1987 AIME Problems/Problem 4
Contents
Problem
Find the area of the region enclosed by the graph of
Solution 1
Since is nonnegative, . Solving this gives us two equations: . Thus, . The maximum and minimum y value is when , which is when and . Since the graph is symmetric about the y-axis, we just need casework upon . , so we break up the condition :
- . Then .
- . Then .
The area of the region enclosed by the graph is that of the quadrilateral defined by the points . Breaking it up into triangles and solving or using the Shoelace Theorem, we get .
Solution 2
Since is the only present "term" in this equation, we know that the area must be symmetrical about the x-axis.
We'll consider the area when and we only consider the portion enclosed with . Then, we'll double that area since the graph is symmetrical.
Now, let us remove the absolute values:
When or : . This rearranges to .
When : . So .
When : . So .
By simple sketching, we see the shape that looks like the image in Solution 1 (graph it out and you'll see). We see that the partial area we seek in this part is the triangle with the vertices , , . This triangle has an area of .
Simply double the area and we get as our final answer. ~hastapasta
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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