Difference between revisions of "2021 AMC 12A Problems/Problem 21"
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<cmath>c^2 = a^2-b^2 = 361\left(\frac{1}{100}-\frac{1}{120}\right) = \frac{361}{600},</cmath> | <cmath>c^2 = a^2-b^2 = 361\left(\frac{1}{100}-\frac{1}{120}\right) = \frac{361}{600},</cmath> | ||
from which <cmath>\frac{c}{a}=\sqrt{\frac{c^2}{a^2}}=\sqrt{\frac{361/600}{361/100}}=\sqrt{\frac 16}.</cmath> | from which <cmath>\frac{c}{a}=\sqrt{\frac{c^2}{a^2}}=\sqrt{\frac{361/600}{361/100}}=\sqrt{\frac 16}.</cmath> | ||
− | + | The answer is <math>1+6=\boxed{\textbf{(A) } 7}.</math> | |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 12:26, 22 April 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 eccentricity of can be written in the form where and are relatively prime positive integers. What is ? (Recall that the eccentricity of an ellipse is the ratio , where is the length of the major axis of and is the is the distance between its two foci.)
Solution 1
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 .
Solution 2 (Three Variables, Three Equations)
Completing the square in the original equation, we have from which
Now, we will find the equation of an ellipse that passes through and in the -plane. By symmetry, the center of must be on the -axis.
The formula of is with the center at and the axes' lengths and Plugging the points and into we have the following three equations in a system, respectively: Clearing fractions gives Since holds for all real numbers we rewrite the system as Applying the Transitive Property to and we get Applying the results of and to we get Substituting this into we get
Substituting the current results into we get
Finally, we obtain from which The answer is
~MRENTHUSIASM
Remark
The graph of is shown below.
Graph in Desmos: https://www.desmos.com/calculator/ptdpdzsgyo
~MRENTHUSIASM
Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic)
~ 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.