Difference between revisions of "2021 AMC 12A Problems/Problem 21"
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==Problem== | ==Problem== | ||
− | The five solutions to the equation <cmath>(z-1)(z^{2}+2z+4)(z^{2}+4z+6</cmath> may be written in the form <math>x_{k}+y_{k}i</math> for <math>1\leq k\leq 5</math>, where <math>x_{k}</math> and <math>y_{k}</math> are real. Let <math>\mathbb{E}</math> be the unique ellipse that passes through the points <math>(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), (x_{4}, y_{4}),</math> and <math>(x_{5}, y_{5})</math>. The excentricity of <math>\mathbb{E}</math> can be written in the form <math>\frac{m}{\sqrt{n}}</math> where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. What is <math>m+n</math>? | + | The five solutions to the equation <cmath>(z-1)(z^{2}+2z+4)(z^{2}+4z+6)=0</cmath> may be written in the form <math>x_{k}+y_{k}i</math> for <math>1\leq k\leq 5</math>, where <math>x_{k}</math> and <math>y_{k}</math> are real. Let <math>\mathbb{E}</math> be the unique ellipse that passes through the points <math>(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), (x_{4}, y_{4}),</math> and <math>(x_{5}, y_{5})</math>. The excentricity of <math>\mathbb{E}</math> can be written in the form <math>\frac{m}{\sqrt{n}}</math> where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. What is <math>m+n</math>? |
<math>\textbf{(A) } 7\qquad\textbf{(B) } 9\qquad\textbf{(C) } 11\qquad\textbf{(D) } 13\qquad\textbf{(E) } 15\qquad</math> | <math>\textbf{(A) } 7\qquad\textbf{(B) } 9\qquad\textbf{(C) } 11\qquad\textbf{(D) } 13\qquad\textbf{(E) } 15\qquad</math> |
Revision as of 14:07, 11 February 2021
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
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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 |
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