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Isoperimetric problems for tilings, corrigendum

Published online by Cambridge University Press:  26 February 2010

L. Fejes Tóth
Affiliation:
Mathematical Institute of the Hungarian Academy of Sciences, Budapest, V. Reāltanode u. 13-15, Hungary.
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Extract

In [1] I made the following conjecture. If we tile a convex polygon of at most six sides with N convex tiles of areas a1, …,aN, then the total perimeter of the tiles is never less than the total perimeter of N regular hexagons of areas a1, …, aN. In order to show that the condition of the convexity of the tiles cannot be omitted I constructed a tiling with an equal number of “pentagons” and “heptagons” of perimeters and areas and , and ā5 and ā7, respectively. In a letter to me R. Schneider kindly pointed out that, in contrast to the value given in [1], the value of the quotient was equal to 3.7263… This being greater than √8√3 = 3–7224… the tiling given in [1] does not yield the desired counter-example. In what follows we shall construct a tiling for which the respective quotient will turn out to be less than √8√3.

Type
Correction
Copyright
Copyright © University College London 1986

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References

1.TÓth, L. Fejes. Isoperimetric problems for tilings. Mathematika, 32 (1985), 1015.CrossRefGoogle Scholar