

A093603


Bisecting a triangular cake using a curved cut of minimal length: decimal expansion of sqrt(Pi/sqrt(3))/2 = d/2, where d^2 = Pi/sqrt(3).


2



6, 7, 3, 3, 8, 6, 8, 4, 3, 5, 4, 4, 2, 9, 9, 1, 8, 0, 3, 0, 9, 5, 4, 0, 1, 1, 8, 7, 7, 3, 0, 8, 2, 1, 6, 6, 7, 7, 2, 1, 6, 7, 7, 0, 1, 8, 2, 7, 0, 0, 3, 9, 7, 3, 0, 9, 9, 8, 0, 1, 6, 6, 1, 3, 7, 3, 7, 9, 7, 9, 0, 1, 8, 2, 6, 2, 9, 5, 5, 0, 3, 2, 0, 0, 8, 2, 8, 3, 1, 5, 0, 3, 0, 7, 7, 5, 9, 6, 1, 5, 3, 8, 6, 4, 6
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OFFSET

0,1


COMMENTS

A minimal dissection. The number d/2 = sqrt(Pi/sqrt(3))/2 = sqrt(Pi)/(2*3^(1/4)) gives the length of the shortest cut that bisects a unitsided equilateral triangle. From A093602, it is plain that d^2 < 2, i.e., (d/2)^2 < 1/2 = square of the bisecting line segment parallel to the triangle's side. d/2 actually is the arc subtending the angle Pi/3 about the center of the circle with radius D/2, where D^2 = 3/d^2. Since Pi/3~1, d~D (see A093604).


REFERENCES

P. Halmos, Problems for Mathematicians Young and Old, Math. Assoc. of Amer. Washington DC 1991.
C. W. Triggs, Mathematical Quickies, Dover NY 1985.


LINKS

G. C. Greubel, Table of n, a(n) for n = 0..5000
Scott Carr, Bisecting an arbitrary triangular cake (with a straight cut of shortest length)


FORMULA

This is sqrt(Pi)/(2*3^(1/4)).


EXAMPLE

0.67338684354429918030954011877308216677216770182700......


MATHEMATICA

RealDigits[Sqrt[Pi]/(2*3^(1/4)), 10, 50][[1]] (* G. C. Greubel, Jan 13 2017 *)


PROG

(PARI) sqrt(Pi/sqrt(3))/2 \\ G. C. Greubel, Jan 13 2017


CROSSREFS

Cf. A093604.
Sequence in context: A277135 A153628 A154972 * A245513 A105739 A105831
Adjacent sequences: A093600 A093601 A093602 * A093604 A093605 A093606


KEYWORD

easy,nonn,cons


AUTHOR

Lekraj Beedassy, May 14 2004


STATUS

approved



