Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1992 AIME Problems/Problem 10"

(→‎Problem)
(→‎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>?
 
−
(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 19:22, 7 July 2015

Problem

Solution

Let $z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i$. Since $0\leq \frac{a}{40},\frac{b}{40}\leq 1$ we have the inequality \[0\leq a,b \leq 40\]which is a square of side length $40$.

Also, $\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i$ so we have $0\leq a,b \leq \frac{a^2+b^2}{40}$, which leads to:\[(a-20)^2+b^2\geq 20^2\] \[a^2+(b-20)^2\geq 20^2\]

We graph them:

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 $40^2-\frac{40^2}{4}-\frac{1}{2}\pi 20^2\approx 571.68$

$\boxed{572}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_10&oldid=71028"