Question

把一个单位立方体的各个角切去，使六个面都变成正八边形。被切去的四面体总体积是多少？

Corners are sliced off a unit cube so that the six faces each become regular octagons. What is the total volume of the removed tetrahedra?

$\dfrac{5\sqrt2-7}{3}$

$\dfrac{10-7\sqrt2}{3}$

$\dfrac{3-2\sqrt2}{3}$

$\dfrac{8\sqrt2-11}{3}$

$\dfrac{6-4\sqrt2}{3}$

答案：B

知识点：正方体体积

难度评级：1840

解答：

切角会从每条棱的两端各去掉长度 $x$ 。每个八边形的边长为 $x\sqrt2$ ，且棱长满足 $1=2x+x\sqrt2$ ，所以 $x=\frac{1}{2+\sqrt2}=\frac{2-\sqrt2}{2}.$

每个被切去的角都是三条互相垂直的棱长为 $x$ 的四面体，体积为 $\tfrac16 x^3$ 。共有 $8$ 个角，所以总体积为 $\begin{aligned} &8\cdot\tfrac16 x^3 \\ &=\tfrac43\left(\tfrac{2-\sqrt2}{2}\right)^3 \\ &=\frac{10-7\sqrt2}{3}. \end{aligned}$

因此，正确答案是 B。

Slicing removes two equal segments of length $x$ from each edge. Each octagon then has side length $x\sqrt2,$ and the edge satisfies $1=2x+x\sqrt2,$ so $x=\frac{1}{2+\sqrt2}=\frac{2-\sqrt2}{2}.$

Each removed corner is a tetrahedron with three mutually perpendicular legs of length $x,$ so its volume is $\tfrac16 x^3.$ There are $8$ corners, giving total volume $\begin{aligned} &8\cdot\tfrac16 x^3 \\ &=\tfrac43\left(\tfrac{2-\sqrt2}{2}\right)^3 \\ &=\frac{10-7\sqrt2}{3}. \end{aligned}$

Thus, the correct answer is B.

2007 AMC 12A 第 20 题

答案：B

难度评级：1840

解答：

其他年份的第 20 题