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Let's say that the radius of the outer circle is 5 and the radius of the largest circle in the Apollonian gasket is 3(both from the center of the outer circle) This would mean 8 circles of radius 2 right?

Apollonian Gasket

What I get for calculating the area within the circles of the Apollonian gasket(assuming we go no further than the circles of radius 2) is this equation:

$$πR_{c}^{2} - (πR_{c}^{2} - (πr_{l}^{2} + 8*πr_{s}^{2})) = Area$$

where $πR_{c}^{2}$ is the area of the outer circle, $πr_{l}^{2}$ is the area of the largest inner circle, and $πr_{s}^{2}$ is the area of the smallest inner circle we are considering.

So with the radii given I get this:

$$25π - (25π - (9π + 32π)) = Area$$

$$25π - (25π - 41π) = Area$$

$$25π - (-16π) = Area$$

$$41π = Area$$

So obviously something is wrong with my area formula because if I have a circle of radius 5, no matter how many circles I have in there, the area cannot be any bigger than $25π$.

So where am I wrong? I mean the term in the outer parentheses is supposed to represent the area of the Apollonian gasket not covered by the inner circles and the term in the inner parentheses is supposed to represent the area of the inner circles.

Caters
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  • I think the 8 circles would have $r=1$? – Karolis Juodelė Dec 03 '16 at 19:21
  • But shouldn't the radius of the smallest inner circle you are considering + the radius of the largest inner circle you are considering + any other radii between those(in my case there aren't any) = the radius of the outer circle? With a radius of 5 for the outer circle and 3 for the largest inner circle, this gives me a radius of 2. – Caters Dec 03 '16 at 19:23
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    No, the small circle is on the side, not centered at the origin, so you have computed it's diameter, not radius. – Karolis Juodelė Dec 03 '16 at 19:27
  • So that means that all my steps are off by $24π$ just because of that simple radius mistake and that the true area of the apollonian gasket just considering the inner circles is $17π$? Wow that is a big difference. It is a good thing my area formula for at least that type of apollonian gasket isn't wrong, otherwise I would have to do the subtraction and summation of all the $πr^2$ formulas all over again. – Caters Dec 03 '16 at 19:39
  • This was asked before, http://math.stackexchange.com/questions/563942/what-is-the-area-of-the-apollonian-gaskets. The measure of the complement to the gasket is zero, hence, the area of the union of inner circles is.... – Moishe Kohan Dec 07 '16 at 04:39

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