A093603 - OEIS

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).

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 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

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 *)