2006 AMC 12B Problems/Problem 24
Contents
Problem
Let be the set of all point in the coordinate plane such that and . What is the area of the subset of for which
Solution
We start out by solving the equality first. We end up with three lines that matter: , , and . We plot these lines below. Note that by testing the point , we can see that we want the area of the pentagon. We can calculate that by calculating the area of the square and then subtracting the area of the 3 triangles. (Note we could also do this by adding the areas of the isosceles triangle in the bottom left corner and the rectangle with the previous triangle's hypotenuse as the longer side.)
Solution 2
We can write the given equation as Note that when , we have which implies . Similary we have when . Then we see what happens at . Clearly at , we see that . By symmetry we have when . So we get a graph like Thus the area of what we are interested in is ~coolmath_2018
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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