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1998 AIME Problems/Problem 3: Difference between revisions

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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>
:<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>\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>.
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>800</math>.


== See also ==
== See also ==
Line 26: Line 26:
[[Category:Intermediate Algebra Problems]]
[[Category:Intermediate Algebra Problems]]
[[Category:Intermediate Geometry Problems]]
[[Category:Intermediate Geometry Problems]]
{{MAA Notice}}

Revision as of 18:37, 4 July 2013

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

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

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 $800$.

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

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