2021 AMC 12A Problems/Problem 21

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.)

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

Solution 1

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 $\boxed{\textbf{(A) } 7}$ .

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: $\begin{align*} \frac{(1-h)^2}{a^2}&=1, \\ \frac{(-1-h)^2}{a^2}+\frac{{\sqrt3}^2}{b^2}&=1, \\ \frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1. \end{align*}$ Clearing fractions gives $\begin{align*} (1-h)^2&=a^2, \\ b^2(-1-h)^2 + 3a^2 &= a^2b^2, \\ b^2(-2-h)^2 + 2a^2 &= a^2b^2. \end{align*}$ Since $t^2=(-t)^2$ holds for all real numbers $t,$ we rewrite the system as $\begin{align*} (1-h)^2&=a^2, \hspace{30.25mm} &(1)\\ b^2(1+h)^2 + 3a^2 &= a^2b^2, &(2)\\ b^2(2+h)^2 + 2a^2 &= a^2b^2. &(3) \end{align*}$ 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. $[asy] /* Made by MRENTHUSIASM */ size(220); import TrigMacros; rr_cartesian_axes(-4,2,-3,3); draw(ellipse((-9/10,0),19/10,19/sqrt(120)),red); pair A = (-9/10,0); pair B = (1,0); pair C = (-1,sqrt(3)); pair D = (-1,-sqrt(3)); pair E = (-2,sqrt(2)); pair F = (-2,-sqrt(2)); pair G = (-9/10+19/sqrt(600),0); pair H = (-9/10-19/sqrt(600),0); dot(A,red+linewidth(4.5)); dot(B,red+linewidth(4.5)); dot(C,red+linewidth(4.5)); dot(D,red+linewidth(4.5)); dot(E,red+linewidth(4.5)); dot(F,red+linewidth(4.5)); dot(G,blue+linewidth(4.5)); dot(H,blue+linewidth(4.5)); label("$\left(-\frac{9}{10},0\right)$",A,S,UnFill); label("$(1,0)$",B,SE,UnFill); label("$\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

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 $r := 1/a$ in the $x$ -dimension and $s := 1/b$ in the $y$ -dimension.

The transformation then sends the points $(-1,\pm\sqrt 3)$ and $(-2,\pm\sqrt 2)$ to the points $(1-2r, \pm s\sqrt 3)$ and $(1-3r, \pm s\sqrt 2)$ , 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 $\begin{align*} 12r^2 + 9s^2 &= 18r^2 + 4s^2 \\ r^2/s^2 &= 5/6. \end{align*}$ Recalling that $r = 1/a$ and $s = 1/b$ , this implies $b^2/a^2 = 5/6$ . 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 $c/a = \sqrt{1/6}$ , 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

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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