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 $6^2 + 2^2 = \boxed{040}$ .

2012 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2012 AIME II Problems/Problem 8

Problem 8

Solution

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