from: category_eng

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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 $p+q+r+s$ ?

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

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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 $E = (frac{27}{8}, frac{15}{8})$ . $27 + 15 + 8 + 8 = oxed{ extbf{(B) }58}$