Question

Let points $A = (0, 0), ~B = (1, 2),$ $C=(3, 3),~ D = (4, 0).$ 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:

Let the cutting line meet $\overline{CD}$ at $G$ . Drop perpendiculars from $B$ , $C$ , and $G$ to the $x$ -axis as in the diagram.

The areas of $\triangle ABF$ , trapezoid $BCEF$ , and $\triangle CDE$ are $1$ , $5$ , and $\frac32$ , respectively, so $[ABCD]=\frac{15}{2}$ .

Thus $\triangle ADG$ has area $\frac{15}{4}$ . Since $AD=4$ , the height of $G$ is $\frac{15}{8}$ .

The line $CD$ has equation $y=-3x+12$ , so $\frac{15}{8}=-3x+12$ , giving $x=\frac{27}{8}$ . Therefore $p+q+r+s=27+8+15+8=58$ .

Thus, B is the correct answer.

2013 AMC 10A Problem 18

Solution:

Problem 18 in Other Years