AMC12 2014 A

AMC12 2014 A · Q17

AMC12 2014 A · Q17. It mainly tests Coordinate geometry, Distance / midpoint.

A $4\times 4\times h$ rectangular box contains a sphere of radius $2$ and eight smaller spheres of radius $1$. The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is $h$?

一个$4\times 4\times h$的长方体盒子内有一个半径为$2$的球体和八个半径为$1$的小球。小球各自与盒子的三个面相切，大球与每个小球相切。$h$是多少？

(A) 2+2\sqrt 7 2+2\sqrt 7

(B) 3+2\sqrt 5 3+2\sqrt 5

(D) 4\sqrt 5 4\sqrt 5

(E) 4\sqrt 7\qquad 4\sqrt 7\qquad

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let one of the corners be $(0, 0, 0)$. We can orient the box such that the center of the small sphere closest to the corner is $(1,1,1)$, and the center of the large sphere is $(2, 2, h/2)$. Since the two spheres are tangent, the distance between their centers is $1+2 = 3$, so $\sqrt{(2-1)^2+(2-1)^2+(h/2-1)^2} = 3$. Solving, $h=2 + 2\sqrt{7}=\boxed{\textbf{(A)}}$

设一个角为$(0, 0, 0)$。我们可以定向盒子，使得最靠近该角的小球中心为$(1,1,1)$，大球中心为$(2, 2, h/2)$。由于两个球相切，它们的中心距离为$1+2 = 3$，所以$\sqrt{(2-1)^2+(2-1)^2+(h/2-1)^2} = 3$。解得$h=2 + 2\sqrt{7}=\boxed{\textbf{(A)}}$

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