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 $x_{k}+y_{k}i$ 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 $\boxed{7\textbf{ (A)}}$ .

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2021 AMC 12A Problems/Problem 21

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