2013 AMC 12A Problem 13

Problem 13 of 25IntermediateGeometry

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?

Show solution

Solution

By the shoelace formula, the area of

ABCD

\tfrac{15}{2}.

Let the line meet

\overline{CD}

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

y = -3(x - 4),

x_G = \tfrac{27}{8}.

Then

p + q + r + s

= 27 + 8 + 15 + 8

= 58.

Thus, the correct answer is B.