2013 AMC 10A Problem 18

First find the area of $ABCD$ by the shoelace formula on $(0,0), (1,2), (3,3), (4,0)$:
$$\frac12\left|0\cdot2 - 1\cdot0 + 1\cdot3 - 3\cdot2 + 3\cdot0 - 4\cdot3 + 4\cdot0 - 0\cdot0\right| = \frac12\left|0 - 3 - 12\right| = \frac{15}{2}.$$
Each piece must have area $\frac{15}{4}$.

Let the cutting line meet $\overline{CD}$ at $P$. The piece on the $D$ side is triangle $APD$, whose base $AD$ lies on the $x$-axis with length $4$. Its area is $\frac12 \cdot 4 \cdot y_P = 2y_P$, so $2y_P = \frac{15}{4}$ and $y_P = \frac{15}{8}$.

Now use that $P$ is on line $CD$. From $C = (3,3)$ to $D = (4,0)$ the slope is $-3$, so the line is $y = -3(x - 4)$, i.e. $x = 4 - \frac{y}{3}$. With $y = \frac{15}{8}$:
$$x = 4 - \frac{15}{24} = 4 - \frac{5}{8} = \frac{27}{8}.$$

So $P = \left(\frac{27}{8}, \frac{15}{8}\right)$, both fractions in lowest terms, and $p + q + r + s = 27 + 8 + 15 + 8 = 58$.

The answer is $\boxed{\textbf{(B)}\ 58}$.

A line through a vertex that bisects the area cuts off a triangle on one side; if that triangle has a base on an axis, its area is controlled by a single coordinate of the unknown point. One area equation plus the line equation of the side it lands on pins the point down. Choosing the piece with the convenient base ($AD$ on the $x$-axis) is what keeps the algebra short.

• Assuming the bisecting line passes through C or the midpoint of CD, instead of using area to locate P.
• Working with the other piece (quadrilateral $ABCP$), which needs a second shoelace computation with an unknown point and invites errors.