Talk:Borromean rings
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Can this arrangement be twisted about into a Mobius Strip? THAT would be weird. Same surface continuity, no two rings connected, yet "one" complex. I can't get my mind around that! bt -- unsigned comment by anonymous IP 68.102.13.50, 06:48, 26 August 2006
[edit] clarification
Why wouldn't you be able to form this figure from regular circles? Or should it say from two-dimensional figures, as the linkages require overlap? -- nae'blis (talk) 20:27, 1 February 2006 (UTC)
- No -- you can't do it with exact geometric circles. Take a close look at the picture. Michael Hardy 21:19, 1 February 2006 (UTC)
- I didn't mean you can't make 2-dimensional pictures of it with exact circles. In the first place, any two of the circles would have to be in two different planes; otherwise they would have common points rather than being linked. That means we need to embed them in a three-dimensional space. I was speaking of the actual circles, not of pictures of them. So is this article. Michael Hardy 23:46, 1 February 2006 (UTC)
- OK, I've found a reference: B. Lindström, "Borromean Circles are Impossible", American Mathematical Monthly, volume 98 (1991), pages 340—341. I'm going to add that to the article.
- But anyway, it seems you had in mind 2-dimensional pictures in which the circles are perfect circles. No one has said that those are impossible, and you already see those in the article. Michael Hardy 23:59, 1 February 2006 (UTC)
I've not seen the Math Monthly article, but I learned about it first from a short note by Ian Agol. The proof is fairly simple but ingenious. I don't know if we need two references, but this has the advantage of being freely available through the Internet (instead of say, through JSTOR). --Chan-Ho (Talk) 01:34, 2 February 2006 (UTC)
