1998 AIME Problem 3

Question

The graph of

y^2 + 2xy + 40|x| = 400

partitions the plane into several regions. What is the area of the bounded region?

Answer

For

x \ge 0

the equation rearranges to

2x(y + 20) = 400 - y^2,

and the right side factors

Each sign case yields two lines; the four rays bound a parallelogram with horizontal sides on

y = 20

and

y = -20

For

x \ge 0

rewrite the equation as

2x(y + 20)

= 400 - y^2

= (20 - y)(20 + y),

so either

y = -20

or

y = 20 - 2x.

For

x \le 0

it becomes

2x(y - 20) = -(y - 20)(y + 20),

so either

y = 20

or

y = -20 - 2x.

The graph therefore consists of two horizontal rays and two rays of slope

-2.

These rays bound a parallelogram: the top edge runs from

(-20, 20)

to

(0, 20)

along

y = 20,

the bottom edge from

(0, -20)

to

(20, -20)

along

y = -20,

and the two slanted edges of slope

-2

connect them.

The parallelogram has horizontal base

20

and height

40

between the lines

y = 20

and

y = -20,

so its area is

20 \cdot 40 = 800.

Problem 3 in other years