2012 AMC 12B Problems/Problem 10: Difference between revisions
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{{AMC12 box|year=2012|ab=B|num-b=9|num-a=11}} | {{AMC12 box|year=2012|ab=B|num-b=9|num-a=11}} | ||
extremely misplaced problem. | |||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
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Revision as of 20:10, 1 April 2024
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
See Also
| 2012 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 | |
extremely misplaced problem. These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
