Question

平面上画出 $100$ 个同心圆，半径分别为 $1, 2, 3, \ldots, 100$ 。半径为 $1$ 的圆的内部涂成红色；每两个相邻圆之间围成的区域涂成红色或绿色，且任意两个相邻区域颜色不同。绿色区域的总面积与最外圆面积之比可写为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

One hundred concentric circles with radii $1, 2, 3, \ldots, 100$ are drawn in a plane. The interior of the circle of radius $1$ is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius $100$ can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：301

知识点：圆环平方差求和

难度评级：1790

解答：

从中心向外，区域颜色依次为红、绿、红、绿、 $\ldots$ ，所以绿色区域是半径 $1$ 与 $2$ 之间、 $3$ 与 $4$ 之间，依此直到 $99$ 与 $100$ 之间的圆环。它们的总面积为即 $\pi\,(1 + 2 + \cdots + 100) = 5050\pi$ 。 $\begin{aligned} &\scriptsize \pi\left[(2^2 - 1^2) + (4^2 - 3^2) + \cdots + (100^2 - 99^2)\right] \\ &\scriptsize = \pi\left[(2 + 1) + (4 + 3) + \cdots + (100 + 99)\right], \end{aligned}$

所求比值为 $\frac{5050\pi}{100^2 \pi} = \frac{101}{200}$ ，因此 $m + n = 101 + 200 = 301$ 。

The regions alternate red, green, red, green, $\ldots$ from the center outward, so the green regions are the annuli between radii $1$ and $2,$ between $3$ and $4,$ and so on up to the annulus between $99$ and $100.$ Their total area is $\begin{aligned} &\scriptsize \pi\left[(2^2 - 1^2) + (4^2 - 3^2) + \cdots + (100^2 - 99^2)\right] \\ &\scriptsize = \pi\left[(2 + 1) + (4 + 3) + \cdots + (100 + 99)\right], \end{aligned}$ which is $\pi\,(1 + 2 + \cdots + 100) = 5050\pi.$

The desired ratio is $\frac{5050\pi}{100^2 \pi} = \frac{101}{200},$ so $m + n = 101 + 200 = 301.$

2003 AIME I 第 2 题

答案：301

难度评级：1790

解答：

其他年份的第 2 题