1992 AIME Problems/Problem 10

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Problem

Consider the region $A$ in the complex plane that consists of all points $z$ such that both $\frac{z}{40}$ and $\frac{40}{\overline{z}}$ have real and imaginary parts between $0$ and $1$, inclusive. What is the integer that is nearest the area of $A$? $If$z=a+bi$with$a$and$b$real, then$z=a-bi$is the conjugate of$z$)

== Solution == Let$ (Error compiling LaTeX. Unknown error_msg)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 <cmath>0\leq a,b \leq 40</cmath>which is a square of side length$40$.

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}i$so we have$0\leq a,b \leq \frac{a^2+b^2}{40}$, 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}$