2009 AMC 10B Problem 17

The five squares have total area $5$, so the shaded part below the line must have area $\frac52$. The shaded part is an awkward shape, so build it from a simpler one.

1. Look at the big right triangle with corners $(c, 0)$, $(3, 0)$ and $(3, 3)$. Its bottom side runs from $c$ to $3$, so it is $3 - c$ long. Its right side runs from $0$ up to $3$, so it is $3$ tall.
2. Its area is $\frac12 \times (3 - c) \times 3 = \frac{3(3 - c)}{2}$.
3. Compare this triangle with the shaded region. The triangle covers the whole shaded region, plus one extra piece: the unit square with corners $(2, 0)$ and $(3, 1)$, which is empty in the figure.
4. That whole square really is inside the triangle: the line starts at $x = c$ (less than $1$) and rises to height $3$ at $x = 3$, so by $x = 2$ it is already above height $1$.
5. So
$$\text{shaded area} = \frac{3(3 - c)}{2} - 1.$$
6. Set this equal to $\frac52$:
$$\frac{3(3 - c)}{2} - 1 = \frac52 \quad\Longrightarrow\quad \frac{3(3 - c)}{2} = \frac72 \quad\Longrightarrow\quad 3(3 - c) = 7.$$
7. So $3 - c = \frac73$ and $c = 3 - \frac73 = \frac23$.

The answer is $\boxed{\textbf{(C)}\ \frac23}$.

A region cut off by a slanted line and a staircase is hard to measure directly. Fill it out to a simple triangle, then take away the pieces that do not belong. Just make sure each piece you take away lies completely inside the triangle.

Test a choice. With $c = \frac23$, the line goes from $\left(\frac23, 0\right)$ to $(3, 3)$. The big triangle has area $\frac12 \times \frac73 \times 3 = \frac72$. Taking away the empty square leaves $\frac72 - 1 = \frac52$, exactly half of $5$.