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

Revision as of 16:53, 31 March 2012

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 $7i-15 \pm \sqrt{(7i-15)^2 + 240}$ = $6+2i,$ $12i - 36.$ The smallest possible value of $\vert zw\vert^2$ is then obviously $6^2 + 2^2 = \boxed{040.}$

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 The complex numbers <math>z</math> and <math>w</math> satisfy the system <cmath> z + \frac{20i}w = 5+i \\ \\
 w+\frac{12i}z = -4+10i </cmath> Find the smallest possible value of <math>\vert zw\vert^2</math>.
 == Solution ==
+Multiplying the two equations together gives us <cmath>zw + 32i - \frac{240}{zw} = -30 + 46i</cmath> and multiplying by <math>zw</math> then gives us a quadratic in <math>zw</math>: <cmath>(zw)^2 + (30-14i)zw - 240 =0.</cmath> Using the quadratic formula, we find the two possible values of <math>zw</math> to be <math>7i-15 \pm \sqrt{(7i-15)^2 + 240}</math> = <math>6+2i,</math> <math>12i - 36.</math> The smallest possible value of <math>\vert zw\vert^2</math> is then obviously <math>6^2 + 2^2 = \boxed{040.}</math>
 == See also ==
 {{AIME box|year=2012|n=II|num-b=7|num-a=9}}

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

Revision as of 16:53, 31 March 2012

Problem 8

Solution

See also