Question

三个互相相切、半径为 $1$ 的球放在一个水平平面上。一个半径为 $2$ 的球放在它们上面。从平面到较大球顶部的距离是多少？

Three mutually tangent spheres of radius $1$ rest on a horizontal plane. A sphere of radius $2$ rests on them. What is the distance from the plane to the top of the larger sphere?

$3 + \dfrac{\sqrt{30}}{2}$

$3 + \dfrac{\sqrt{69}}{3}$

$3 + \dfrac{\sqrt{123}}{4}$

$\dfrac{52}{9}$

$3 + 2\sqrt{2}$

答案：B

知识点：球重心勾股定理

难度评级：2150

解答：

设三个单位球的球心为 $A$ ， $B$ ， $C$ ，它们在离平面高度 $1$ 处形成边长为 $2$ 的等边三角形；设 $E$ 为大球球心，位于 $\triangle ABC$ 的重心 $D$ 正上方。

从顶点到重心的距离为 $AD = \tfrac{2\sqrt3}{3}$ ，且 $AE = 1 + 2 = 3$ ，所以 $\begin{aligned} DE &= \sqrt{3^2 - \left(\tfrac{2\sqrt3}{3}\right)^2} \\ &= \sqrt{9 - \tfrac{4}{3}} \\ &= \dfrac{\sqrt{69}}{3}. \end{aligned}$

因为 $D$ 位于平面上方 $1$ 个单位，而大球顶部在 $E$ 上方 $2$ 个单位，所以总高度为 $1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}.$

所以正确答案是 B。

Let the centers of the three unit spheres be $A,$ $B,$ $C,$ forming an equilateral triangle of side $2$ at height $1$ above the plane, and let $E$ be the center of the large sphere directly above the centroid $D$ of $\triangle ABC.$

The distance from a vertex to the centroid is $AD = \tfrac{2\sqrt3}{3},$ and $AE = 1 + 2 = 3,$ so $\begin{aligned} DE &= \sqrt{3^2 - \left(\tfrac{2\sqrt3}{3}\right)^2} \\ &= \sqrt{9 - \tfrac{4}{3}} \\ &= \dfrac{\sqrt{69}}{3}. \end{aligned}$

Since $D$ is $1$ unit above the plane and the top of the large sphere is $2$ units above $E,$ the total height is $1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}.$

Thus, the correct answer is B.

2004 AMC 12A 第 22 题

答案：B

难度评级：2150

解答：

其他年份的第 22 题