Difference between revisions of "1992 AIME Problems/Problem 10"
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== Solution == | == Solution == | ||
| − | + | Let <math>z=a+bi</math>. | |
| + | <math>\frac{z}{40}=\frac{a}{40}+\frac{b}{40}i</math> | ||
| + | |||
| + | Therefore, we have the inequality | ||
| + | |||
| + | <math>0\leq a,b \leq 40</math> | ||
| + | |||
| + | <math>\frac{40}{\overline{z}}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i</math> | ||
| + | |||
| + | <math>0\leq a,b \leq \frac{a^2+b^2}{40}</math> | ||
| + | |||
| + | We graph them: | ||
| + | |||
| + | {{image}} | ||
| + | |||
| + | Doing a little geometry, the area of the intersection of those three graphs is <math>200\pi -400\approx 228.30</math> | ||
| + | |||
| + | <math>\boxed{228}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=1992|num-b=9|num-a=11}} | {{AIME box|year=1992|num-b=9|num-a=11}} | ||
[[Category:Intermediate Complex Numbers Problems]] | [[Category:Intermediate Complex Numbers Problems]] | ||
Revision as of 20:56, 11 November 2007
Problem
Consider the region 
in the complex plane that consists of all points 
such that both 
and 
have real and imaginary parts between 
and 
, inclusive. What is the integer that is nearest the area of ?
Solution
Let 
.
Therefore, we have the inequality
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 
See also
| 1992 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
