Difference between revisions of "1992 AIME Problems/Problem 10"
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(Undo revision 71028 by Starwars123 (talk)) |
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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>? | ||
+ | (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</math>) | ||
== Solution == | == Solution == |
Revision as of 12:37, 8 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
?
(If
with
and
real, then
is the conjugate of
)
Solution
Let . Since
we have the inequality
which is a square of side length
.
Also, so we have
, which leads to:
We graph them:

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