2009 AMC 10B Problem 17

The figure is a staircase of five unit squares with total area $5$, so each side of the line must have area $\frac52$. Look at the shaded region below the slanted line.

Extend the picture: the triangle with vertices $(c, 0)$, $(3, 0)$, $(3, 3)$ has legs $3 - c$ and $3$, so its area is $\frac{3(3 - c)}{2}$. This triangle contains the whole shaded region plus one extra piece: the unit square with corners $(2, 0)$ and $(3, 1)$, which is not part of the figure. (The line passes above that square, since at $x = 2$ its height is already more than $1$.) Therefore
$$\text{shaded} = \frac{3(3 - c)}{2} - 1 = \frac52 \quad\Longrightarrow\quad 3(3 - c) = 7 \quad\Longrightarrow\quad c = 3 - \frac73 = \frac23.$$

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

Awkward regions bounded by a line and a staircase become easy when you complete them to a triangle and subtract the pieces that do not belong. Verify that the subtracted piece lies entirely on one side of the line; otherwise the subtraction is wrong.

Shoelace on the shaded polygon $(c,0), (3,3), (3,1), (2,1), (2,0)$ gives area $\frac{7 - 3c}{2}$; setting this equal to $\frac52$ yields $c = \frac23$.

• Forgetting that the unit square with corners (2,0) and (3,1) is not part of the figure, and solving 3(3 - c)/2 = 5/2 to get c = 4/3 (not a choice, but a sign of the slip).
• Miscounting the total area (for example using $6$) so that each half should be $3$.