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Topic: 10 Points in a Square (Read 472 times) |
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FiBsTeR
Senior Riddler
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10 Points in a Square
« on: Jun 14th, 2008, 6:44pm » |
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10 points are placed in or on a square of side length 4. Find the least upper bound on the smallest distance between any two of these points. (Source: AoPS) My current upper bound is 1.66525..., though I have no idea if I'm anywhere near the answer. Any thoughts?
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« Last Edit: Jun 14th, 2008, 6:46pm by FiBsTeR » |
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Eigenray
wu::riddles Moderator Uberpuzzler
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Re: 10 Points in a Square
« Reply #1 on: Jun 14th, 2008, 11:32pm » |
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It is approximately 1.6851181759356137310752870426..., or 4 times the smallest positive root of 1180129x^18 - 11436428x^17 + 98015844x^16 - 462103584x^15 + 1145811528x^14 - 1398966480x^13 + 227573920x^12 + 1526909568x^11 - 1038261808x^10 - 2960321792x^9 + 7803109440x^8 - 9722063488x^7 + 7918461504x^6 - 4564076288x^5 + 1899131648x^4 - 563649536x^3 + 114038784x^2 - 14172160x + 819200. No, I didn't just come up with that. Up through 9 points were known by 1964. The optimal pattern for 10 points was found by Schluter in 1971, but not proved optimal until 1990 by de Groot, Peikert, and Wurtz. See: C. de Groot, M. Monagan, R. Peikert, and D.Wurtz, Packing circles in a square: a review and new results, in P. Kall (ed.), System Modelling and Optimization (Proc. 15th IFIP Conf., Zurich, 1991), pp. 45–54, Lecture Notes in Control and Information Sciences, Vol. 180, Springer-Verlag, Berlin, 1992. This has optimal solutions for up through 20 points, and several references. With the exception of n<10, n=14,16,25,36, all optimality proofs have been by computer. Edit: actually the paper is freely available here. See here for up to 50 points. Note the patterns for n=1,4,9,16,25,36, but not 49!
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« Last Edit: Jun 14th, 2008, 11:56pm by Eigenray » |
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FiBsTeR
Senior Riddler
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Re: 10 Points in a Square
« Reply #2 on: Jun 15th, 2008, 6:19am » |
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Thank you, Eigenray! It's just like those middle schoolers to post problems they don't have solutions to.
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