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%I #22 Jan 13 2017 02:28:07
%S 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,
%T 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,
%U 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
%N 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).
%C 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).
%D P. Halmos, Problems for Mathematicians Young and Old, Math. Assoc. of Amer. Washington DC 1991.
%D C. W. Triggs, Mathematical Quickies, Dover NY 1985.
%H G. C. Greubel, <a href="/A093603/b093603.txt">Table of n, a(n) for n = 0..5000</a>
%H Scott Carr, <a href="http://datagenetics.com/blog/may22016/index.html">Bisecting an arbitrary triangular cake</a> (with a straight cut of shortest length)
%F This is sqrt(Pi)/(2*3^(1/4)).
%e 0.67338684354429918030954011877308216677216770182700......
%t RealDigits[Sqrt[Pi]/(2*3^(1/4)), 10, 50][[1]] (* _G. C. Greubel_, Jan 13 2017 *)
%o (PARI) sqrt(Pi/sqrt(3))/2 \\ _G. C. Greubel_, Jan 13 2017
%Y Cf. A093604.
%K easy,nonn,cons
%O 0,1
%A _Lekraj Beedassy_, May 14 2004
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