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 $E = (\frac{27}{8}, \frac{15}{8})$ . $27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}$

2013 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 12 Problems and Solutions

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2013 AMC 10A Problems/Problem 18

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