1992 AIME Problems/Problem 10

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+biabz=a-biz$)

== Solution == Let$ (Error compiling LaTeX. Unknown error_msg)z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i0\leq \frac{a}{40},\frac{b}{40}\leq 140$.

Also,$ (Error compiling LaTeX. Unknown error_msg)\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i0\leq a,b \leq \frac{a^2+b^2}{40}$, which leads to:<cmath>(a-20)^2+b^2\geq 20^2</cmath> <cmath>a^2+(b-20)^2\geq 20^2</cmath>

We graph them:

<center>[[Image:AIME_1992_Solution_10.png]]</center>

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$ (Error compiling LaTeX. Unknown error_msg)40^2-\frac{40^2}{4}-\frac{1}{2}\pi 20^2\approx 571.68$$ (Error compiling LaTeX. Unknown error_msg)\boxed{572}$