2017 AMC 12A Problem 25

The vertices

V

of a centrally symmetric hexagon in the complex plane are given by

V=\left\{\begin{gathered} \sqrt2 i,\;-\sqrt2 i, \\ \tfrac{1}{\sqrt8}(1+i), \\ \tfrac{1}{\sqrt8}(-1+i), \\ \tfrac{1}{\sqrt8}(1-i), \\ \tfrac{1}{\sqrt8}(-1-i) \end{gathered}\right\}.

For each

j,

1\le j\le12,

an element

z_j

is chosen from

V

at random, independently of the other choices. Let

P=\prod_{j=1}^{12}z_j

be the product of the

12

numbers selected. What is the probability that

P=-1?

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Solution

Let

A=\{\sqrt2 i,-\sqrt2 i\}

(each of magnitude

\sqrt2

) and

B

be the other four elements (each of magnitude

\dfrac12

). Since

|P|=(\sqrt2)^{\#A}\left(\tfrac12\right)^{\#B}=1

forces

\#A=8

and

\#B=4,

exactly

8

factors must come from

A

and

4

from

B.

A product of

8

elements of

A

equals

\pm16

(real), and a product of

4

elements of

B

equals one of

\pm\tfrac{1}{16},\pm\tfrac{i}{16}.

Their product is one of

\pm1,\pm i,

each equally likely, so exactly

\tfrac14

of these configurations give

P=-1.

The chance of landing in the

8

-from-

A

4

-from-

B

pattern is

\binom{12}{4}\left(\tfrac13\right)^8\left(\tfrac23\right)^4=\dfrac{880}{3^{10}}.

Multiplying by

\tfrac14

gives

\begin{aligned} P&=\dfrac{1}{4}\cdot\dfrac{880}{3^{10}} \\ &=\dfrac{220}{3^{10}}=\dfrac{2^2\cdot5\cdot11}{3^{10}}. \end{aligned}

Thus, the correct answer is E.