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Crystal Math

Source: AMC 12 2017A/25

February 8, 2017
AMCAMC 12probability2017 AMC 12A

Problem Statement

The vertices VV of a centrally symmetric hexagon in the complex plane are given by V={2i,2i,18(1+i),18(1+i),18(1i),18(1i)}.V=\left\{ \sqrt{2}i,-\sqrt{2}i, \frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}. For each jj, 1j121\leq j\leq 12, an element zjz_j is chosen from VV at random, independently of the other choices. Let P=j=112zjP={\prod}_{j=1}^{12}z_j be the product of the 1212 numbers selected. What is the probability that P=1P=-1?
<spanclass=latexbold>(A)</span>511310<spanclass=latexbold>(B)</span>52112310<spanclass=latexbold>(C)</span>51139<spanclass=latexbold>(D)</span>57112310<spanclass=latexbold>(E)</span>22511310<span class='latex-bold'>(A) </span> \dfrac{5\cdot11}{3^{10}} \qquad <span class='latex-bold'>(B) </span> \dfrac{5^2\cdot11}{2\cdot3^{10}} \qquad <span class='latex-bold'>(C) </span> \dfrac{5\cdot11}{3^{9}} \qquad <span class='latex-bold'>(D) </span> \dfrac{5\cdot7\cdot11}{2\cdot3^{10}} \qquad <span class='latex-bold'>(E) </span> \dfrac{2^2\cdot5\cdot11}{3^{10}}