2012 AMC 10A Problem 21

Compute the midpoints:
$$E = \left(\tfrac12, 0, \tfrac32\right), \quad F = \left(\tfrac12, 0, 0\right), \quad G = (0, 1, 0), \quad H = \left(0, 1, \tfrac32\right).$$

Look at the sides. $\overrightarrow{FE} = (0, 0, \tfrac32)$ and $\overrightarrow{GH} = (0, 0, \tfrac32)$ are equal, so $EFGH$ is a parallelogram with $EF = GH = \tfrac32$. Its other side is $\overrightarrow{FG} = \left(-\tfrac12, 1, 0\right)$, of length $\sqrt{\tfrac14 + 1} = \tfrac{\sqrt5}{2}$.

Since $\overrightarrow{FE} \cdot \overrightarrow{FG} = 0$, adjacent sides are perpendicular and the parallelogram is a rectangle. Its area is
$$\frac32 \cdot \frac{\sqrt5}{2} = \frac{3\sqrt5}{4}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{3\sqrt{5}}{4}}$.

This is the midpoint (Varignon) parallelogram of the skew quadrilateral $B, A, C, D$: $EF$ and $GH$ are midlines of triangles $ABD$ and $ACD$, so both are parallel to $AD$ and half its length; $FG$ and $EH$ are midlines parallel to $BC$. Here $AD$ lies along the $z$-axis and $BC$ lies in the $xy$-plane, so they are perpendicular and the parallelogram is a rectangle with sides $\frac{AD}{2} = \frac32$ and $\frac{BC}{2} = \frac{\sqrt5}{2}$. Midpoints of a quadrilateral's sides always give a parallelogram; check whether the diagonals of the original figure are perpendicular or equal to upgrade it.

• Computing the four midpoints correctly but then assuming EFGH is a square or using a diagonal as a side, or mislabeling H as the midpoint of BC.
• Using the full lengths $AD = 3$ and $BC = \sqrt5$ instead of their halves, giving $3\sqrt5$.