2008 AMC 10A Problem 21

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

The cube's symmetry (a half-turn about the axis through $B$ and $D$ swaps $A$ and $C$) forces the cross-section to be a rhombus, and a rhombus is easiest to measure by its diagonals, which here are two "famous" cube lengths. Coordinates make the distance computations mechanical and confirm the planarity. Slicing a cube through opposite vertices and edge midpoints is a classic configuration worth remembering.

Split the rhombus along $AC$ into two congruent triangles with base $\sqrt3$. The distance from $B$ to line $AC$ is half of $BD$, namely $\frac{\sqrt2}{2}$, so each triangle has area $\frac12\cdot\sqrt3\cdot\frac{\sqrt2}{2} = \frac{\sqrt6}{4}$, and the total is $\frac{\sqrt6}{2}$.

• Treating the cross-section as a square or rectangle and multiplying two sides, rather than using the diagonals of a rhombus.
• Taking $BD = 1$ (the edge length) instead of $\sqrt2$; $B$ and $D$ sit on opposite edges, so they differ in two coordinates.