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| | == Problem == | | == Problem == |
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| − | == Solution ==
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| − | Let <math>z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i</math>. Since <math>0\leq \frac{a}{40},\frac{b}{40}\leq 1</math> we have the inequality <cmath>0\leq a,b \leq 40</cmath>which is a square of side length <math>40</math>.
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| − | Also, <math>\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i</math> so we have <math>0\leq a,b \leq \frac{a^2+b^2}{40}</math>, which leads to:<cmath>(a-20)^2+b^2\geq 20^2</cmath>
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| − | <cmath>a^2+(b-20)^2\geq 20^2</cmath>
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| − | We graph them:
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| − | <center>[[Image:AIME_1992_Solution_10.png]]</center>
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| − | We want the area outside the two circles but inside the square. Doing a little geometry, the area of the intersection of those three graphs is <math>40^2-\frac{40^2}{4}-\frac{1}{2}\pi 20^2\approx 571.68</math>
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| | <math>\boxed{572}</math> | | <math>\boxed{572}</math> |
