Difference between revisions of "2012 AMC 12B Problems/Problem 10"
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==Solution 2== | ==Solution 2== | ||
Given the equations <math>x^2 + y^2 =25</math> and <math>(x-4)^2 + 9y^2 = 81</math>, we can substitute <math>y^2=25-x^2</math> from the first equation and plug it in to the 2nd equation, giving us <math>(x-4)^2+9(25-x^2)=81</math>. After rearranging, <math>8x^2+8x-160=0</math> or <math>x^2+x-20=0</math>. The solutions are <math>x=-5</math> and <math>x=4</math>. This gives us the points <math>(-5,0),(4,3)</math>,and <math>(4,-3)</math>. The area of the triangle formed by these points is <math>27=\fbox{B}</math> | Given the equations <math>x^2 + y^2 =25</math> and <math>(x-4)^2 + 9y^2 = 81</math>, we can substitute <math>y^2=25-x^2</math> from the first equation and plug it in to the 2nd equation, giving us <math>(x-4)^2+9(25-x^2)=81</math>. After rearranging, <math>8x^2+8x-160=0</math> or <math>x^2+x-20=0</math>. The solutions are <math>x=-5</math> and <math>x=4</math>. This gives us the points <math>(-5,0),(4,3)</math>,and <math>(4,-3)</math>. The area of the triangle formed by these points is <math>27=\fbox{B}</math> | ||
| − | ~dragnin | + | ~dragnin ~minor edits by KevinChen_Yay |
== See Also == | == See Also == | ||
Revision as of 21:58, 12 January 2024
Contents
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
