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

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

Solution 2 (Three Variables, Three Equations)

Completing the square in the original equation, we get $(z-1)\left((z+1)^2+3\right)\left((z+2)^2+2\right)=0,$ from which $z=1,-1\pm\sqrt{3}i,-2\pm\sqrt{2}i.$ Now, we will find the equation of an ellipse $\mathbb{E}$ that passes through $(1,0),(-1,\pm\sqrt3),$ and $(-2,\pm\sqrt2).$ By symmetry, the center of $\mathbb{E}$ must be on the $x$ -axis.

The formula of $\mathbb{E}$ is $\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1,$ with the center at $(h,0)$ and the axes' lengths $2a$ and $2b.$ Plugging the points $(1,0),(-1,\sqrt3),$ and $(-2,\sqrt2)$ in, respectively, we get the following system of three 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 $t^2=(-t)^2$ for all real numbers $t,$ we rewrite the system as $\begin{align*} (1-h)^2&=a^2, \ \ \ \ \ \ \ \ \ \ \ \ &\text{(1)}\\ b^2(1+h)^2 + 3a^2 &= a^2b^2, &\text{(2)}\\ b^2(2+h)^2 + 2a^2 &= a^2b^2. &\text{(3)} \end{align*}$ Applying the Transitive Property in $\text{(2)}$ and $\text{(3)},$ we get $\begin{align*} b^2(1+h)^2 + 3a^2 &= b^2(2+h)^2 + 2a^2 \\ a^2 &= b^2\left((2+h)^2-(1+h)^2\right) \\ a^2 &= b^2(2h+3). \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(4)} \end{align*}$ Applying the results of $\text{(1)}$ and $\text{(4)}$ on $\text{(2)},$ we get $\begin{align*} b^2(1+h)^2 + 3\underbrace{b^2(2h+3)}_\text{by (4)} &= \underbrace{(1-h)^2}_\text{by (1)}b^2 \\ (1+h)^2+3(2h+3)&=(1-h)^2 \\ 1+2h+h^2+6h+9&=1-2h+h^2 \\ 10h&=-9 \\ h&=-\frac{9}{10}. \end{align*}$ Substituting this into $\text{(1)},$ we get $a^2=\frac{361}{100}.$

Substituting the current results into $\text{(4)},$ we get $b^2=\frac{361}{120}.$

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

The graph of $\mathbb{E}$ can be found in Desmos: https://www.desmos.com/calculator/etqgd838jn

~MRENTHUSIASM

Video Solution by OmegaLearn (Using Ellipse properties & Quadratic)

https://youtu.be/eIYFQSeIRzM

~ pi_is_3.14

@@ Line 58: / Line 58: @@
 Our answer is <math>1+6=\boxed{\textbf{(A) } 7}.</math>
-The graph of <math>\mathbb{E}</math> can be found in Desmos: https://www.desmos.com/calculator/3rml1nvlyi
+The graph of <math>\mathbb{E}</math> can be found in Desmos: https://www.desmos.com/calculator/etqgd838jn
 ~MRENTHUSIASM

2021 AMC 12A (Problems • Answer Key • Resources)
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