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to construct the whole gasket. And if any four mutually tangent circles in the gasket have integer curvatures, then all circles in the gasket will have integer curvatures. We can characterise each gasket by giving the curvature of the outermost circle (this will be negative) and the curvatures of the
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To obtain an integer packing one needs four integer curvatures that satisfy DCE, not just three. The first three in the tables happen to produce the fourth, but how were these first 3 obtained to begin with? Are all possible packings accounted for? Also, DCE solutions yield the same packing as .
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Thanks for that, Gandalf61. My guess is that all possible packings could also be accounted for by proceeding in some fashion from the simplest packing and reflecting each curvature in turn. Perhaps a tree structure is the result? On the
Hausdorff dimension calculation can you say what values for
1502:? The figure shown in the article is similar to the one shown on the Ford Circle page, but has more circles and is, I think, more clear. If there are no objections, I will (eventually) create the picture and add a description somewhere in the section on integral packings.
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748:. To avoid this, it would help if you gave more references that discuss integer gaskets - the Lagarias, Mallows and Wilks paper that you reference is about extensions of Descartes' circle theorem, and only mentions integer gaskets in passing.
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1537:"(in the general construction, these three circles have to be different sizes, and they must have a common tangent)" I'm not an expert, but shouldn't that be "three circles may be different sizes"Â ?
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I added a huge section on
Integral Apollonian Circle Packing, and included the reference for what you are thinking about. Feedback on the section is appreciated.
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in order to make clear exactly which are the 4th & 5th circles. As a clueless non-editor, I don't dare attempt such a potentially space-disrupting change.
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but I couldn't add the link to the
External Links section due to potential conflict of interest. Can someone review it and add if appropriate? Thanks!
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I'd love to see this article worked in with a nicely-explained example: The Local-Global
Conjecture for Apollonian circle packings is false
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How does one move from the one characterization to the other? This could be explained a little better. Otherwise, very nice article.
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any topic like this just cries out for illustrations so us folks that aren't so hot at math can understand what's being discussed.
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If we know the curvatures of any three mutually tangent circles in an
Apollonian gasket then we can use repeated applications of
1168:{\displaystyle k_{4}=k_{1}+k_{2}+k_{3}\pm 2{\sqrt {k_{1}k_{2}+k_{2}k_{3}+k_{3}k_{1}}}=2+3+6\pm 2{\sqrt {36}}=-1{\text{ or }}23}
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To answer your second question, and generate the same gasket because, applying
Descartes' theorem to , we have
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to delete these "External links modified" talk page sections if they want to de-clutter talk pages, but see the
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The problem remains that there is currently no diagram accompanying the text that identifies C1, C2, C3, etc.
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If you have discovered URLs which were erroneously considered dead by the bot, you can report them with
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are also integers, and it then follows that all the circles in the gasket will have integer curvature.
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https://web.archive.org/web/20060914030236/http://local.wasp.uwa.edu.au/~pbourke/papers/apollony/
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before doing mass systematic removals. This message is updated dynamically through the template
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https://www.quantamagazine.org/two-students-unravel-a-widely-believed-math-conjecture-20230810/
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is a square number. This gives us a way of systematically generating all integer gaskets. I
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I'm surprised not to see a picture of the special case (0,0,1,1), and it's relationship to
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to be integers, then there are a finite number of possibile triplets for each value of
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If you found an error with any archives or the URLs themselves, you can fix them with
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1498:. Is there any objection to adding one, for example, as in figure 2 of this article
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https://web.archive.org/web/20110502081052/http://closet.zao.se:80/emilk/circles.html
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is the curvature of the bounding circle) then the curvatures of the next two circles
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two largest circles inside the gasket - so we have a triplet where 0 <
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Add: The Local-Global
Conjecture for Apollonian circle packings is false
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https://commons.m.wikimedia.org/File:Apollonian_gasket_construction.svg
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Article should explain from the start that C1, C2, C3 can be any size
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It is possible that someone might suggest that this new section is
652:. Therefore, up to a scaling factor, the gasket must be (-1,2,2,3).
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Each gasket is completely described by the curvatures of its first
267:, before the second comment (Matt me's) above (06:02, 9 Oct 2004,
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column on 11 October 2004. The text of the entry was as follows:
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I have now made changes in the article to correct these errors.
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When you have finished reviewing my changes, please set the
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yeah, this page def needs a pic, or at least a link to one.
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for additional information. I made the following changes:
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http://www.math.ucsd.edu/~ronspubs/03_02_appolonian.pdf
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http://local.wasp.uwa.edu.au/~pbourke/papers/apollony/
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Hi. Can apollonian gaskets be divided in types like :
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Special case (0,0,1,1) and reference to Ford
Circles
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1319:http://www.jasondavies.com/apollonian-gasket/
720:{\displaystyle {\frac {a}{b}}=2{\sqrt {3}}-3}
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