MathGrit
ProblemsTechniquesPricing
Sign inGet started
Back to problems

Hexagon Vertices Random Product Probability

1750AdeptComplex NumbersCombinatoricsProbability

AMC 12A · 2017 · Problem 25

The vertices $V$ of a centrally symmetric hexagon in the complex plane are given by $$V=\left\{ \sqrt{2}i,-\sqrt{2}i, \frac{1}{\sqrt{8}}(1+i),\frac{1}{\sqrt{8}}(-1+i),\frac{1}{\sqrt{8}}(1-i),\frac{1}{\sqrt{8}}(-1-i) \right\}.$$ For each $j$, $1\leq j\leq 12$, 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 $P=-1$?

Answer choices

A
\dfrac{5\cdot11}{3^{10}}
B
\dfrac{5^2\cdot11}{2\cdot3^{10}}
C
\dfrac{5\cdot11}{3^{9}}
D
\dfrac{5\cdot7\cdot11}{2\cdot3^{10}}
E
\dfrac{2^2\cdot5\cdot11}{3^{10}}
0 students attempted0% solvedRating 1750

Related practice paths

AMC 12 PracticeAdvanced high school contest practice and review.AMC 10 vs AMC 12Choose the right practice path and difficulty level.AIME Practice StrategyHow to improve accuracy on high-difficulty problems.

Ready to check your answer?

Create an account to submit answers, save history, and track your rating.

Progressive Hints5

Unlock hints one at a time — each reveals a little more without spoiling the solution.

Step-by-Step Solutions1

Multiple solution approaches with detailed walkthroughs, unlocked after you solve the problem.

AI-Powered Grading

Instant feedback on your answer — handles fractions, decimals, and equivalent forms.

Curated problem bank

Supported tracks for AMC, AIME, MATHCOUNTS, and olympiad-style training, plus global problem sources like UKMT, Euclid, and Kangaroo.