2008 AMC 10A Problem 21

Give each corner of the cube coordinates, so every length becomes a quick calculation. Let the cube have corners at $0$ and $1$ in each direction.

1. Put $A = (1, 0, 0)$. The opposite corner is $C = (0, 1, 1)$.
2. Two opposite edges that miss both $A$ and $C$ are the upright edge from $(0,0,0)$ to $(0,0,1)$ and the upright edge from $(1,1,0)$ to $(1,1,1)$. Their midpoints are
$$B = (0, 0, \tfrac12), \qquad D = (1, 1, \tfrac12).$$
3. The sides. Use the distance formula (the Pythagorean theorem in three directions):
$$AB = \sqrt{1^2 + 0^2 + \left(\tfrac12\right)^2} = \sqrt{\tfrac54} = \frac{\sqrt5}{2}.$$
In the same way, $BC$, $CD$ and $DA$ each have one difference of $1$, one of $0$ and one of $\frac12$, so they all equal $\frac{\sqrt5}{2}$. Four equal sides make $ABCD$ a rhombus.
4. The diagonals. $AC$ goes from one corner of the cube to the opposite corner:
$$AC = \sqrt{1^2 + 1^2 + 1^2} = \sqrt3.$$
$BD$ goes from $(0,0,\frac12)$ to $(1,1,\frac12)$:
$$BD = \sqrt{1^2 + 1^2 + 0^2} = \sqrt2.$$
5. The area of a rhombus is half the product of its diagonals:
$$\frac{1}{2} \times \sqrt3 \times \sqrt2 = \frac{\sqrt6}{2}.$$

The answer is $\boxed{\textbf{(A)}\ \frac{\sqrt{6}}{2}}$.

Coordinates turn a hard-to-picture 3D shape into a few distance calculations. Once you see the four sides are equal, you know it is a rhombus, and a rhombus is easiest to measure by its diagonals. Here the diagonals are two lengths worth knowing in a unit cube: corner to opposite corner is $\sqrt3$, and across a face is $\sqrt2$.

Cut the rhombus along $AC$ into two equal triangles, each with base $\sqrt3$. The diagonals of a rhombus cross at right angles and cut each other in half, so the height of each triangle is half of $BD$, which is $\frac{\sqrt2}{2}$. Each triangle has area $\frac12 \times \sqrt3 \times \frac{\sqrt2}{2} = \frac{\sqrt6}{4}$, and the two together make $\frac{\sqrt6}{2}$.

Place the cube as $[0,1]^3$ with $A = (1,0,0)$ and the opposite vertex $C = (0,1,1)$. The two opposite edges avoiding $A$ and $C$ can be taken as the vertical edges above $(0,0)$ and $(1,1)$, with midpoints
$$B = (0,0,\tfrac12),\qquad D = (1,1,\tfrac12).$$

Compute the four sides:
$$AB = \sqrt{1 + 0 + \tfrac14} = \tfrac{\sqrt5}{2},\quad BC = \sqrt{0 + 1 + \tfrac14} = \tfrac{\sqrt5}{2},$$
and by the same arithmetic $CD = DA = \frac{\sqrt5}{2}$. So $ABCD$ is a rhombus (it is planar because $B + D = A + C$, i.e. the diagonals share a midpoint).

Its diagonals are the space diagonal $AC = \sqrt3$ and
$$BD = \sqrt{1 + 1 + 0} = \sqrt2.$$

A rhombus has area half the product of its diagonals:
$$[ABCD] = \frac{1}{2}\cdot\sqrt3\cdot\sqrt2 = \frac{\sqrt6}{2}.$$

The answer is $\boxed{\textbf{(A)}\ \frac{\sqrt{6}}{2}}$.