2013 AMC 10A Problem 18

Problem 18 of 25IntermediateAlgebra Geometry

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?

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Solution

Let the cutting line meet

\overline{CD}

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

\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.