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

(Adding problem section)
(Remove extra problem section)
Line 1: Line 1:
 
==Problem==
 
 
== Problem 8 ==
 
== Problem 8 ==
 
The complex numbers <math>z</math> and <math>w</math> satisfy the system <cmath> z + \frac{20i}w = 5+i </cmath>
 
The complex numbers <math>z</math> and <math>w</math> satisfy the system <cmath> z + \frac{20i}w = 5+i </cmath>

Revision as of 17:01, 9 August 2018

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 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. AMC logo.png