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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 (list; constant; graph; refs; listen; history; text; internal format)
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 unit-sided 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
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
KEYWORD
easy,nonn,cons
AUTHOR
Lekraj Beedassy, May 14 2004
STATUS
approved

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Last modified May 1 16:12 EDT 2024. Contains 372175 sequences. (Running on oeis4.)