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

m (→‎Solution 2 (Three Variables, Three Equations))
m (→‎Video Solution by MRENTHUSIASM (English & Chinese))
 
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==Problem==
 
==Problem==
−
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_ki</math> for <math>1\le k\le 5,</math> where <math>x_k</math> and <math>y_k</math> are real. Let <math>\mathcal 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 eccentricity of <math>\mathcal E</math> can be written in the form <math>\sqrt{\frac mn}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>? (Recall that the eccentricity of an ellipse <math>\mathcal E</math> is the ratio <math>\frac ca</math>, where <math>2a</math> is the length of the major axis of <math>\mathcal E</math> and <math>2c</math> is the is the distance between its two foci.)
+
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_ki</math> for <math>1\le k\le 5,</math> where <math>x_k</math> and <math>y_k</math> are real. Let <math>\mathcal 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 eccentricity of <math>\mathcal E</math> can be written in the form <math>\sqrt{\frac mn}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>? (Recall that the eccentricity of an ellipse <math>\mathcal E</math> is the ratio <math>\frac ca</math>, where <math>2a</math> is the length of the major axis of <math>\mathcal E</math> and <math>2c</math> is the is the distance between its two foci.)
  
−
<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</math>
  
−
==Solution 1==
+
==Solution 1 (Perpendicular Bisectors)==
  
−
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>.
+
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>\left(-1,\pm\sqrt 3\right)</math>, and <math>\left(-2,\pm\sqrt 2\right)</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>.
  
−
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.
+
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>\left(-r,\pm\sqrt 3\right)</math>, and <math>\left(-2r,\pm\sqrt 2\right)</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 = \left(-r,\sqrt 3\right)</math>, and <math>P_3 = \left(-2r,\sqrt 2\right)</math> lies on the <math>x</math>-axis.
  
−
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>.
+
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}\left(x + \tfrac r2\right)</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>.
  
−
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 - \left(\tfrac ba\right)^2} = \sqrt{\tfrac 16}</math> and the requested answer is <math>\boxed{\textbf{(A) } 7}</math>.
  
 
==Solution 2 (Three Variables, Three Equations)==
 
==Solution 2 (Three Variables, Three Equations)==
Line 19: Line 19:
 
Now, we will find the equation of an ellipse <math>\mathcal E</math> that passes through <math>(1,0),\left(-1,\pm\sqrt3\right),</math> and <math>\left(-2,\pm\sqrt2\right)</math> in the <math>xy</math>-plane. By symmetry, the center of <math>\mathcal E</math> must be on the <math>x</math>-axis.
 
Now, we will find the equation of an ellipse <math>\mathcal E</math> that passes through <math>(1,0),\left(-1,\pm\sqrt3\right),</math> and <math>\left(-2,\pm\sqrt2\right)</math> in the <math>xy</math>-plane. By symmetry, the center of <math>\mathcal E</math> must be on the <math>x</math>-axis.
  
−
The formula of <math>\mathcal E</math> is <cmath>\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1, \hspace{44.5mm} (\bigstar)</cmath> with the center <math>(h,0)</math> and the axes' lengths <math>2a</math> and <math>2b.</math> Plugging the points <math>(1,0),\left(-1,\sqrt3\right),</math> and <math>\left(-2,\sqrt2\right)</math> into <math>(\bigstar),</math> respectively, we have the following system:
+
The formula of <math>\mathcal E</math> is <cmath>\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1, \hspace{44.5mm} (\bigstar)</cmath> with the center <math>(h,0)</math> and the axes' lengths <math>2a</math> and <math>2b.</math>
 +
 
 +
Plugging the points <math>(1,0),\left(-1,\sqrt3\right),</math> and <math>\left(-2,\sqrt2\right)</math> into <math>(\bigstar),</math> respectively, we have the following system of equations:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\frac{(1-h)^2}{a^2}&=1, \\
 
\frac{(1-h)^2}{a^2}&=1, \\
Line 25: Line 27:
 
\frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1.
 
\frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Clearing fractions gives
+
Since <math>t^2=(-t)^2</math> holds for all real numbers <math>t,</math> we clear fractions and simplify:
−
<cmath>\begin{align*}
 
−
(1-h)^2&=a^2, \\
 
−
b^2(-1-h)^2 + 3a^2 &= a^2b^2, \\
 
−
b^2(-2-h)^2 + 2a^2 &= a^2b^2.
 
