# 1998 AIME Problems/Problem 3

## Problem

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

## Solution

The equation given can be rewritten as: $40|x| = - y^2 - 2xy + 400$

We can split the equation into a piecewise equation by breaking up the absolute value: $40x = -y^2 - 2xy + 400\quad\quad x\ge 0$ $40x = y^2 + 2xy - 400 \quad \quad x < 0$

Factoring the first one: (alternatively, it is also possible to complete the square) $40x + 2xy = -y^2 + 400$ $2x(20 + y)= (20 - y)(20 + y)$

Hence, either $y = -20$, or $2x = 20 - y \Longrightarrow y = -2x + 20$.

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

## Solution 2

The equation can be rewritten as: $(x+y)^2=(|x|-20)^2$. Do casework as above.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 