Difference between revisions of "2012 AIME II Problems/Problem 8"
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== Solution == | == 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{(15-7i)^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 | + | 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{(15-7i)^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 == | == See Also == | ||
{{AIME box|year=2012|n=II|num-b=7|num-a=9}} | {{AIME box|year=2012|n=II|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:34, 29 February 2020
Problem 8
The complex numbers 
and 
satisfy the system ![\[z + \frac{20i}w = 5+i\]](http://latex.artofproblemsolving.com/a/5/6/a5690e0eaacffab8de4e2d90581025dec061fe5c.png)
![\[w+\frac{12i}z = -4+10i\]](http://latex.artofproblemsolving.com/9/4/f/94fda964f36c3465d4b09a2f2983cde550239cc6.png)
Find the smallest possible value of 
.
Solution
Multiplying the two equations together gives us ![\[zw + 32i - \frac{240}{zw} = -30 + 46i\]](http://latex.artofproblemsolving.com/5/f/8/5f87ccc5184e1f0c230daba70a62bd32355770d1.png)
and multiplying by 
then gives us a quadratic in : ![\[(zw)^2 + (30-14i)zw - 240 =0.\]](http://latex.artofproblemsolving.com/e/b/2/eb2b9c78149a8b0b64bb3ae792f7e1607b6e33da.png)
Using the quadratic formula, we find the two possible values of to be 
= 
The smallest possible value of is then obviously 
.
See Also
| 2012 AIME II (Problems • Answer Key • Resources) | ||
| 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. 
