2005 AIME I Problem 10

Question

Triangle

ABC

lies in the Cartesian plane and has area

70.

The coordinates of

B

and

C

are

(12, 19)

and

(23, 20),

respectively, and the coordinates of

A

are

(p, q).

The line containing the median to side

\overline{BC}

has slope

-5.

Find the largest possible value of

p + q.

Answer

Answer: 47

Concepts:coordinate geometry shoelace formula median (geometry)

Difficulty rating: 2560

Small hint:

The median to

\overline{BC}

passes through its midpoint

\left(\frac{35}{2}, \frac{39}{2}\right),

so

A

lies on the line

y = -5x + 107

Big hint:

Write

A = (p, -5p + 107)

and set the shoelace area equal to

70;

then

p + q = 107 - 4p

is larger for the smaller

p

Solution:

The median to

\overline{BC}

passes through the midpoint

M = \left(\frac{35}{2}, \frac{39}{2}\right)

of

\overline{BC}.

The line through

M

with slope

-5

is

y = -5x + 107,

and

A

lies on this line, so

A = (p, -5p + 107)

and

q = -5p + 107.

By the shoelace formula with

B = (12, 19)

and

C = (23, 20),

\begin{aligned} [ABC] &= \small \frac{1}{2}\left|{-p} + 12\bigl(20 - q\bigr) + 23\bigl(q - 19\bigr)\right| \\ &= \frac{1}{2}\left|980 - 56p\right| = 70, \end{aligned}

so

|56p - 980| = 140,

giving

p = 15

or

p = 20.

Since

\begin{aligned} p + q &= p + (-5p + 107) \\ &= 107 - 4p, \end{aligned}

the smaller value

p = 15

gives the larger sum

107 - 60 = 47.