2021 AMC 12A Problems/Problem 21
Contents
- 1 Problem
- 2 Solution 1 (Perpendicular Bisectors)
- 3 Solution 2 (Three Variables, Three Equations)
- 4 Solution 3 (Alternate Version of Solution 2)
- 5 Solution 4 (Four Variables, Three Equations)
- 6 Solution 5 (Transformations)
- 7 Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic)
- 8 Video Solution by MRENTHUSIASM (English & Chinese)
- 9 Video Solution
- 10 See also
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 (Perpendicular Bisectors)
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 and the axes' lengths and
Plugging the points and into respectively, we have the following system of equations: Since holds for all real numbers we clear fractions and simplify: Applying the Transitive Property to and we isolate Substituting and into we solve for Substituting this into we get
Substituting the current results into we get
Finally, we obtain from which The answer is
The graph of is shown below. Note that the foci are at as shown in the blue points. ~MRENTHUSIASM
Solution 3 (Alternate Version of Solution 2)
Starting from this system of equations from Solution 2: Let and . Therefore, the system can be rewritten as: Subtracting from and , we get Plugging the former into the latter and simplifying yields . Hence . Since , we get , so the eccentricity is .
Therefore, the answer is .
~wzs26843545602
Solution 4 (Four Variables, Three Equations)
The five roots are
So, we express this conic in the form Note that this conic cannot have the term since the roots are symmetric about the -axis.
Now we have equations from which
So, the conic can be written in the form If it is written in the form of then
Therefore, the desired eccentricity is and the answer is
~bluesoul
Solution 5 (Transformations)
After calculating the points that lie on , we try to find a transformation that sends to the unit circle. Scaling about works, since is already on the unit circle and such a transformation will preserve the ellipse's symmetry about the -axis. If and are the lengths of the major and minor axes, respectively, then the ellipse will be scaled by a factor of in the -dimension and in the -dimension.
The transformation then sends the points and to the points and , respectively. These points are on the unit circle, so This yields from which Recalling that and , this implies . From this, we get so , giving an answer of .
~building
Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic)
~ pi_is_3.14
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=PQdz8IBAZig&t=8s
~MRENTHUSIASM
Video Solution
~MathProblemSolvingSkills.com
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.