−
\end{align*}</cmath>
 
−
Since <math>t^2=(-t)^2</math> holds for all real numbers <math>t,</math> we rewrite the system as
 
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
(1-h)^2&=a^2, \hspace{30.25mm} &(1)\\
 
(1-h)^2&=a^2, \hspace{30.25mm} &(1)\\
Line 60: Line 56:
 
The answer is <math>1+6=\boxed{\textbf{(A) } 7}.</math>
 
The answer is <math>1+6=\boxed{\textbf{(A) } 7}.</math>
  
 +
The graph of <math>\mathcal E</math> is shown below. Note that the foci are at <math>(h\pm c,0)=\left(-\frac{9}{10}\pm\frac{19\sqrt6}{60},0\right),</math> as shown in the blue points.
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(220);
 +
 +
int xMin = -4;
 +
int xMax = 2;
 +
int yMin = -3;
 +
int yMax = 3;
 +
 +
//Draws the horizontal gridlines
 +
void horizontalLines()
 +
{
 +
  for (int i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
 +
  }
 +
}
 +
 +
//Draws the vertical gridlines
 +
void verticalLines()
 +
{
 +
  for (int i = xMin+1; i < xMax; ++i)
 +
  {
 +
    draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
 +
  }
 +
}
 +
 +
//Draws the horizontal ticks
 +
void horizontalTicks()
 +
{
 +
  for (int i = yMin+1; i < yMax; ++i)
 +
  {
 +
    draw((-1/8,i)--(1/8,i), black+linewidth(1));
 +
  }
 +
}
 +
 +
//Draws the vertical ticks
 +
void verticalTicks()
 +
{
 +
  for (int i = xMin+1; i < xMax; ++i)
 +
  {
 +
    draw((i,-1/8)--(i,1/8), black+linewidth(1));
 +
  }
 +
}
 +
 +
horizontalLines();
 +
verticalLines();
 +
horizontalTicks();
 +
verticalTicks();
 +
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
 +
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
 +
label("$x$",(xMax,0),(2,0));
 +
label("$y$",(0,yMax),(0,2));
 +
 +
draw(ellipse((-9/10,0),19/10,19/sqrt(120)),red);
 +
 +
pair A = (-9/10,0);
 +
pair B = (1,0);
 +
pair C = (-1,sqrt(3));
 +
pair D = (-1,-sqrt(3));
 +
pair E = (-2,sqrt(2));
 +
pair F = (-2,-sqrt(2));
 +
pair G = (-9/10+19/sqrt(600),0);
 +
pair H = (-9/10-19/sqrt(600),0);
 +
 +
dot(A,red+linewidth(4.5));
 +
dot(B,red+linewidth(4.5));
 +
dot(C,red+linewidth(4.5));
 +
dot(D,red+linewidth(4.5));
 +
dot(E,red+linewidth(4.5));
 +
dot(F,red+linewidth(4.5));
 +
dot(G,blue+linewidth(4.5));
 +
dot(H,blue+linewidth(4.5));
 +
 +
label("$\left(-\frac{9}{10},0\right)$",A,(0,-2),UnFill);
 +
label("$(1,0)$",B,(1.5,-2),UnFill);
 +
label("$\left(-1,\sqrt3\right)$",C,N,UnFill);
 +
label("$\left(-1,-\sqrt3\right)$",D,S,UnFill);
 +
label("$\left(-2,\sqrt2\right)$",E,NW,UnFill);
 +
label("$\left(-2,-\sqrt2\right)$",F,SW,UnFill);
 +
</asy>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Solution 3==
+
== Solution 3 (Alternate Version of Solution 2) ==
  
