2021 AMC 12A Spring Problem 21

Question

The five solutions to the equation

\begin{aligned} &(z - 1)(z^2 + 2z + 4) \\ &\quad {}\cdot (z^2 + 4z + 6) = 0 \end{aligned}

may be written in the form

x_k + y_k i

for

1 \le k \le 5,

where

x_k

and

y_k

are real. Let

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

E

can be written in the form

\sqrt{\tfrac{m}{n}},

where

m

and

n

are relatively prime positive integers. What is

m + n?

(Recall that the eccentricity of an ellipse

E

is the ratio

\tfrac{c}{a},

where

2a

is the length of the major axis of

E

and

2c

is the distance between its two foci.)

7

9

11

13

15

Answer: A

Concepts:ellipse complex number system of equations

Difficulty rating: 2450

Small hint:

The five roots are

(1, 0),

(-1, \pm\sqrt3),

and

(-2, \pm\sqrt2);

they are symmetric about the

x

-axis

Big hint:

Fit

Ax^2 + Cy^2 + Dx + F = 0

through the points, then read off

a^2

and

b^2

to find

e^2 = 1 - \tfrac{b^2}{a^2}

Solution:

The roots are

z = 1,

z = -1 \pm i\sqrt3,

and

z = -2 \pm i\sqrt2,

giving the points

(1, 0),

(-1, \pm\sqrt3),

and

(-2, \pm\sqrt2).

By symmetry about the

x

-axis, the ellipse has the form

Ax^2 + Cy^2 + Dx + F = 0.

Substituting the points yields

5x^2 + 6y^2 + 9x - 14 = 0.

Completing the square gives

5\left(x + \tfrac{9}{10}\right)^2 + 6y^2 = \tfrac{361}{20},

so

a^2 = \tfrac{361}{100}

(along

x

) and

b^2 = \tfrac{361}{120}.

Then

e^2 = 1 - \dfrac{b^2}{a^2} = 1 - \dfrac{100}{120} = \dfrac16,

so

e = \sqrt{\tfrac16}.

With

m = 1

and

n = 6,

we get

m + n = 7.

Thus, the correct answer is A.

2021 AMC 12A Spring Problem 21

Answer: A

Difficulty rating: 2450

Small hint:

Big hint:

Solution:

Problem 21 in other years