Question

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 $P=-1?$

$\dfrac{5\cdot11}{3^{10}}$

$\dfrac{5^2\cdot11}{2\cdot3^{10}}$

$\dfrac{5\cdot11}{3^9}$

$\dfrac{5\cdot7\cdot11}{2\cdot3^{10}}$

$\dfrac{2^2\cdot5\cdot11}{3^{10}}$

Answer: E

Concepts:complex number binomial probability

Difficulty rating: 2650

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.

2017 AMC 12A Problem 25

Answer: E

Difficulty rating: 2650

Solution:

Problem 25 in Other Years