2014 AMC 10A Problem 16

Every point in the picture is a corner or a midpoint, so coordinates will be small and easy. Put $D$ at the origin.

1. Coordinates. $D = (0, 0)$, $C = (1, 0)$, $B = (1, 2)$, $A = (0, 2)$. The midpoints are $F = \left(\tfrac12, 0\right)$, $G = (0, 1)$, $E = (1, 1)$, and $H$, the middle of $GE$, is $\left(\tfrac12, 1\right)$.
2. The shape. The shaded region has top corner $H$, bottom corner $F$, and two side corners. The left side corner is where line $DH$ crosses line $AF$. Call it $X$.
3. Line $DH$. From $(0,0)$ to $\left(\tfrac12, 1\right)$ the line rises $1$ while moving $\tfrac12$ across, so its slope is $2$: $y = 2x$.
4. Line $AF$. From $(0, 2)$ down to $\left(\tfrac12, 0\right)$ it falls $2$ while moving $\tfrac12$ across, so its slope is $-4$: $y = 2 - 4x$.
5. Where they cross. Set $2x = 2 - 4x$. Then $6x = 2$, so $x = \tfrac13$ and $y = \tfrac23$. So $X = \left(\tfrac13, \tfrac23\right)$.
6. The other side corner. The picture is the same on the left and right of the line $x = \tfrac12$, so the right side corner is $Y = \left(\tfrac23, \tfrac23\right)$.
7. Area. The line from $H$ to $F$ is straight up and down, with length $1$. The line from $X$ to $Y$ is flat, with length $\tfrac23 - \tfrac13 = \tfrac13$. These two diagonals are perpendicular, and a four-sided shape with perpendicular diagonals has area half their product:
$$\frac12 \cdot 1 \cdot \frac13 = \frac16.$$

The answer is $\boxed{\textbf{(E)}\ \frac16}$.

Figures full of midpoints in a rectangle are perfect for coordinates: each point has simple fractions as coordinates and each line has an easy equation. Using the left-right symmetry means you only find one crossing point. Seeing that the shape is a kite (perpendicular diagonals) makes the area one multiplication.

Subtract triangles instead. Triangle $DHC$ has base $DC = 1$ and height $1$, so its area is $\frac12$. The shaded kite is triangle $DHC$ with two triangles removed: $DXF$ on the left and $CYF$ on the right. Triangle $DXF$ has base $DF = \frac12$ and height $\frac23$ (the height of $X$), so its area is $\frac12 \cdot \frac12 \cdot \frac23 = \frac16$. Triangle $CYF$ is its mirror image, also $\frac16$. So the shaded area is $\frac12 - \frac16 - \frac16 = \frac16$.

Place $D = (0,0)$, $C = (1,0)$, $B = (1,2)$, $A = (0,2)$. Then $F = \left(\tfrac12, 0\right)$, $G = (0,1)$, $E = (1,1)$ and $H = \left(\tfrac12, 1\right)$.

The shaded region is bounded by segments of $\overline{AF}$, $\overline{DH}$, $\overline{CH}$, $\overline{BF}$, with vertices $H$ (top), $F$ (bottom), and the two side corners $X = AF \cap DH$ and $Y = BF \cap CH$.

Line $AF$ through $(0,2)$ and $\left(\tfrac12,0\right)$: $y = 2 - 4x$. Line $DH$ through $(0,0)$ and $\left(\tfrac12,1\right)$: $y = 2x$. Solving $2x = 2 - 4x$ gives $x = \tfrac13$, $y = \tfrac23$, so $X = \left(\tfrac13, \tfrac23\right)$. By the left-right symmetry of the figure, $Y = \left(\tfrac23, \tfrac23\right)$.

The quadrilateral $HXFY$ has perpendicular diagonals: $\overline{HF}$ is vertical with length $1$, and $\overline{XY}$ is horizontal with length $\tfrac13$. Its area is half their product:
$$\frac12 \cdot 1 \cdot \frac13 = \frac16.$$

The answer is $\boxed{\textbf{(E)}\ \frac16}$.