Talk:Problem of Apollonius: Difference between revisions
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::::::::: Oh, I misunderstood then. Ok, yes, looking at the J solution, it explicitly checks for the cases where all circles are interior tangent and logically, something like <code>;(_1^#:i.8) <@apollonius 0 _3 2, 0 0 1,: 0 3 2</code> should treat all the cases, but that's not working for me, and I am going to have to do some debugging to figure out why. (Note, by the way, that I did not write the J implementation here.) --[[User:Rdm|Rdm]] 20:27, 16 September 2011 (UTC)
::::::::: If I turn the <code>while.</code> to <code>whilst.</code> in math/misc/amoeba, then: <lang j> ;(_1^#:i.8) <@apollonius 0 _3 2, 0 0 1,: 0 3 2
0 0 1</
==Turbines==
In general there are eight solutions to this problem. [http://en.wikipedia.org/wiki/File:Apollonius8ColorMultiplyV2.svg see] for a picture showing the eight solutions for a configuration similar to the one depicted in the task description.
Circles are of course passé, in modern geometry they are replaced by an abstract object called a turbine. A turbine is made of modern points, which are like old fashioned points but have an added direction property. A turbine is the set of points which are equidistant from an origin point. If the points point towards the origin it looks like a turbine. If the points point at 90deg to the direction to the origin (tangential) it looks like a circle. If the origin is directed to a particular point then the structure is called a clock. If all the points are tangential in the same direction it is called a cycle.
If we say that two cycles touch only if their directions are the same at this point then if we replace the three circles with three cycles then the problem has a unique cycle as a solution. Of course there are eight ways to replace the three circles with three cycles each of the eight solutions (converted to a cycle) in the picture will solve one of these arrangements.
--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 11:46, 26 July 2013 (UTC)
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