Question

八个半径为 $1$ 的球，每个八分体中一个，都与坐标平面相切。以原点为球心、能包含这八个球的最小球的半径是多少？

Eight spheres of radius $1,$ one per octant, are each tangent to the coordinate planes. What is the radius of the smallest sphere, centered at the origin, that contains these eight spheres?

$\sqrt{2}$

$\sqrt{3}$

$1 + \sqrt{2}$

$1 + \sqrt{3}$

$3$

答案：D

知识点：立体几何球距离公式

难度评级：1660

解答：

在某个八分体中，与三个坐标平面相切、半径为 $1$ 的球球心可取为 $(1, 1, 1)$ ，到原点距离为 $\sqrt{1^2 + 1^2 + 1^2} = \sqrt3$

这个球上离原点最远的点距离为 $\sqrt3 + 1$ ，所以包含八个球的最小球半径为 $1 + \sqrt3$ 。

所以正确答案是 D。

A sphere of radius $1$ tangent to the three coordinate planes in one octant has its center at a point like $(1, 1, 1),$ at distance $\sqrt{1^2 + 1^2 + 1^2} = \sqrt3$ from the origin.

The farthest point of that sphere from the origin is at distance $\sqrt3 + 1,$ so the containing sphere has radius $1 + \sqrt3.$

Thus, the correct answer is D.

2005 AMC 12B 第 16 题

答案：D

难度评级：1660

解答：

其他年份的第 16 题