2018 AMC 12A Problem 22

Problem 22 of 25HarderAlgebra Geometry

The solutions to the equations

z^2 = 4 + 4\sqrt{15}\,i

and

z^2 = 2 + 2\sqrt{3}\,i,

where

i = \sqrt{-1},

form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form

p\sqrt{q} - r\sqrt{s},

where

p,

q,

r,

and

s

are positive integers and neither

q

nor

s

is divisible by the square of any prime number. What is

p + q + r + s?

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Solution

Writing

z = a + bi

with

(a + bi)^2 = 4 + 4\sqrt{15}\,i

gives

a^2 - b^2 = 4

and

2ab = 4\sqrt{15}.

Then

a^4 - 4a^2 - 60 = 0,

(a^2 - 10)(a^2 + 6) = 0,

yielding

a = \pm\sqrt{10},

b = \pm\sqrt{6}.

The vertices from the first equation are

\pm(\sqrt{10} + \sqrt{6}\,i).

The same method on

z^2 = 2 + 2\sqrt{3}\,i

gives

\pm(\sqrt{3} + i).

Applying the shoelace formula to

(\sqrt{10}, \sqrt{6}),

(\sqrt{3}, 1),

(-\sqrt{10}, -\sqrt{6}),

(-\sqrt{3}, -1)

gives area

6\sqrt{2} - 2\sqrt{10}.

Thus

p + q + r + s

= 6 + 2 + 2 + 10

= 20.

Thus, the correct answer is A.