2006 AMC 10A 第 24 题

Question

连接一个单位立方体相邻面的中心，形成一个正八面体。这个八面体的体积是多少？

Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?

\dfrac{1}{8}

\dfrac{1}{6}

\dfrac{1}{4}

\dfrac{1}{3}

\dfrac{1}{2}

这个八面体可以看成两个底面相贴的正方形棱锥。

The octahedron is two square pyramids glued at their bases

相邻面中心之间的距离为

\frac{\sqrt2}{2}

；求底面积和高。

Adjacent face-centers are

\frac{\sqrt2}{2}

apart; find the base area and height

六个面中心形成一个正八面体，可看成两个全等的正方形棱锥共用底面。相邻面中心距离为

\frac{\sqrt2}{2}

，所以正方形底面面积为

\left(\frac{\sqrt2}{2}\right)^2 = \frac12

。

每个棱锥高为

\frac12

，体积为

\frac13 \cdot \frac12 \cdot \frac12 = \frac{1}{12}

。八面体体积为

2 \cdot \frac{1}{12} = \frac16

。

所以正确答案是 B。

The six face centers form a regular octahedron, viewed as two congruent square pyramids sharing a base. Adjacent face centers are

\frac{\sqrt2}{2}

apart, so the square base has area

\left(\frac{\sqrt2}{2}\right)^2 = \frac12.

Each pyramid has height

\frac12,

so its volume is

\frac13 \cdot \frac12 \cdot \frac12 = \frac{1}{12}.

The octahedron has volume

2 \cdot \frac{1}{12} = \frac16.

Thus, the correct answer is B.

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