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?
$\mathrm{(A)}\ \frac{5\sqrt{2}-7}{3}\qquad \mathrm{(B)}\ \frac{10-7\sqrt{2}}{3}\qquad \mathrm{(C)}\ \frac{3-2\sqrt{2}}{3}\qquad \mathrm{(D)}\ \frac{8\sqrt{2}-11}{3}\qquad \mathrm{(E)}\ \frac{6-4\sqrt{2}}{3}$
[2007 AMC 12A Problem 20.png](https://artofproblemsolving.com/wiki/index.php/File:2007_AMC_12A_Problem_20.png)
Since the sides of a [regular polygon](https://artofproblemsolving.com/wiki/index.php/Regular_polygon) are equal in length, we can call each side $x$. Examine one edge of the unit cube: each contains two slanted diagonal edges of an octagon and one straight edge. The diagonal edges form $45-45-90 \triangle$ [right triangles](https://artofproblemsolving.com/wiki/index.php/Right_triangle), making the distance on the edge of the cube $\frac{x}{\sqrt{2}}$. Thus, $2 \cdot \frac{x}{\sqrt{2}} + x = 1$, and $x = \frac{1}{\sqrt{2} + 1} \cdot \left(\frac{\sqrt{2} - 1}{\sqrt{2} - 1}\right) = \sqrt{2} - 1$. 
Each of the cut off corners is a [pyramid](https://artofproblemsolving.com/wiki/index.php/Pyramid), whose volume can be calculated by $V = \frac 13 Bh$. Use the base as one of the three [congruent](https://artofproblemsolving.com/wiki/index.php?title=Congruency_(geometry)&action=edit&redlink=1) [isosceles triangles](https://artofproblemsolving.com/wiki/index.php/Isosceles_triangle), with the height being one of the edges of the pyramid that sits on the edges of the cube. The height is $\frac x{\sqrt{2}} = 1 - \frac{1}{\sqrt{2}}$. The base is a $45-45-90 \triangle$ with leg of length $1 - \frac{1}{\sqrt{2}}$, making its area $\frac 12\left(1 - \frac 1{\sqrt{2}}\right)^2 = \frac{3 - 2\sqrt{2}}4$. Plugging this in, we get that the volume of one of the tetrahedra is $\frac 13 \left(1 - \frac{1}{\sqrt{2}}\right)\left(\frac{3 - 2\sqrt{2}}4\right) = \frac{10 - 7\sqrt{2}}{24}$. Since there are 8 removed corners, we get an answer of $\frac{10 - 7\sqrt{2}}{3} \Longrightarrow \boxed{\mathrm{B}}$