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Find the area of the region of points $z$ in the complex plane such that $$\begin{align*}|z+4-4i| &\leq 4\sqrt 2\\ |z-4-4i| &\geq 4\sqrt 2.\end{align*}$$

Source: Art of Problem Solving (AoPS) Intermediate Algebra.

Note: My solution is below. I request verification but also simpler solutions. The problem is marked intermediate algebra, but to solve it, I had to use trig which is not officially required for this course, implying there is a simpler solution.

Solution: The area is $16(4 - \pi)$.

Proof: The equations define two circles, which we'll call $P'$ and $Q'$:

$$\begin{align*}P'&: \text{Center: } -4 + 4i \quad \text{ Radius: } 4 \sqrt 2\\ Q'&: \text{Center: } 4 + 4i \quad \quad \text{ Radius: } 4 \sqrt 2.\end{align*}$$

We want the area of $P' - P' \cap Q'$. To simplify things, we'll dilate everything down by $4 \sqrt 2$, giving us $$\begin{align*}P&: \text{Center: } -\frac{\sqrt 2}2 + \frac{\sqrt 2}2i \quad \text{ Radius: } 1\\ Q&: \text{Center: } \frac{\sqrt 2}2 + \frac{\sqrt 2}2i \quad \quad \text{ Radius: } 1.\end{align*}$$

These are two unit circles with distance between their centers $d = \sqrt 2$. As shown here, letting $\phi = 2 \arccos \frac d 2 = \frac \pi 2$, the area of the intersection is $$\phi - \sin \phi = \frac {\pi} 2 - 1.$$ Therefore, $$\text{Area } [P - P \cap Q] = 1 - (\frac \pi 2 - 1) = 2 - \frac \pi 2$$ and $$\text{Area } [P' - P' \cap Q'] = \left (4 \sqrt 2 \right )^2 (2 - \frac \pi 2) = 16(4 - \pi).$$

SRobertJames
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  • How do you even define the area of a region in the plane if geometry is not to be used? – David K Jul 11 '23 at 01:38
  • @DavidK Fair enough - I edited the question to say "trig is not required." And if we can't find a "simpler" solution: is my solution correct? – SRobertJames Jul 11 '23 at 01:39
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    You don't need trig to recognize an isosceles right triangle. Hence you get the areas of the triangles in your other post by the formula $\frac12ab$ (which is actually $\frac12a^2$ when $a=b$). – David K Jul 11 '23 at 01:40
  • @DavidK Aha! $PX = 1, PM = \frac {\sqrt 2} 2,$ so $MX = \frac {\sqrt 2} 2$ by Pythagorean Theorem, and $A_{\triangle PMX} = \frac 1 4$. And since $\theta + \theta + \frac \pi 2 = \pi$, we have $\theta = \frac \pi 4,$ and $A_{\text{sector } PSX} = \frac \pi 4$. Very nice. – SRobertJames Jul 11 '23 at 03:15

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Write $w=z-4i$, which amounts to a translation, so we intersect the the disk $\mathcal D$ centered in $-4$ with radius $4\sqrt 2$ with the exterior of the disk $\mathcal D'$ centered in $4$ with same radius: $$ \begin{align*} |w+4| &\leq 4\sqrt 2\ ,\\ |w-4| &\geq 4\sqrt 2\ . \end{align*} $$ Picture proof:

area computation for the intersection of a disk with the exterior of the other disk

Let $A,B,C,D$ be the points on the axes at distance $4$ from the origin, as shown in the picture. The square $ABCD$ has sides equal to $4\sqrt 2$. We need the area of the disk $\mathcal D$ centered in $A$ and radius $AB=AD=4\sqrt 2$ after removing the $BD$-reflected disk $\mathcal D'$ from it. The needed area is the brown region, $\frac 34\cdot \pi AB^2$, plus the square area $AB^2$, minus the orange area $\frac 14\cdot \pi CD^2$, this is: $$ \left(\frac 34-\frac 14\right)\pi(4\sqrt 2)^2+(4\sqrt 2)^2 = \frac 12\cdot\pi\cdot 16\cdot 2+ 16\cdot 2 = 16(\pi+2) \ . $$

dan_fulea
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