−
After calculating the 5 points that lie on <math>\mathcal E</math>, we try to find a transformation that sends <math>\mathcal E</math> to the unit circle. Scaling about <math>(1, 0)</math> works, since <math>(1, 0)</math> is already on the unit circle and such a transformation will preserve the ellipse's symmetry about the <math>x</math>-axis. If <math>2a</math> and <math>2b</math> are the lengths of the major and minor axes, respectively, then the ellipse will be scaled by a factor of <math>r := 1/a</math> in the <math>x</math>-dimension and <math>s := 1/b</math> in the <math>y</math>-dimension.
+
Starting from this system of equations from Solution 2:
 +
<cmath>\begin{align*}
 +
\frac{(1-h)^2}{a^2}&=1, \\
 +
\frac{(-1-h)^2}{a^2}+\frac{{\sqrt3}^2}{b^2}&=1, \\
 +
\frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1.
 +
\end{align*}</cmath>
 +
Let <math>A=a^{-2}</math> and <math>B=b^{-2}</math>. Therefore, the system can be rewritten as:
 +
<cmath>\begin{align*}
 +
(h^2-2h+1)A&=1, &(1)\\
 +
(h^2+2h+1)A+3B&=1, &(2)\\
 +
(h^2+4h+4)A+2B&=1. &(3)
 +
\end{align*}</cmath>
 +
Subtracting <math>(1)</math> from <math>(2)</math> and <math>(3)</math>, we get
 +
<cmath>4hA+3B=0\quad\text{and}\quad 3A-6hA+3B=0.</cmath>
 +
Plugging the former into the latter and simplifying yields <math>6A=5B</math>. Hence <math>a^2:b^2=6:5</math>. Since <math>c^2=a^2-b^2</math>, we get <math>a^2=6c^2</math>, so the eccentricity is <math>\frac ca=\sqrt{\frac16}</math>.
  
−
The transformation then sends the points <math>(-1,\pm\sqrt 3)</math> and <math>(-2,\pm\sqrt 2)</math> to the points <math>(1-2r, \pm s\sqrt 3)</math> and <math>(1-3r, \pm s\sqrt 2)</math>, respectively. These points are on the unit circle, so
+
Therefore, the answer is <math>1+6=\boxed{\textbf{(A) }7}</math>.
 +
 
 +
~wzs26843545602
 +
 
 +
==Solution 4 (Four Variables, Three Equations)==
 +
 
 +
The five roots are <math>1,-1+i\sqrt{3},-1-i\sqrt{3},-2+i\sqrt{2},-2-i\sqrt{2}.</math>
 +
 
 +
So, we express this conic in the form <math>ax^2+by^2+cx+z=0.</math> Note that this conic cannot have the <math>ky</math> term since the roots are symmetric about the <math>x</math>-axis.
 +
 
 +
Now we have equations
 +
<cmath>\begin{align*}
 +
a+c+z&=0, \\
 +
a+3b-c+z&=0, \\
 +
4a+2b-2c+z&=0,
 +
\end{align*}</cmath>
 +
from which <math>a:b:c=5:6:9.</math>
 +
 
 +
So, the conic can be written in the form <math>5x^2+6y^2+9x=14.</math> If it is written in the form of <math>\frac{(x-m)^2}{r^2}+\frac{y^2}{s^2}=1,</math> then <math>r^2:s^2=6:5.</math>
 +
 
 +
Therefore, the desired eccentricity is <math>\sqrt{\frac{\sqrt{6-5}}{6}}=\sqrt{\frac{1}{6}},</math> and the answer is <math>1+6=\boxed{\textbf{(A) }7}.</math>
 +
 
 +
~bluesoul
 +
 
 +
==Solution 5 (Transformations)==
 +
 
 +
After calculating the <math>5</math> points that lie on <math>\mathcal E</math>, we try to find a transformation that sends <math>\mathcal E</math> to the unit circle. Scaling about <math>(1, 0)</math> works, since <math>(1, 0)</math> is already on the unit circle and such a transformation will preserve the ellipse's symmetry about the <math>x</math>-axis. If <math>2a</math> and <math>2b</math> are the lengths of the major and minor axes, respectively, then the ellipse will be scaled by a factor of <math>r := \frac1a</math> in the <math>x</math>-dimension and <math>s := \frac1b</math> in the <math>y</math>-dimension.
 +
 
 +
The transformation then sends the points <math>\left(-1,\pm\sqrt 3\right)</math> and <math>\left(-2,\pm\sqrt 2\right)</math> to the points <math>\left(1-2r, \pm s\sqrt 3\right)</math> and <math>\left(1-3r, \pm s\sqrt 2\right)</math>, respectively. These points are on the unit circle, so
 
