2012 AMC 10A Problem 15

Place $A = (0, 0)$ and $B = (1, 0)$, with the squares below the $x$-axis: two side by side occupying $0 \le x \le 2$, $-1 \le y \le 0$, and one more below the left square occupying $0 \le x \le 1$, $-2 \le y \le -1$.

One segment runs from $A$ to the far corner $(2, -1)$; its line is $y = -\tfrac{1}{2}x$. The other runs from $B$ to the bottom corner $(0, -2)$; its line has slope $\frac{0 - (-2)}{1 - 0} = 2$, so it is $y = 2x - 2$.

Point $C$ is where they meet:
$$-\tfrac12 x = 2x - 2 \quad\Longrightarrow\quad 2 = \tfrac52 x \quad\Longrightarrow\quad x = \tfrac45, \qquad y = -\tfrac25.$$

Triangle $ABC$ has base $AB = 1$ and height equal to the distance from $C$ to line $AB$, which is $\tfrac25$. Its area is
$$\frac12 \cdot 1 \cdot \frac25 = \frac15.$$

The answer is $\boxed{\textbf{(B)}\ \frac15}$.

A figure built from unit squares is already a coordinate grid, so lines through grid vertices have simple equations and their intersection is a two-line solve. With a horizontal base, the area only needs the vertical distance of the third vertex. Read the figure carefully to get the right endpoints before writing any equations.

Without coordinates: let $C$ be $h$ below $AB$ and $x$ to the right of $A$. Along $AC$, the drop is half the run, so $h = \frac{x}{2}$; along $BC$, the drop is twice the run, so $h = 2(1 - x)$. Then $\frac{x}{2} = 2 - 2x$ gives $x = \frac45$, $h = \frac25$, area $\frac15$.

• Misreading which vertices the segments join (e.g. assuming the segments are diagonals of single squares) and getting a height of 1/2 or 1/3.
• Using the slant lengths $AC$ or $BC$ as a height, which produces an answer involving $\sqrt{5}$.