Question

两个不同的正四面体的所有顶点都在同一个单位立方体的顶点中。两个四面体交集形成区域的体积是多少？

Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?

$\dfrac{1}{12}$

$\dfrac{\sqrt2}{12}$

$\dfrac{\sqrt3}{12}$

$\dfrac{1}{6}$

$\dfrac{\sqrt2}{6}$

答案：D

知识点：立体几何体积正方体相似

难度评级：2380

解答：

两个正四面体使用立方体两组交替的四个顶点。每个四面体边长为 $\sqrt2$ 。

边长为 $s$ 的正四面体体积为 $\dfrac{\sqrt2}{12}s^3$ ，所以一个大四面体体积为 $\dfrac{\sqrt2}{12}(\sqrt2)^3=\dfrac13$ 。

一个四面体的每个面都会从另一个四面体切去一个相似比为 $\dfrac12$ 的角四面体，每个切去部分体积为原四面体的 $\dfrac18$ 。

共有四个这样的角部分被切去，所以交集体积是一个大四面体体积的 $1-4\cdot\dfrac18=\dfrac12$ 。故交集体积为 $\dfrac12\cdot\dfrac13=\dfrac16$ 。

所以正确答案是 D。

The two regular tetrahedra use the two alternating sets of four vertices of the cube. Each has edge length $\sqrt2$ , a face diagonal of the cube.

The volume of a regular tetrahedron with edge length $s$ is $\dfrac{\sqrt2}{12}s^3$ . Thus one large tetrahedron has volume $\dfrac{\sqrt2}{12}(\sqrt2)^3=\dfrac13$ .

Intersect one tetrahedron with the other. Each face of the first cuts from the second a corner tetrahedron similar to the original with scale factor $\dfrac12$ , so each cut-off piece has $\dfrac18$ of the large tetrahedron's volume.

There are four such corner pieces, so the intersection has $1-4\cdot\dfrac18=\dfrac12$ of the volume of one large tetrahedron. Hence the intersection volume is $\dfrac12\cdot\dfrac13=\dfrac16$ .

Thus, D is the correct answer.

2011 AMC 10A 第 24 题

答案：D

难度评级：2380

解答：

其他年份的第 24 题