2012 AMC 10A Problem 15

Side $AB$ is the top edge of one unit square, so it has length $1$. The area is $\frac12 \times 1 \times (\text{height})$, where the height is how far $C$ is below line $AB$. So we only need that height.

1. Put the grid on coordinates. Let $A = (0, 0)$ and $B = (1, 0)$. The two top squares fill $0 \le x \le 2$ and $-1 \le y \le 0$, and the third square sits under the left one, filling $0 \le x \le 1$ and $-2 \le y \le -1$.
2. The first segment goes from $A = (0, 0)$ to the right-hand corner $(2, -1)$. It drops $1$ while moving $2$ to the right, so at any point on it, the drop is half the distance across. Its equation is $y = -\frac12 x$.
3. The second segment goes from $B = (1, 0)$ down to the bottom-left corner $(0, -2)$. It drops $2$ while moving $1$ to the left, so its slope is $2$. It passes through $(1, 0)$, so its equation is $y = 2x - 2$.
4. $C$ is where the two lines meet, so set the two $y$-values equal:
$$-\tfrac12 x = 2x - 2 \quad\Longrightarrow\quad 2 = \tfrac52 x \quad\Longrightarrow\quad x = \tfrac45.$$
Then $y = -\frac12 \cdot \frac45 = -\frac25$.
5. So $C$ is $\frac25$ below line $AB$, and the area is
$$\frac12 \times 1 \times \frac25 = \frac15.$$

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

A picture made of unit squares is already a coordinate grid, so every corner has whole-number coordinates and every segment between corners has an easy equation. Where two lines cross, set their equations equal. With a flat base of known length, the area only needs the vertical distance of the third corner.

Skip the equations. Say $C$ is $h$ below $AB$ and $x$ to the right of $A$. Along segment $AC$ the drop is half the distance across, so $h = \frac{x}{2}$. Along segment $BC$, the point $C$ is $1 - x$ to the left of $B$, and the drop is twice that, so $h = 2(1 - x)$. Setting them equal, $\frac{x}{2} = 2 - 2x$, gives $x = \frac45$ and $h = \frac25$. The area is $\frac12 \cdot 1 \cdot \frac25 = \frac15$.