1998 AIME Problems/Problem 3: Difference between revisions

Revision as of 14:07, 21 August 2017

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 $\boxed{800}$ .

@@ Line 3: / Line 3: @@
 == Solution ==
+The equation given can be rewritten as:
 :<math>40|x| = - y^2 - 2xy + 400</math>

1998 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1998 AIME Problems/Problem 3: Difference between revisions

Revision as of 14:07, 21 August 2017

Problem

Solution

See also