2012 AMC 12B Problems/Problem 10
Problem
What is the area of the polygon whose vertices are the points of intersection of the curves and
Solution 1
The first curve is a circle with radius centered at the origin, and the second curve is an ellipse with center and end points of and . Finding points of intersection, we get , , and , forming a triangle with height of and base of So the area of this triangle is
Solution 2
Given the equations and , we can substitute from the first equation and plug it in to the 2nd equation, giving us . After rearranging, or . The solutions are and . This gives us the points ,and . The area of the triangle formed by these points is
~dragnin
~minor edits by KevinChen_Yay
Solution 3
Using our algebra skills we find the points of intersection to be ,and . Using the shoelace theorem, we can easily find the area of the triangle to be .
~PeterDoesPhysics
More information on the shoelace theorem: https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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All AMC 12 Problems and Solutions |
extremely misplaced problem. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.