2006 AMC 10A Problem 17

The figure is built on a small rectangle with evenly spaced points, so coordinates make every point easy to find.

1. Place the rectangle. Put $H$ at $(0, 0)$, $E$ at $(3, 0)$, $D$ at $(3, 2)$ and $A$ at $(0, 2)$. The points on the top edge are $B = (1, 2)$ and $C = (2, 2)$; on the bottom edge, $G = (1, 0)$ and $F = (2, 0)$.
2. The four lines. Each drawn segment goes $2$ across while going $2$ up or down, so its slope is $1$ or $-1$:
- $AF$: from $(0,2)$ to $(2,0)$, so $y = 2 - x$.
- $BE$: from $(1,2)$ to $(3,0)$, so $y = 3 - x$.
- $CH$: from $(0,0)$ to $(2,2)$, so $y = x$.
- $DG$: from $(1,0)$ to $(3,2)$, so $y = x - 1$.
3. The four corners. Set the equations equal in pairs:
- $AF$ and $CH$: $x = 2 - x$, so $x = 1$. Point $Z = (1, 1)$.
- $BE$ and $CH$: $x = 3 - x$, so $x = 1.5$. Point $W = (1.5, 1.5)$.
- $BE$ and $DG$: $x - 1 = 3 - x$, so $x = 2$. Point $X = (2, 1)$.
- $AF$ and $DG$: $x - 1 = 2 - x$, so $x = 1.5$. Point $Y = (1.5, 0.5)$.
4. The shape. Lines with slopes $1$ and $-1$ meet at right angles, so all four corners are right angles. The diagonal $ZX$ goes straight across from $(1,1)$ to $(2,1)$: length $1$. The diagonal $WY$ goes straight down from $(1.5, 1.5)$ to $(1.5, 0.5)$: length $1$. Equal diagonals crossing at right angles in the middle make a square.
5. Area. A square (like any rhombus) has area half the product of its diagonals: $\frac12 \times 1 \times 1 = \frac12$.

The answer is $\boxed{\textbf{(A)}\ \frac{1}{2}}$.

When a figure is a rectangle with equally spaced points, coordinates turn "find the inner shape" into a few short equations. Here the numbers were chosen so every line is at $45^\circ$, which makes the corners land on halves. For a square tilted like a diamond, the diagonals are often easier to measure than the sides.

The square's side is the distance between the parallel lines $y = x$ and $y = x - 1$. They are $1$ apart going straight across, and at $45^\circ$ the straight-across gap is the diagonal of a small square whose side is the true distance. So the side is $\frac{1}{\sqrt2}$ and the area is $\left(\frac{1}{\sqrt2}\right)^2 = \frac12$.

Place $H$ at the origin with $E = (3, 0)$, $D = (3, 2)$, $A = (0, 2)$. The trisection points are $B = (1, 2)$, $C = (2, 2)$ on top and $G = (1, 0)$, $F = (2, 0)$ on the bottom.

Each drawn segment rises or falls $2$ over a horizontal run of $2$, so every one has slope $\pm 1$:
$$AF:\ y = 2 - x,\qquad BE:\ y = 3 - x,\qquad CH:\ y = x,\qquad DG:\ y = x - 1.$$
Intersect pairs to locate the four vertices:
- $Z = AF \cap CH$: $x = 2 - x$, so $Z = (1, 1)$.
- $W = BE \cap CH$: $x = 3 - x$, so $W = \left(\tfrac32, \tfrac32\right)$.
- $X = BE \cap DG$: $x - 1 = 3 - x$, so $X = (2, 1)$.
- $Y = AF \cap DG$: $x - 1 = 2 - x$, so $Y = \left(\tfrac32, \tfrac12\right)$.

Lines of slope $1$ and $-1$ are perpendicular, so $WXYZ$ is a rectangle, and by the left-right symmetry of the figure it is a square. Its diagonals are $ZX$, horizontal from $(1,1)$ to $(2,1)$ with length $1$, and $WY$, vertical from $\left(\tfrac32,\tfrac32\right)$ to $\left(\tfrac32,\tfrac12\right)$ with length $1$. A square with diagonal $1$ has area $\tfrac12 \cdot 1 \cdot 1 = \tfrac12$.

The answer is $\boxed{\textbf{(A)}\ \frac{1}{2}}$.