Question

六个半径为 $1$ 的球摆放成它们的球心位于边长为 $2$ 的正六边形的六个顶点。这六个球都内切于一个大球，大球球心为该正六边形的中心。第八个球外切于这六个小球，并内切于大球。第八个球的半径是多少？

Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2.$ The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?

$\sqrt2$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$\sqrt3$

$2$

答案：B

知识点：球立体几何勾股定理

难度评级：1970

解答：

六个半径为 $1$ 的小球球心构成边长为 $2$ 的正六边形，所以每个小球球心到大球球心的距离为 $2$ 。因此大球半径为 $3$ 。

设第八个球半径为 $r$ ，球心到大球球心的距离为 $x$ 。内切给出 $x+r=3$ ，所以 $x=3-r$ 。

由图中的直角三角形， $(r+1)^2=2^2+(3-r)^2$ 。化简得 $2r+1=13-6r$ ，所以 $r=\frac32$ 。

所以正确答案是 B。

The centers of the six radius- $1$ spheres form a regular hexagon of side length $2$ , so each is $2$ units from the large sphere's center. Hence the large sphere has radius $3$ .

Let the eighth sphere have radius $r$ , and let its center be distance $x$ from the large sphere's center. Internal tangency gives $x+r=3$ , so $x=3-r$ .

Using the right triangle between the large center, a small-sphere center, and the eighth-sphere center, $(r+1)^2=2^2+(3-r)^2$ . Thus $2r+1=13-6r$ , so $r=\frac32$ .

Thus, B is the correct answer.

2013 AMC 10A 第 22 题

答案：B

难度评级：1970

解答：

其他年份的第 22 题