1992 AIME Problems/Problem 10: Difference between revisions

Revision as of 21:51, 11 November 2007

Problem

Consider the region $A$ in the complex plane that consists of all points $z$ such that both $\frac{z}{40}$ and $\frac{40}{z}$ have real and imaginary parts between $0$ and $1$ , inclusive. What is the integer that is nearest the area of $A$ ?

Solution

Let $z=a+bi$ .

$\frac{z}{40}=\frac{a}{40}+\frac{b}{40}i$

Therefore, we have the inequality

$0\leq a,b \leq 40$

$\frac{40}{\overline{z}}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i$

$0\leq a,b \leq \frac{a^2+b^2}{40}$

We graph them:

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Doing a little geometry, the area of the intersection of those three graphs is $200\pi -400\approx 228.30$

$\boxed{228}$

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 == Problem ==
-Consider the region <math>A^{}_{}</math> in the complex plane that consists of all points <math>z^{}_{}</math> such that both <math>\frac{z^{}_{}}{40}</math> and <math>\frac{40^{}_{}}{\overline{z}}</math> have real and imaginary parts between <math>0^{}_{}</math> and <math>1^{}_{}</math>, inclusive. What is the integer that is nearest the area of <math>A^{}_{}</math>?
+Consider the region <math>A</math> in the complex plane that consists of all points <math>z</math> such that both <math>\frac{z}{40}</math> and <math>\frac{40}{z}</math> have real and imaginary parts between <math>0</math> and <math>1</math>, inclusive. What is the integer that is nearest the area of <math>A</math>?
 == Solution ==

1992 AIME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1992 AIME Problems/Problem 10: Difference between revisions

Revision as of 21:51, 11 November 2007

Problem

Solution

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