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?

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

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

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

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

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

Solution:

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 $8\cdot\tfrac16 x^3=\tfrac43\left(\tfrac{2-\sqrt2}{2}\right)^3=\frac{10-7\sqrt2}{3}.$

Thus, the correct answer is B.

2007 AMC 12A Problem 20

Solution:

Problem 20 in Other Years