Question

对一个实心正方体的每个顶点，取由该顶点和与它相接的三条棱的中点所确定的四面体。切去这八个四面体后，正方体剩余的部分称为截半立方体。截半立方体与原正方体的体积之比最接近下列哪一项？

For each vertex of a solid cube, consider the tetrahedron determined by the vertex and the midpoints of the three edges that meet at that vertex. The portion of the cube that remains when these eight tetrahedra are cut away is called a cuboctahedron. The ratio of the volume of the cuboctahedron to the volume of the original cube is closest to which of these?

$75\%$

$78\%$

$81\%$

$84\%$

$87\%$

答案：D

知识点：solid geometry tetrahedron volume volume ratio

难度评级：1960

小提示：

为正方体选取一个便于计算的边长，再求一个被切去的角四面体的体积

Choose a convenient side length for the cube and compute one removed corner tetrahedron

大提示：

每个被切去的四面体都有三条两两垂直的棱，其长度等于正方体边长的一半

Each removed tetrahedron has three mutually perpendicular edges equal to half the cube’s side

解答：

取正方体边长为 $2$ ，则其体积为 $8$ 。每个角四面体有三条长度为 $1$ 且两两垂直的棱，所以体积为 $\frac16$ 。八个被切去的四面体总体积为 $\frac43$ ，剩余体积为 $\frac{20}{3}$ 。所求比值为 $\frac{\frac{20}{3}}{8}=\frac56=83\tfrac13\%\text{，}$ 最接近 $84\%$ 。

因此正确答案是 D。

Take the cube’s side length to be $2.$ Its volume is $8.$ Each corner tetrahedron has three perpendicular edges of length $1,$ so its volume is $\frac16.$ The eight removed tetrahedra have total volume $\frac43,$ leaving $\frac{20}{3}.$ The ratio is $\frac{\frac{20}{3}}{8}=\frac56=83\tfrac13\%,$ which is closest to $84\%.$

Thus the correct answer is D.

1992 AMC 12 第 19 题

答案：D

难度评级：1960

小提示：

大提示：

解答：

其他年份的第 19 题