)}{2}>2023 \Rightarrow n(n+1) > 4046$。由于$60(61) > 60^2=3600$，高于$60$的唯一选项是$\boxed{\textbf{(E) } 64}$。

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