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2007 AMC 12A Problem 20

Problem 20 of 25HarderGeometry

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?

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Solution

Slicing removes two equal segments of length xx from each edge. Each octagon then has side length x2,x\sqrt2, and the edge satisfies 1=2x+x2,1=2x+x\sqrt2, so x=12+2=222.x=\frac{1}{2+\sqrt2}=\frac{2-\sqrt2}{2}. Each removed corner is a tetrahedron with three mutually perpendicular legs of length x,x, so its volume is 16x3.\tfrac16 x^3. There are 88 corners, giving total volume 816x3=43(222)3=10723. \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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Concepts: cube geometry · volume

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.