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)

$40x + 2xy  = -y^2 + 400$
$2x(20 + y)=  (20 - y)(20 + y)$
AIME 1998-3.png

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.

See also

1998 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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