Question

六个全等圆围成一个环，每个圆都与相邻的两个圆外切。这六个圆都内切于半径为 $30$ 的圆 $\mathcal{C}$ 。设 $K$ 为在 $\mathcal{C}$ 内且在这六个圆外的区域面积。求 $\lfloor K\rfloor$ 。（记号 $\lfloor K\rfloor$ 表示小于或等于 $K$ 的最大整数。）

Six congruent circles form a ring with each circle externally tangent to the two circles adjacent to it. All six circles are internally tangent to a circle $\mathcal{C}$ with radius $30.$ Let $K$ be the area of the region inside $\mathcal{C}$ and outside all of the six circles in the ring. Find $\lfloor K\rfloor.$ (The notation $\lfloor K\rfloor$ denotes the greatest integer that is less than or equal to $K.$ )

答案

答案：942

知识点：相切圆正多边形圆面积

难度评级：2010

解答：

设六个圆的公共半径为 $r$ 。相邻圆外切，所以它们的圆心相距 $2r$ ，六个圆心构成边长为 $2r$ 的正六边形。因为正六边形的外接圆半径等于它的边长，所以每个圆心到 $\mathcal{C}$ 的圆心 $O$ 的距离都是 $2r$ 。与 $\mathcal{C}$ 内切意味着从 $O$ 到每个小圆圆心的距离再加上 $r$ 等于 $30$ ，所以 $3r = 30$ ， $r = 10$ 。

因此 $\begin{aligned} K &= \pi\left(30^2 - 6 \cdot 10^2\right) \\ &= 300\pi \approx 942.48, \end{aligned}$ 所以 $\lfloor K\rfloor = 942$ 。

Let $r$ be the common radius of the six circles. Adjacent circles are externally tangent, so their centers are $2r$ apart, and the six centers form a regular hexagon with side $2r.$ Since a regular hexagon's circumradius equals its side length, each center is at distance $2r$ from the center $O$ of $\mathcal{C}.$ Internal tangency to $\mathcal{C}$ means the distance from $O$ to each small center plus $r$ equals $30,$ so $3r = 30$ and $r = 10.$

Therefore $\begin{aligned} K &= \pi\left(30^2 - 6 \cdot 10^2\right) \\ &= 300\pi \approx 942.48, \end{aligned}$ and $\lfloor K\rfloor = 942.$

2005 AIME I 第 1 题

答案：942

难度评级：2010

解答：

其他年份的第 1 题