<cmath>(1-2r)^2 + 3s^2 = 1 \quad \text{and} \quad (1-3r)^2 + 2s^2 = 1.</cmath>
 
<cmath>(1-2r)^2 + 3s^2 = 1 \quad \text{and} \quad (1-3r)^2 + 2s^2 = 1.</cmath>
−
This yields
+
This yields <cmath>4r^2 + 3s^2 = 4r \quad \text{and} \quad 9r^2 + 2s^2 = 6r,</cmath> from which
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
4r^2 + 3s^2 = 4r \quad \text{and} \quad 9r^2 + 2s^2 = 6r& \\
+
12r^2 + 9s^2 &= 18r^2 + 4s^2 \\
−
\implies \enskip 12r^2 + 9s^2 = 18r^2 + 4s^2& \\
+
\frac{r^2}{s^2} &= \frac56.
−
\implies \enskip r^2/s^2 = 5/6.&
 
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Recalling that <math>r = 1/a</math> and <math>s = 1/b</math>, this implies <math>b^2/a^2 = 5/6</math>. From this, we get
+
Recalling that <math>r = \frac1a</math> and <math>s = \frac1b</math>, this implies <math>\frac{b^2}{a^2} = \frac56</math>. From this, we get
 
<cmath>\frac{c^2}{a^2} = \frac{a^2-b^2}{a^2} = 1 - \frac{b^2}{a^2} = \frac{1}{6},</cmath>
 
<cmath>\frac{c^2}{a^2} = \frac{a^2-b^2}{a^2} = 1 - \frac{b^2}{a^2} = \frac{1}{6},</cmath>
−
so <math>c/a = \sqrt{1/6}</math>, giving an answer of <math>1 + 6 = \boxed{\textbf{(A) } 7}</math>.
+
so <math>\frac ca = \sqrt{\frac16}</math>, giving an answer of <math>1 + 6 = \boxed{\textbf{(A) } 7}</math>.
  
 
~building
 
~building
  
−
==Remark==
+
== Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic) ==
−
The graph of <math>\mathcal E</math> is shown below.
+
https://youtu.be/eIYFQSeIRzM
  
−
[[File:2021 AMC 12A Problem 21.png|center]]
+
~ pi_is_3.14
  
−
Graph in Desmos: https://www.desmos.com/calculator/ptdpdzsgyo
+
== Video Solution by MRENTHUSIASM (English & Chinese) ==
 +
https://www.youtube.com/watch?v=PQdz8IBAZig&t=8s
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
== Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic) ==
+
== Video Solution ==
−
https://youtu.be/eIYFQSeIRzM
+
https://youtu.be/zAIcLfye_Mc
  
−
~ pi_is_3.14
+
~MathProblemSolvingSkills.com
  
 
==See also==
 
==See also==

Latest revision as of 17:33, 1 August 2022

Problem

The five solutions to the equation\[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ 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 eccentricity of $\mathcal E$ can be written in the form $\sqrt{\frac mn}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\frac ca$, where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)

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

Solution 1 (Perpendicular Bisectors)

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 $(1,0)$, $\left(-1,\pm\sqrt 3\right)$, and $\left(-2,\pm\sqrt 2\right)$; 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 $(r,0)$, $\left(-r,\pm\sqrt 3\right)$, and $\left(-2r,\pm\sqrt 2\right)$ 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 $P_1 = (r,0)$, $P_2 = \left(-r,\sqrt 3\right)$, and $P_3 = \left(-2r,\sqrt 2\right)$ 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\[y = \tfrac{\sqrt 3}2 + \tfrac{2r}{\sqrt 3}x\qquad\text{and}\qquad y = \tfrac{\sqrt 2}2 + \tfrac{3r}{\sqrt 2}\left(x + \tfrac r2\right)\]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 $\tfrac ca = \sqrt{1 - \left(\tfrac ba\right)^2} = \sqrt{\tfrac 16}$ and the requested answer is $\boxed{\textbf{(A) } 7}$.

Solution 2 (Three Variables, Three Equations)

Completing the square in the original equation, we have \[(z-1)\left((z+1)^2+3\right)\left((z+2)^2+2\right)=0,\] from which $z=1,-1\pm\sqrt{3}i,-2\pm\sqrt{2}i.$

Now, we will find the equation of an ellipse $\mathcal E$ that passes through $(1,0),\left(-1,\pm\sqrt3\right),$ and $\left(-2,\pm\sqrt2\right)$ in the $xy$-plane. By symmetry, the center of $\mathcal E$ must be on the $x$-axis.

