Question

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 $\left(\dfrac{p}{q}, \dfrac{r}{s}\right),$ where these fractions are in lowest terms. What is $p + q + r + s?$

$54$

$58$

$62$

$70$

$75$

Solution:

By the shoelace formula, the area of $ABCD$ is $\tfrac{15}{2}.$ Let the line meet $\overline{CD}$ at $G.$ Triangle $ADG$ must have area $\tfrac{15}{4}.$

Since $AD = 4$ lies on the $x$ -axis, $\tfrac12\cdot 4\cdot y_G = \tfrac{15}{4}$ gives $y_G = \tfrac{15}{8}.$ Line $CD$ is $y = -3(x - 4),$ so $x_G = \tfrac{27}{8}.$

Then $p + q + r + s = 27 + 8 + 15 + 8 = 58.$

Thus, the correct answer is B.

2013 AMC 12A Problem 13

Solution:

Problem 13 in Other Years