Difference between revisions of "1998 AIME Problems/Problem 3"
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== Problem == | == Problem == | ||
− | The graph of <math> y^2 + 2xy + 40|x| | + | 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> | + | :<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 | + | :<math>40x = -y^2 - 2xy + 400\quad\quad x\ge 0 </math> |
− | :<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> | + | :<math>40x + 2xy + = -y^2 + 400</math> |
− | :<math> 2x(20 + y) | + | :<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> | + | Hence, either <math>y = -20</math>, or <math>2x = 20 - y \Longrightarrow y = -2x + 20</math>. |
− | Similarily, for the second one, we get <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 == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 19:37, 4 July 2013
Problem
The graph of partitions the plane into several regions. What is the area of the bounded region?
Solution
We can split the equation into a piecewise equation by breaking up the absolute value:
Factoring the first one: (alternatively, it is also possible to complete the square)
Hence, either , or
.
Similarily, for the second one, we get or
. If we graph these four equations, we see that we get a parallelogram with base 20 and height 40. Hence the answer is
.
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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