The formula of $\mathcal E$ is \[\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1, \hspace{44.5mm} (\bigstar)\] with the center $(h,0)$ and the axes' lengths $2a$ and $2b.$

Plugging the points $(1,0),\left(-1,\sqrt3\right),$ and $\left(-2,\sqrt2\right)$ into $(\bigstar),$ respectively, we have the following system of equations: \begin{align*} \frac{(1-h)^2}{a^2}&=1, \\ \frac{(-1-h)^2}{a^2}+\frac{{\sqrt3}^2}{b^2}&=1, \\ \frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1. \end{align*} Since $t^2=(-t)^2$ holds for all real numbers $t,$ we clear fractions and simplify: \begin{align*} (1-h)^2&=a^2, \hspace{30.25mm} &(1)\\ b^2(1+h)^2 + 3a^2 &= a^2b^2, &(2)\\ b^2(2+h)^2 + 2a^2 &= a^2b^2. &(3) \end{align*} Applying the Transitive Property to $(2)$ and $(3),$ we isolate $a^2:$ \begin{align*} b^2(1+h)^2 + 3a^2 &= b^2(2+h)^2 + 2a^2 \\ a^2 &= b^2\left((2+h)^2-(1+h)^2\right) \\ a^2 &= b^2(2h+3). \hspace{26.75mm} (*) \end{align*} Substituting $(1)$ and $(*)$ into $(2),$ we solve for $h:$ \begin{align*} b^2(1+h)^2 + 3\underbrace{b^2(2h+3)}_{\text{by }(*)} &= \underbrace{(1-h)^2}_{\text{by }(1)}b^2 \\ (1+h)^2+3(2h+3)&=(1-h)^2 \\ 1+2h+h^2+6h+9&=1-2h+h^2 \\ 10h&=-9 \\ h&=-\frac{9}{10}. \end{align*} Substituting this into $(1),$ we get $a^2=\frac{361}{100}.$

Substituting the current results into $(*),$ we get $b^2=\frac{361}{120}.$

Finally, we obtain \[c^2 = a^2-b^2 = 361\left(\frac{1}{100}-\frac{1}{120}\right) = \frac{361}{600},\] from which \[\frac{c}{a}=\sqrt{\frac{c^2}{a^2}}=\sqrt{\frac{361/600}{361/100}}=\sqrt{\frac 16}.\] The answer is $1+6=\boxed{\textbf{(A) } 7}.$

The graph of $\mathcal E$ is shown below. Note that the foci are at $(h\pm c,0)=\left(-\frac{9}{10}\pm\frac{19\sqrt6}{60},0\right),$ as shown in the blue points. [asy] /* Made by MRENTHUSIASM */ size(220); int xMin = -4; int xMax = 2; int yMin = -3; int yMax = 3; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-1/8,i)--(1/8,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-1/8)--(i,1/8), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); draw(ellipse((-9/10,0),19/10,19/sqrt(120)),red); pair A = (-9/10,0); pair B = (1,0); pair C = (-1,sqrt(3)); pair D = (-1,-sqrt(3)); pair E = (-2,sqrt(2)); pair F = (-2,-sqrt(2)); pair G = (-9/10+19/sqrt(600),0); pair H = (-9/10-19/sqrt(600),0); dot(A,red+linewidth(4.5)); dot(B,red+linewidth(4.5)); dot(C,red+linewidth(4.5)); dot(D,red+linewidth(4.5)); dot(E,red+linewidth(4.5)); dot(F,red+linewidth(4.5)); dot(G,blue+linewidth(4.5)); dot(H,blue+linewidth(4.5)); label("$\left(-\frac{9}{10},0\right)$",A,(0,-2),UnFill); label("$(1,0)$",B,(1.5,-2),UnFill); label("$\left(-1,\sqrt3\right)$",C,N,UnFill); label("$\left(-1,-\sqrt3\right)$",D,S,UnFill); label("$\left(-2,\sqrt2\right)$",E,NW,UnFill); label("$\left(-2,-\sqrt2\right)$",F,SW,UnFill); [/asy] ~MRENTHUSIASM

