1998 AIME Problems/Problem 3

Problem

The graph of $y^2 + 2xy + 40|x| \displaystyle = 400$ partitions the plane into several regions. What is the area of the bounded region?

$\displaystyle 40|x| = - y^2 - 2xy + 400$

We can split the equation into a piecewise equation by breaking up the absolute value:

$40x = -y^2 - 2xy + 400 \displaystyle \quad \displaystyle \quad x\ge 0$

$\displaystyle 40x = y^2 + 2xy - 400 \quad \quad x < 0$

Factoring the first one: (alternatively, it is also possible to complete the square)

$\displaystyle 40x + 2xy + = -y^2 + 400$

$2x(20 + y)\displaystyle = (20 - y)(20 + y)$

Similarily, for the second one, we get $\displaystyle y = 20$ or $y = -2x - 20 \displaystyle$ . If we graph these four equations, we see that we get a parallelogram with base 20 and height 40. Hence the answer is $\displaystyle 800$ .