Difference between revisions of "2012 AMC 12B Problems/Problem 10"
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<math>\textbf{(A)}\ 24\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 42</math> | <math>\textbf{(A)}\ 24\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 42</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | The first curve is a circle with radius 5 centered at the origin, and the second curve is an ellipse with center (4,0) and end points of (-5,0) and (13,0). Finding points of intersection, we get (-5,0) (4,3) and (4,-3), forming a triangle with height of 9 and base of 6. So | + | The first curve is a circle with radius <math>5</math> centered at the origin, and the second curve is an ellipse with center <math>(4,0)</math> and end points of <math>(-5,0)</math> and <math>(13,0)</math>. Finding points of intersection, we get <math>(-5,0)</math>, <math>(4,3)</math>, and <math>(4,-3)</math>, forming a triangle with height of <math>9</math> and base of <math>6.</math> So the area of this triangle is <math>9 \cdot 6 \cdot 0.5 =27 \textbf{ (B)}.</math> |
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| + | ==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> | ||
| + | |||
| + | ~dragnin | ||
| + | |||
| + | ~minor edits by [https://artofproblemsolving.com/wiki/index.php/User:Kevinchen_yay KevinChen_Yay] | ||
| + | |||
| + | ==Solution 3== | ||
| + | Using our algebra skills we find the points of intersection to be <math>(-5,0),(4,3)</math>,and <math>(4,-3)</math>. Using the shoelace theorem, we can easily find the area of the triangle to be <math>27=\fbox{B}</math>. | ||
| + | |||
| + | ~PeterDoesPhysics | ||
| + | |||
| + | More information on the shoelace theorem: | ||
| + | https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem | ||
== See Also == | == See Also == | ||
{{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]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 08:56, 7 August 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
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 |
| 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.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
