Question

三个两两相切、半径为 $1$ 的球放在水平平面上。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：立体几何球重心勾股定理

难度评级：2180

解答：

三个小球球心形成边长为 $2$ 的等边三角形，每个球心距平面 $1$ 。设其重心为 $D$ ，则它到每个顶点的距离为 $\dfrac{2\sqrt{3}}{3}$ 。

大球球心 $E$ 位于 $D$ 正上方，且 $E$ 到每个小球球心的距离为 $1 + 2 = 3$ ，所以 $\begin{aligned} DE &= \sqrt{3^2 - \left(\dfrac{2\sqrt{3}}{3}\right)^2} \\ &= \sqrt{9 - \dfrac{4}{3}} = \dfrac{\sqrt{69}}{3}. \end{aligned}$

大球球心 $E$ 位于 $D$ 正上方，所以从平面到大球顶部的距离为小球球心高度 $1$ 、高度差和大球半径 $2$ 之和： $1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}.$

所以正确答案是 B。

The three small centers form an equilateral triangle of side $2,$ each $1$ unit above the plane. Its centroid $D$ is at distance $\dfrac{2\sqrt{3}}{3}$ from each vertex.

The large sphere's center $E$ sits directly above $D,$ and the distance between $E$ and a small center is $1 + 2 = 3.$ Thus $\begin{aligned} DE &= \sqrt{3^2 - \left(\dfrac{2\sqrt{3}}{3}\right)^2} \\ &= \sqrt{9 - \dfrac{4}{3}} = \dfrac{\sqrt{69}}{3}. \end{aligned}$

Adding the $1$ unit from the plane to $D$ and the $2$ units from $E$ to the top of the large sphere gives $1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}.$

Thus, the correct answer is B.

2004 AMC 10A 第 25 题

答案：B

难度评级：2180

解答：

其他年份的第 25 题