2016 AMC 12B 第 23 题

Question

三维空间中，由不等式

|x|+|y|+|z|\le1

和

|x|+|y|+|z-1|\le1

所定义的区域体积是多少？

What is the volume of the region in three-dimensional space defined by the inequalities

|x|+|y|+|z|\le1

and

|x|+|y|+|z-1|\le1?

\dfrac16

\dfrac13

\dfrac12

\dfrac23

1

集合

|x|+|y|+|z|\le1

是一个正八面体，对角线长为

2

，体积为

\tfrac43

。

The set

|x|+|y|+|z|\le1

is a regular octahedron with diagonals of length

2

and volume

\tfrac43

第二个不等式是该八面体向上平移

1

；两者交集是线性尺寸为一半的相似八面体。

The second inequality is that octahedron translated up by

1;

their overlap is a similar octahedron with half the linear size

区域

|x|+|y|+|z|\le1

是一个正八面体，顶点为

(\pm1,0,0),(0,\pm1,0),(0,0,\pm1)

，体积为

2\cdot\tfrac13\cdot(\sqrt2)^2\cdot1=\tfrac43

。第二个区域是同一个八面体向上平移

1

后得到的。两者的交集是另一个对角线长为

1

的正八面体，其线性尺寸是原八面体的一半，所以体积为

\left(\tfrac12\right)^3\cdot\tfrac43=\tfrac16

。

所以正确答案是 A。

The region

|x|+|y|+|z|\le1

is a regular octahedron with vertices at

(\pm1,0,0),(0,\pm1,0),(0,0,\pm1),

whose volume is

2\cdot\tfrac13\cdot(\sqrt2)^2\cdot1=\tfrac43.

The second region is the same octahedron shifted up by

1.

Their intersection is bounded by another regular octahedron with diagonals of length

1,

half the linear dimensions of the first, so its volume is

\left(\tfrac12\right)^3\cdot\tfrac43=\tfrac16.

Thus, the correct answer is A.

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