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Jan 13, 2012 at 19:30 comment added Will Jagy @S. Carnahan Thanks, I asked Igor the same thing, or tried to, I don't see that his argument applies to any possible incomplete graph on $n$ vertices. I think he was just showing me the induction step in the answer in his book, which is about the complete graphs.
Jan 13, 2012 at 7:34 comment added S. Carnahan In $\mathb{R}^{n-1}$, you have enough room to suitably perturb the simplex by some small $\epsilon$ without changing the sizes of the discs. If I'm not mistaken, you may also surround such a formation of discs with a $n+1$st disc turned inside-out (i.e., you can get a valid formation by a Möbius transformation).
Jan 13, 2012 at 6:21 history edited Will Jagy CC BY-SA 3.0
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Jan 13, 2012 at 3:43 history answered Will Jagy CC BY-SA 3.0