2006 AMC 10A Problem 17

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

The numbers $2$ and $3$ with trisection points were chosen so that every segment is a $45^\circ$ line, which is why coordinates are so quick here: the intersections are half-integers. Whenever a figure is built from a rectangle with equally spaced points, coordinates turn "find the area of the inner region" into a few one-line equations, and the diagonal formula $\tfrac12 d_1 d_2$ finishes.

Synthetic: triangle $ACH$ has legs $AH = AC = 2$, so $\angle ACH = 45^\circ$, and in the same way every drawn segment makes $45^\circ$ with the sides of the rectangle. The parallel lines $CH$ and $BE$ are $1$ unit apart measured horizontally, so their perpendicular distance, which is the side of the square, is $\tfrac{1}{\sqrt 2}$. Area $\left(\tfrac{1}{\sqrt2}\right)^2 = \tfrac12$.

• Assuming the square has side $1$ (the spacing of the trisection points) and answering $1$, or confusing the diagonal length $1$ with the side length.
• Guessing the shape is a rhombus that is not a square and trying to compute angles; the perpendicular slopes settle it immediately.