# Difference between revisions of "2012 AIME II Problems/Problem 8"

## Problem 8

The complex numbers $z$ and $w$ satisfy the system $$z + \frac{20i}w = 5+i$$ $$w+\frac{12i}z = -4+10i$$ Find the smallest possible value of $\vert zw\vert^2$.

## Solution

Multiplying the two equations together gives us $$zw + 32i - \frac{240}{zw} = -30 + 46i$$ and multiplying by $zw$ then gives us a quadratic in $zw$: $$(zw)^2 + (30-14i)zw - 240 =0.$$ Using the quadratic formula, we find the two possible values of $zw$ to be $7i-15 \pm \sqrt{(15-7i)^2 + 240}$ = $6+2i,$ $12i - 36.$ The smallest possible value of $\vert zw\vert^2$ is then obviously $6^2 + 2^2 = \boxed{040}$.

## See Also

 2012 AIME II (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Invalid username
Login to AoPS