1989 AMC 12 第 26 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

26.

连接一个立方体相邻各面的中心,形成一个正八面体。该八面体与立方体的体积之比为

A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is

312\frac{\sqrt3}{12}

616\frac{\sqrt6}{16}

16\frac16

28\frac{\sqrt2}{8}

14\frac14

答案:C
知识点:正方体多面体体积
难度评级:1940
小提示:

将立方体的边长缩放为 22,并把它的中心置于原点

Scale the cube to side length 22 and place its center at the origin

大提示:

该八面体的顶点为 (±1,0,0),(0,±1,0),(0,0,±1)(\pm1,0,0),(0,\pm1,0),(0,0,\pm1)

The octahedron has vertices (±1,0,0),(0,±1,0),(0,0,±1)(\pm1,0,0),(0,\pm1,0),(0,0,\pm1)

解答:

取边长为 22 的立方体,则其体积为 88。各面的中心为 (±1,0,0),(0,±1,0),(0,0,±1)(\pm1,0,0),(0,\pm1,0),(0,0,\pm1)。在每个卦限中,八面体截出一个三条互相垂直的棱长均为单位长度的四面体,其体积为 16\frac{1}{6}。所以八面体的体积为 8(16)=438(\frac{1}{6})=\frac{4}{3},所求体积比为 438=16\frac{\frac{4}{3}}{8}=\frac{1}{6}

所以正确答案是 C

Take a cube of side 2,2, so its volume is 8.8. The face centers are (±1,0,0),(0,±1,0),(0,0,±1).(\pm1,0,0),(0,\pm1,0),(0,0,\pm1). In each octant, the octahedron cuts out a tetrahedron with three perpendicular unit edges and volume 16.\frac{1}{6}. Thus its volume is 8(16)=43,8(\frac{1}{6})=\frac{4}{3}, and the ratio is 438=16.\frac{\frac{4}{3}}{8}=\frac{1}{6}.

Thus the correct answer is C.

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