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Difference between revisions of "2021 AMC 12A Problems/Problem 21"

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==Solution==
 
==Solution==
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The solutions to this equation are <math>z = 1</math>, <math>z = -1 \pm i\sqrt 3</math>, and <math>z = -2\pm i\sqrt 2</math>. Consider the five points <math>(1,0)</math>, <math>(-1,\pm\sqrt 3)</math>, and <math>(-2,\pm\sqrt 2)</math>; these are the five points which lie on <math>\mathcal E</math>. Note that since these five points are symmetric about the <math>x</math>-axis, so must <math>\mathcal E</math>.
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Now let <math>r:= b/a</math> denote the ratio of the length of the minor axis of <math>\mathcal E</math> to the length of its major axis. Remark that if we perform a transformation of the plane which scales every <math>x</math>-coordinate by a factor of <math>r</math>, <math>\mathcal E</math> is sent to a circle <math>\mathcal E'</math>. Thus, the problem is equivalent to finding the value of <math>r</math> such that <math>(r,0)</math>, <math>(-r,\pm\sqrt 3)</math>, and <math>(-2r,\pm\sqrt 2)</math> all lie on a common circle; equivalently, it suffices to determine the value of <math>r</math> such that the circumcenter of the triangle formed by the points <math>P_1 = (r,0)</math>, <math>P_2 = (-r,\sqrt 3)</math>, and <math>P_3 = (-2r,\sqrt 2)</math> lies on the <math>x</math>-axis.
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Recall that the circumcenter of a triangle <math>ABC</math> is the intersection point of the perpendicular bisectors of its three sides. The equations of the perpendicular bisectors of the segments <math>\overline{P_1P_2}</math> and <math>\overline{P_1P_3}</math> are<cmath>y = \tfrac{\sqrt 3}2 + \tfrac{2r}{\sqrt 3}x\qquad\text{and}\qquad y = \tfrac{\sqrt 2}2 + \tfrac{3r}{\sqrt 2}(x + \tfrac r2)</cmath>respectively. These two lines have different slopes for <math>r\neq 0</math>, so indeed they will intersect at some point <math>(x_0,y_0)</math>; we want <math>y_0 = 0</math>. Plugging <math>y = 0</math> into the first equation yields <math>x = -\tfrac{3}{4r}</math>, and so plugging <math>y=0</math> into the second equation and simplifying yields<cmath>-\tfrac{1}{3r} = x + \tfrac r2 = -\tfrac{3}{4r} + \tfrac{r}{2}.</cmath>Solving yields <math>r=\sqrt{\tfrac 56}</math>.
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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>.
  
 
==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 14:59, 11 February 2021

Problem

The five solutions to the equation \[(z-1)(z^{2}+2z+4)(z^{2}+4z+6)=0\] may be written in the form $x_{k}+y_{k}i$ for $1\leq k\leq 5$, where $x_{k}$ and $y_{k}$ are real. Let $\mathbb{E}$ be the unique ellipse that passes through the points $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), (x_{4}, y_{4}),$ and $(x_{5}, y_{5})$. The excentricity of $\mathbb{E}$ can be written in the form $\frac{m}{\sqrt{n}}$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$?

$\textbf{(A) } 7\qquad\textbf{(B) } 9\qquad\textbf{(C) } 11\qquad\textbf{(D) } 13\qquad\textbf{(E) } 15\qquad$

Solution

The solutions to this equation are $z = 1$, $z = -1 \pm i\sqrt 3$, and $z = -2\pm i\sqrt 2$. Consider the five points $(1,0)$, $(-1,\pm\sqrt 3)$, and $(-2,\pm\sqrt 2)$; these are the five points which lie on $\mathcal E$. Note that since these five points are symmetric about the $x$-axis, so must $\mathcal E$.

Now let $r:= b/a$ denote the ratio of the length of the minor axis of $\mathcal E$ to the length of its major axis. Remark that if we perform a transformation of the plane which scales every $x$-coordinate by a factor of $r$, $\mathcal E$ is sent to a circle $\mathcal E'$. Thus, the problem is equivalent to finding the value of $r$ such that $(r,0)$, $(-r,\pm\sqrt 3)$, and $(-2r,\pm\sqrt 2)$ all lie on a common circle; equivalently, it suffices to determine the value of $r$ such that the circumcenter of the triangle formed by the points $P_1 = (r,0)$, $P_2 = (-r,\sqrt 3)$, and $P_3 = (-2r,\sqrt 2)$ lies on the $x$-axis.

Recall that the circumcenter of a triangle $ABC$ is the intersection point of the perpendicular bisectors of its three sides. The equations of the perpendicular bisectors of the segments $\overline{P_1P_2}$ and $\overline{P_1P_3}$ are\[y = \tfrac{\sqrt 3}2 + \tfrac{2r}{\sqrt 3}x\qquad\text{and}\qquad y = \tfrac{\sqrt 2}2 + \tfrac{3r}{\sqrt 2}(x + \tfrac r2)\]respectively. These two lines have different slopes for $r\neq 0$, so indeed they will intersect at some point $(x_0,y_0)$; we want $y_0 = 0$. Plugging $y = 0$ into the first equation yields $x = -\tfrac{3}{4r}$, and so plugging $y=0$ into the second equation and simplifying yields\[-\tfrac{1}{3r} = x + \tfrac r2 = -\tfrac{3}{4r} + \tfrac{r}{2}.\]Solving yields $r=\sqrt{\tfrac 56}$.

Finally, recall that the lengths $a$, $b$, and $c$ (where $c$ is the distance between the foci of $\mathcal E$) satisfy $c = \sqrt{a^2 - b^2}$. Thus the eccentricity of $\mathcal E$ is $\tfrac ca = \sqrt{1 - (\tfrac ba)^2} = \sqrt{\tfrac 16}$ and the requested answer is $\boxed{7\textbf{ (A)}}$.

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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