2012 AMC 10B Problem 23

1. Which piece is cut off? On the bottom face take two opposite corners. On the top face, a corner next to neither of them sits directly above one of the other two bottom corners (there are two such corners, and they give the same shape). These three points are exactly the three corners joined by an edge to one bottom corner, call it $K$. So the cut slices off the small corner piece at $K$.
2. The cut face. Each side of the cut triangle is a diagonal of a face of the cube, of length $\sqrt2$. So the cut face is an equilateral triangle with side $\sqrt2$.
3. Which way is up? Put the block on the table with the cut face down. $K$ is $1$ away from each of the three cut corners, and the corner of the cube opposite $K$ (call it $K'$) is $\sqrt2$ away from each of them. A point that is the same distance from all three corners of an equilateral triangle lies on the line straight up through the triangle's center. So $K$, the center of the cut face and $K'$ all lie on one line that stands straight up from the table, and $K'$ is the top of the block.
4. The distance from $K$ to $K'$ is the long diagonal of the cube, $\sqrt{1^2 + 1^2 + 1^2} = \sqrt3$.
5. How far is $K$ from the cut face? Use the volume of the cut-off piece. It is a pyramid with a right-angled corner and three edges of length $1$, so its volume is $\frac13 \times \frac12 \times 1 = \frac16$.
6. Now use the cut face as its base. The base area is $\frac{\sqrt3}{4}(\sqrt2)^2 = \frac{\sqrt3}{2}$. From volume $= \frac13 \times \text{base} \times \text{height}$:
$$\frac16 = \frac13 \cdot \frac{\sqrt3}{2} \cdot h, \qquad h = \frac{1}{\sqrt3}.$$
7. $K$ is on the far side of the cut, so the block's height is the full diagonal minus this piece:
$$\sqrt3 - \frac{1}{\sqrt3} = \frac{3}{\sqrt3} - \frac{1}{\sqrt3} = \frac{2}{\sqrt3} = \frac{2\sqrt3}{3}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{2\sqrt3}{3}}$.

Slicing off a cube's corner through its three neighboring corners gives an equilateral cut that faces straight along the cube's long diagonal. So the height of the leftover block is measured along that diagonal. Finding a pyramid's height from its volume and base area is a handy trick whenever the height is awkward to measure directly.

With coordinates, let the cube's corners have coordinates $0$ or $1$ and let $K = (1,0,0)$. The cut goes through $(0,0,0)$, $(1,1,0)$ and $(1,0,1)$, which all satisfy $x - y - z = 0$. The distance from a point to this plane is $\frac{|x - y - z|}{\sqrt3}$. For the opposite corner $(0,1,1)$ it is $\frac{2}{\sqrt3} = \frac{2\sqrt3}{3}$, and every other remaining corner is only $\frac{1}{\sqrt3}$ away, so $(0,1,1)$ is the top.