# 1992 AIME Problems/Problem 10

## 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 $A$?

## 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: To find the area outside the two circles but inside the square, we want to find the unique area of the two circles. We can do this by adding the area of the two circles and then subtracting out their overlap. There are two methods of finding the area of overlap:

1. Consider that the area is just the quarter-circle with radius $20$ minus an isosceles right triangle with base length $20$, and then doubled (to consider the entire overlapped area)

2. Consider that the circles can be converted into polar coordinates, and their equations are $r = 40sin\theta$ and $r = 40cos\theta$. Using calculus with the appropriate bounds, we can compute the overlapped area.

Using either method, we compute the overlapped area to be $200\pi + 400$, and so the area of the intersection of those three graphs is $40^2-(200\pi + 400) \Rightarrow 1200 - 200\pi \approx 571.68$ $\boxed{572}$

## See also

 1992 AIME (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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