2011 AMC 12A 第 15 题

Question

一个半径为

2

的半球的圆形底面放在一个高为

6

的正方形棱锥的底面上。该半球与棱锥的另外四个面相切。棱锥底面边长是多少？

The circular base of a hemisphere of radius

2

rests on the base of a square pyramid of height

6.

The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?

3\sqrt{2}

\dfrac{13}{3}

4\sqrt{2}

6

\dfrac{13}{2}

答案：A

知识点：立体几何棱锥距离公式

难度评级：1870

小提示：

取通过顶点以及两条相对底边中点的截面

Take the cross-section through the apex and the midpoints of two opposite base edges

大提示：

在这个截面中，斜边直线从

\left(\tfrac{s}{2}, 0\right)

到

(0, 6)

，且它到中心的距离为

2

In that cross-section, the slant line runs from

\left(\tfrac{s}{2}, 0\right)

to

(0, 6),

and its distance from the center is

2

解答：

设底面边长为

s

，底面中心在原点，顶点高度为

6

。用通过顶点和两条相对底边中点的竖直平面截取。侧面在截面中表现为从

\left(\tfrac{s}{2}, 0\right)

到

(0, 6)

的直线。

这条直线是

\tfrac{2}{s}x + \tfrac16 y = 1

。半球与该面相切，所以原点到这条直线的距离等于半径

2

：

\dfrac{1}{\sqrt{\tfrac{4}{s^2} + \tfrac{1}{36}}} = 2\text{。}

因此

\tfrac{4}{s^2} + \tfrac{1}{36} = \tfrac14

，所以

\tfrac{4}{s^2} = \tfrac{2}{9}

，

s^2 = 18

，得

s = 3\sqrt2

。

因此，正确答案是 A。

Let the base have side

s,

centered at the origin, with apex at height

6.

Cut with the vertical plane through the apex and the midpoints of two opposite base edges. The slant face appears as the line from

\left(\tfrac{s}{2}, 0\right)

to

(0, 6).

This line is

\tfrac{2}{s}x + \tfrac16 y = 1.

The hemisphere is tangent to the face, so the distance from the origin to this line is the radius

2:

\dfrac{1}{\sqrt{\tfrac{4}{s^2} + \tfrac{1}{36}}} = 2.

Then

\tfrac{4}{s^2} + \tfrac{1}{36} = \tfrac14,

so

\tfrac{4}{s^2} = \tfrac{2}{9}

and

s^2 = 18,

giving

s = 3\sqrt2.

Thus, the correct answer is A.

2011 AMC 12A 第 15 题

答案：A

难度评级：1870

小提示：

大提示：

解答：

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