2007 AMC 12A Problem 20

Question

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}

Cutting length

x

from each end of an edge leaves an octagon side of

x\sqrt2

From

1=2x+x\sqrt2

find

x

; each corner tetrahedron has volume

\tfrac16 x^3

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.

Problem 20 in other years