1998 AIME Problems/Problem 3

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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?

Solution

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

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)

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

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

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

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 $\displaystyle 800$ .

1998 AIME (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1998 AIME Problems/Problem 3

From AoPSWiki

Problem

Solution

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