Difference between revisions of "1998 AIME Problems/Problem 3"
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== Solution == | == Solution == | ||
+ | The equation given can be rewritten as: | ||
+ | |||
:<math>40|x| = - y^2 - 2xy + 400</math> | :<math>40|x| = - y^2 - 2xy + 400</math> | ||
Line 20: | Line 22: | ||
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>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 == |
Latest revision as of 20:13, 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.