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Difference between revisions of "1998 AIME Problems/Problem 3"

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== Problem ==
 
== Problem ==
−
The graph of <math> y^2 + 2xy + 40|x| \displaystyle = 400</math> partitions the plane into several regions.  What is the area of the bounded region?
+
The graph of <math> y^2 + 2xy + 40|x|= 400</math> partitions the plane into several regions.  What is the area of the bounded region?
  
 
== Solution ==
 
== Solution ==
−
:<math>\displaystyle 40|x| = - y^2 - 2xy + 400</math>
+
The equation given can be rewritten as:
 +
 
 +
:<math>40|x| = - y^2 - 2xy + 400</math>
  
 
We can split the equation into a piecewise equation by breaking up the [[absolute value]]:
 
We can split the equation into a piecewise equation by breaking up the [[absolute value]]:
  
−
:<math>40x = -y^2 - 2xy + 400 \displaystyle \quad \displaystyle \quad x\ge 0 </math>
+
:<math>40x = -y^2 - 2xy + 400\quad\quad x\ge 0 </math>
−
:<math>\displaystyle 40x = y^2 + 2xy - 400 \quad \quad x < 0</math>
+
:<math>40x = y^2 + 2xy - 400 \quad \quad x < 0</math>
  
 
Factoring the first one: (alternatively, it is also possible to [[completing the square|complete the square]])
 
Factoring the first one: (alternatively, it is also possible to [[completing the square|complete the square]])
  
−
:<math>\displaystyle 40x + 2xy + = -y^2 + 400</math>
+
:<math>40x + 2xy  = -y^2 + 400</math>
−
:<math> 2x(20 + y)\displaystyle =  (20 - y)(20 + y)</math>
+
:<math> 2x(20 + y)=  (20 - y)(20 + y)</math>
  
 
[[Image:AIME_1998-3.png|right|thumb|300px]]  
 
[[Image:AIME_1998-3.png|right|thumb|300px]]  
  
−
Hence, either <math>\displaystyle y = -20</math>, or <math>\displaystyle 2x = 20 - y \Longrightarrow y = -2x + 20</math>.
+
Hence, either <math>y = -20</math>, or <math>2x = 20 - y \Longrightarrow y = -2x + 20</math>.
 +
 
 +
Similarily, for the second one, we get <math>y = 20</math> or <math> y = -2x - 20</math>. If we graph these four equations, we see that we get a parallelogram with base 20 and height 40. Hence the answer is <math>\boxed{800}</math>.
  
−
Similarily, for the second one, we get <math>\displaystyle y = 20</math> or <math> y = -2x - 20 \displaystyle</math>. If we graph these four equations, we see that we get a parallelogram with base 20 and height 40. Hence the answer is <math>\displaystyle 800</math>.
+
== Solution 2 ==
 +
The equation can be rewritten as: <math>(x+y)^2=(|x|-20)^2</math>. Do casework as above.
  
 
== See also ==
 
== See also ==
Line 26: Line 31:
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:13, 23 February 2018

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 (Problems • Answer Key • Resources)
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

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

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