Difference between revisions of "1992 AIME Problems/Problem 10"
Starwars123 (talk | contribs) (→See also) |
Starwars123 (talk | contribs) (→Problem) |
||
| Line 1: | Line 1: | ||
== 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 ![\[0\leq a,b \leq 40\]](http://latex.artofproblemsolving.com/3/2/2/322dfb47bbc99f5371df34c835fbbfb8a0c07cba.png)
which is a square of side length 
.
Also, 
so we have 
, which leads to:![\[(a-20)^2+b^2\geq 20^2\]](http://latex.artofproblemsolving.com/a/2/a/a2abd8f5678c0cb9cebe1baf8ab5462cb46d6efc.png)
![\[a^2+(b-20)^2\geq 20^2\]](http://latex.artofproblemsolving.com/8/7/3/873e2f6de32b7d6432c3e28734d5ace419fd85ab.png)
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 
