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：

其他年份的第 15 题