2007 AMC 12A 第 20 题

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}

从每条边的两端各切去长度

x

后，留下的八边形边长为

x\sqrt2

。

Cutting length

x

from each end of an edge leaves an octagon side of

x\sqrt2

由

1=2x+x\sqrt2

求

x

；每个角上的四面体体积为

\tfrac16 x^3

。

From

1=2x+x\sqrt2

find

x

; each corner tetrahedron has volume

\tfrac16 x^3

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}\text{。}

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}\text{。}

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.

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