I searched for the internet, but found nothing relavant to the area.
The areas in each intermediate step form a bounded increasing sequence, so there is a limit. But wil it eventually fill in almost everywhere?
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It sums to twice the area of the outer circle (i.e. the circles fill up everywhere). I come across this result in this paper which talks about how to perform sum over the radii appear in the apollonian gasket. – achille hui Nov 12 '13 at 15:50
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Not twice the area (which makes no sense) but the area of the outer circle. The reason is that the residual set has zero area. – Moishe Kohan Nov 12 '13 at 20:09
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@studiosus, but how to prove that? The series for area seems complicated. – Ash GX Nov 13 '13 at 01:06
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@studiosus the sum I mentioned include the outer circle. – achille hui Nov 13 '13 at 04:27
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@achillehui: Oh, sorry, I did not understand the way you count. – Moishe Kohan Nov 13 '13 at 05:36
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The Hausdorff dimension of the residual set $R$ of the Appolonian gasket is approximately $1.3$ (definitely less than $1.4$), see the discussion and references here. In particular, the 2-dimensional Lebesgue measure of $R$ is zero (this was known, I think, long before 1982). Hence the total area of the inner circles of the gasket is the same as the area of the outer circle.
Moishe Kohan
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