Difference between revisions of "1992 AIME Problems/Problem 10"
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== Problem == | == 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}{\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>? | ||
+ | <math>If </math>z=a+bi<math> with </math>a<math> and </math>b<math> real, then </math>z=a-bi<math> is the conjugate of </math>z$) | ||
== Solution == | == Solution == |
Revision as of 20:12, 7 July 2015
Problem
Consider the region in the complex plane that consists of all points
such that both
and
have real and imaginary parts between
and
, inclusive. What is the integer that is nearest the area of
?
z=a+bi
a
b
z=a-bi
z$)
Solution
Let . Since
we have the inequality
which is a square of side length
.
Also, so we have
, which leads to:
We graph them:
![AIME 1992 Solution 10.png](https://wiki-images.artofproblemsolving.com//1/1b/AIME_1992_Solution_10.png)
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