Difference between revisions of "2021 AMC 12A Problems/Problem 21"
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Finally, recall that the lengths <math>a</math>, <math>b</math>, and <math>c</math> (where <math>c</math> is the distance between the foci of <math>\mathcal E</math>) satisfy <math>c = \sqrt{a^2 - b^2}</math>. Thus the eccentricity of <math>\mathcal E</math> is <math>\tfrac ca = \sqrt{1 - (\tfrac ba)^2} = \sqrt{\tfrac 16}</math> and the requested answer is <math>\boxed{7\textbf{ (A)}}</math>. | Finally, recall that the lengths <math>a</math>, <math>b</math>, and <math>c</math> (where <math>c</math> is the distance between the foci of <math>\mathcal E</math>) satisfy <math>c = \sqrt{a^2 - b^2}</math>. Thus the eccentricity of <math>\mathcal E</math> is <math>\tfrac ca = \sqrt{1 - (\tfrac ba)^2} = \sqrt{\tfrac 16}</math> and the requested answer is <math>\boxed{7\textbf{ (A)}}</math>. | ||
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+ | == Video Solution by OmegaLearn(Using Ellipse properties and Quadratic Formula ) == | ||
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+ | https://youtu.be/eIYFQSeIRzM | ||
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+ | ~ pi_is_3.14 | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=20|num-a=22}} | {{AMC12 box|year=2021|ab=A|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:32, 11 February 2021
Contents
Problem
The five solutions to the equation may be written in the form for , where and are real. Let be the unique ellipse that passes through the points and . The excentricity of can be written in the form where and are positive integers and is not divisible by the square of any prime. What is ?
Solution
The solutions to this equation are , , and . Consider the five points , , and ; these are the five points which lie on . Note that since these five points are symmetric about the -axis, so must .
Now let denote the ratio of the length of the minor axis of to the length of its major axis. Remark that if we perform a transformation of the plane which scales every -coordinate by a factor of , is sent to a circle . Thus, the problem is equivalent to finding the value of such that , , and all lie on a common circle; equivalently, it suffices to determine the value of such that the circumcenter of the triangle formed by the points , , and lies on the -axis.
Recall that the circumcenter of a triangle is the intersection point of the perpendicular bisectors of its three sides. The equations of the perpendicular bisectors of the segments and arerespectively. These two lines have different slopes for , so indeed they will intersect at some point ; we want . Plugging into the first equation yields , and so plugging into the second equation and simplifying yieldsSolving yields .
Finally, recall that the lengths , , and (where is the distance between the foci of ) satisfy . Thus the eccentricity of is and the requested answer is .
Video Solution by OmegaLearn(Using Ellipse properties and Quadratic Formula )
~ pi_is_3.14
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.