Question

一个四个面都是等边三角形的四面体内切一个球，并外接一个球。对四个面中的每一个面，都有一个球与该面在其中心处外切，并与外接球相切。随机选择外接球内部一点 $P$ 。 $P$ 落在这五个小球之一内部的概率最接近

A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point $P$ is selected at random inside the circumscribed sphere. The probability that $P$ lies inside one of the five small spheres is closest to

$0$

$0.1$

$0.2$

$0.3$

$0.4$

答案：C

知识点：立体几何球几何概率体积

难度评级：2380

小提示：

对正四面体，外接球半径是内切球半径的 $3$ 倍。

For a regular tetrahedron the circumradius is $3$ times the inradius

大提示：

五个小球中每个都与内切球体积相同。

Each of the five small spheres has the same volume as the inscribed sphere

解答：

设 $O$ 为内切球和外接球的共同球心。从 $O$ 把正四面体分成四个全等部分可知，外接球半径是内切球半径的 $3$ 倍，所以外接球的体积是内切球体积 $V$ 的 $27$ 倍。

若其余四个小球之一的半径为 $s$ ，则它的球心到 $O$ 的距离为 $r+s$ ；从 $O$ 量起，这段距离同时又等于 $3r-s$ ，所以 $s=r$ 。这五个球的内部互不相交（中心球与其余四球分别相切），所以并集体积为 $5V$ 。概率为 $\dfrac{5V}{27V} = \dfrac{5}{27} \approx 0.185$ ，最接近 $0.2$ 。

所以正确答案是 C。

Let $O$ be the common center of the inscribed and circumscribed spheres. Splitting the tetrahedron into four congruent pieces from $O$ shows the circumradius is $3$ times the inradius, so the circumscribed sphere has $27$ times the inscribed sphere’s volume $V.$

If one of the four other small spheres has radius $s,$ its center is $r+s$ from $O$ and also $3r-s$ from $O,$ so $s=r.$ These five spheres have disjoint interiors (the central sphere is tangent to each of the other four), so their union has volume $5V.$ The probability is $\dfrac{5V}{27V} = \dfrac{5}{27} \approx 0.185,$ closest to $0.2.$

Thus, the correct answer is C.

1999 AMC 12 第 29 题

答案：C

难度评级：2380

小提示：

大提示：

解答：

其他年份的第 29 题