2012 AMC 10B Problem 23

The discarded tetrahedron has a right isosceles triangle of leg $1$ as one base and height $1$, so its volume is $\dfrac13\cdot\dfrac12\cdot1=\dfrac16$. The cut face is an equilateral triangle of side $\sqrt2$, so its area is $\dfrac{\sqrt3}{4}(\sqrt2)^2=\dfrac{\sqrt3}{2}$. If $h$ is the height from the opposite vertex to this cut face, then $\dfrac13\cdot\dfrac{\sqrt3}{2}\cdot h=\dfrac16$, so $h=\dfrac{\sqrt3}{3}$. The full cube diagonal has length $\sqrt3$. After the tetrahedron is removed and the cut face is placed down, the height is $\sqrt3-\dfrac{\sqrt3}{3}=\dfrac{2\sqrt3}{3}$. Thus, D is the correct answer.