Difference between revisions of "1998 AIME Problems/Problem 3"
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Hence, either <math>y = -20</math>, or <math>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>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>. |
== See also == | == See also == | ||
Revision as of 19:37, 16 September 2016
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
