2012 AMC 10B Problem 23

Use the cube with vertices at all $(x, y, z)$ with coordinates $0$ or $1$. On the bottom face take the nonadjacent vertices $(0,0,0)$ and $(1,1,0)$; on the top face, $(1,0,1)$ is adjacent to neither. These three points are exactly the three neighbors of the corner $(1,0,0)$, so the discarded tetrahedron is that corner.

The plane through the three points is $x - y - z = 0$ (each point satisfies it). Placed cut-face down, the object's height is the largest distance from this plane to any remaining vertex. The distance from $(x,y,z)$ to the plane is $\frac{|x - y - z|}{\sqrt{1 + 1 + 1}}$; the farthest vertex is $(0, 1, 1)$, the corner opposite the removed one:
$$\frac{|0 - 1 - 1|}{\sqrt3} = \frac{2}{\sqrt3} = \frac{2\sqrt3}{3}.$$

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

Cutting off a corner of a cube through its three neighboring vertices creates an equilateral triangular face that is perpendicular to the space diagonal from that corner. "Height when resting on a face" means the greatest distance to the plane of that face, and the vertex farthest along the perpendicular direction is the opposite corner of the cube. Coordinates make the distance computation mechanical.

Without coordinates: the removed tetrahedron has volume $\frac16$ and its cut face is an equilateral triangle of side $\sqrt2$, area $\frac{\sqrt3}{2}$. Its height above the cut face is $\frac{3V}{\text{area}} = \frac{1/2}{\sqrt3/2} = \frac{1}{\sqrt3}$. The removed corner and the opposite corner are $\sqrt3$ apart along a line perpendicular to the cut face, so the object's height is $\sqrt3 - \frac{1}{\sqrt3} = \frac{2}{\sqrt3}$.

• Answering 1 by imagining the object still resting on an original face, or taking the whole space diagonal √3 as the height without removing the discarded corner.
• Choosing $\frac{\sqrt3}{3}$, which is the height of the discarded tetrahedron, not of the remaining solid.