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.)

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

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

~ pi_is_3.14

Video Solution by MRENTHUSIASM (English & Chinese)

~MRENTHUSIASM

 2021 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 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