Difference between revisions of "2013 AMC 10A Problems/Problem 18"

Problem

Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $(\frac{p}{q}, \frac{r}{s})$, where these fractions are in lowest terms. What is $p+q+r+s$?

$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$

Solution

First, various area formulas (shoelace, splitting, etc) allow us to find that $[ABCD] = \frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\frac{15}{4}$.

Call the point where the line through $A$ intersects $\overline{CD}$ $E$. We know that $[ADE] = \frac{15}{4} = \frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \frac{15}{4}$, so $h = \frac{15}{8}$. This gives that the y coordinate of E is $\frac{15}{8}$.

Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \frac{27}{8}$.

From this, we know that $E = (\frac{27}{8}, \frac{15}{8})$. $27 + 15 + 8 + 8 = 58$, $\textbf{(A)}$.