Difference between revisions of "1998 AIME Problems/Problem 3"
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== Problem == | == Problem == | ||
| + | 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 == | ||
| + | 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]]: | ||
| + | |||
| + | :<math>40x = -y^2 - 2xy + 400\quad\quad x\ge 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]]) | ||
| + | |||
| + | :<math>40x + 2xy = -y^2 + 400</math> | ||
| + | :<math> 2x(20 + y)= (20 - y)(20 + y)</math> | ||
| + | |||
| + | [[Image:AIME_1998-3.png|right|thumb|300px]] | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | == Solution 2 == | ||
| + | The equation can be rewritten as: <math>(x+y)^2=(|x|-20)^2</math>. Do casework as above. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1998|num-b=2|num-a=4}} | |
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
| + | [[Category:Intermediate Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:13, 23 February 2018
Contents
Problem
The graph of 
partitions the plane into several regions. What is the area of the bounded region?
Solution
The equation given can be rewritten as:
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 
.
Solution 2
The equation can be rewritten as: 
. 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. 
