Thread: Problem 47
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Old March 1st, 2008, 09:50 PM
TwistedOne151 TwistedOne151 is offline
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3. This proof isn't entirely rigorous, but contains all the major steps:

Let D be the point where C_1 and C_2 are tangent. Now let us define a Möbius transform z'=M(z) that maps D to infinity, and let us indicate images under M by primes. Now, M maps C_1 and C_2 to parallel lines C_1' and C_2'. Thus c_1',c_2,'\ldots,c_k' are circles tangent to the lines C_1' and C_2' and to the adjacent circles in the sequence. As they are tangent to both C_1' and C_2', they are all the same size, and their centers lie on the line C_3' parallel to C_1' and C_2' and halfway between them. We see also that a_1',a_2,'\ldots,a_{k-1}' must lie on C_3' as well. Thus if we have the circle C_3=M^{-1}(C_3') in the pre-image, we see a_1,a_2,\ldots,a_{k-1} all lie on C_3.

--Kevin C.
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