Solution 3 (Alternate Version of Solution 2)

Starting from this system of equations from Solution 2: \begin{align*} \frac{(1-h)^2}{a^2}&=1, \\ \frac{(-1-h)^2}{a^2}+\frac{{\sqrt3}^2}{b^2}&=1, \\ \frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1. \end{align*} Let $A=a^{-2}$ and $B=b^{-2}$. Therefore, the system can be rewritten as: \begin{align*} (h^2-2h+1)A&=1, &(1)\\ (h^2+2h+1)A+3B&=1, &(2)\\ (h^2+4h+4)A+2B&=1. &(3) \end{align*} Subtracting $(1)$ from $(2)$ and $(3)$, we get \[4hA+3B=0\quad\text{and}\quad 3A-6hA+3B=0.\] Plugging the former into the latter and simplifying yields $6A=5B$. Hence $a^2:b^2=6:5$. Since $c^2=a^2-b^2$, we get $a^2=6c^2$, so the eccentricity is $\frac ca=\sqrt{\frac16}$.

Therefore, the answer is $1+6=\boxed{\textbf{(A) }7}$.

~wzs26843545602

Solution 4 (Four Variables, Three Equations)

The five roots are $1,-1+i\sqrt{3},-1-i\sqrt{3},-2+i\sqrt{2},-2-i\sqrt{2}.$

So, we express this conic in the form $ax^2+by^2+cx+z=0.$ Note that this conic cannot have the $ky$ term since the roots are symmetric about the $x$-axis.

Now we have equations \begin{align*} a+c+z&=0, \\ a+3b-c+z&=0, \\ 4a+2b-2c+z&=0, \end{align*} from which $a:b:c=5:6:9.$

So, the conic can be written in the form $5x^2+6y^2+9x=14.$ If it is written in the form of $\frac{(x-m)^2}{r^2}+\frac{y^2}{s^2}=1,$ then $r^2:s^2=6:5.$

Therefore, the desired eccentricity is $\sqrt{\frac{\sqrt{6-5}}{6}}=\sqrt{\frac{1}{6}},$ and the answer is $1+6=\boxed{\textbf{(A) }7}.$

~bluesoul

Solution 5 (Transformations)

After calculating the $5$ points that lie on $\mathcal E$, we try to find a transformation that sends $\mathcal E$ to the unit circle. Scaling about $(1, 0)$ works, since $(1, 0)$ is already on the unit circle and such a transformation will preserve the ellipse's symmetry about the $x$-axis. If $2a$ and $2b$ are the lengths of the major and minor axes, respectively, then the ellipse will be scaled by a factor of $r := \frac1a$ in the $x$-dimension and $s := \frac1b$ in the $y$-dimension.

The transformation then sends the points $\left(-1,\pm\sqrt 3\right)$ and $\left(-2,\pm\sqrt 2\right)$ to the points $\left(1-2r, \pm s\sqrt 3\right)$ and $\left(1-3r, \pm s\sqrt 2\right)$, respectively. These points are on the unit circle, so \[(1-2r)^2 + 3s^2 = 1 \quad \text{and} \quad (1-3r)^2 + 2s^2 = 1.\] This yields \[4r^2 + 3s^2 = 4r \quad \text{and} \quad 9r^2 + 2s^2 = 6r,\] from which \begin{align*} 12r^2 + 9s^2 &= 18r^2 + 4s^2 \\ \frac{r^2}{s^2} &= \frac56. \end{align*} Recalling that $r = \frac1a$ and $s = \frac1b$, this implies $\frac{b^2}{a^2} = \frac56$. From this, we get \[\frac{c^2}{a^2} = \frac{a^2-b^2}{a^2} = 1 - \frac{b^2}{a^2} = \frac{1}{6},\] so $\frac ca = \sqrt{\frac16}$, giving an answer of $1 + 6 = \boxed{\textbf{(A) } 7}$.

~building

Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic)

https://youtu.be/eIYFQSeIRzM

~ pi_is_3.14

Video Solution by MRENTHUSIASM (English & Chinese)

https://www.youtube.com/watch?v=PQdz8IBAZig&t=8s

~MRENTHUSIASM

Video Solution

https://youtu.be/zAIcLfye_Mc

~MathProblemSolvingSkills.com

See also

2021 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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