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A307238
This is claimed to be the minimal cut length required to cut a unit circle into 4 pieces of equal area after making certain assumptions about the cuts (compare A307234).
3
3, 9, 4, 5, 7, 0, 2, 9, 6, 7, 2, 6, 7, 1, 8, 5, 7, 1, 3, 8, 4, 2, 8, 9, 9, 5, 5, 2, 1, 1, 1, 7, 9, 9, 1, 8, 8, 8, 7, 4, 8, 3, 5, 4, 0, 1, 0, 7, 4, 7, 4, 1, 5, 2, 4, 2, 6, 8, 1, 6, 9, 6, 7, 1, 3, 1, 8, 7, 4, 3, 2, 9, 8, 3, 8, 1, 6, 2, 0, 0, 8, 4, 8, 7, 8, 5, 1, 4, 7, 7, 3, 8, 6, 0, 2, 1
OFFSET
1,1
COMMENTS
It is assumed that:
all cut edges must be straight-line segments or circular arcs,
the angle between any two cut edges sharing the same point is 120 degrees,
the sum of the curvatures of three cut edges meeting at a point is 0, and
cut edges meeting the unit circle must be perpendicular to the circle.
EXAMPLE
3.945702967267185713842899552111799188874835401074741524...
MATHEMATICA
p[x_]:=Sin[x]/(Sin[Pi/3]+Sin[Pi/3-x]); q[x_]:=Sin[Pi/3-x]/(Sin[Pi/3]+Sin[Pi/3-x]); R[x_]:=q[x]/Tan[x/2]; S[x_]:=(Pi/3 - x -p[x]*Sin[Pi/3 -x] + R[x]^2*(x-Sin[x]))/2; d := FindRoot[S[x] - Pi/8, {x, 0.1, 0.5}, WorkingPrecision -> 150]; RealDigits[2*(p[x] + 2*x*R[x])/.d, 10, 100][[1]] (* G. C. Greubel, Jul 02 2019 *)
PROG
(PARI)
default(realprecision, 100);
p(t)=sin(t)/(sin(Pi/3)+sin(Pi/3-t));
q(t)=sin(Pi/3-t)/(sin(Pi/3)+sin(Pi/3-t));
R(t)=q(t)/tan(t/2);
S(t)=( Pi/3 - t - p(t)*sin(Pi/3-t) + R(t)^2*(t-sin(t)) )/2;
d = solve(t=0.1, 0.5, S(t)-Pi/8);
2*(p(d)+2*d*R(d))
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Zhao Hui Du, Mar 30 2019
EXTENSIONS
Terms a(32) onward added by G. C. Greubel, Jul 02 2019
Edited by N. J. A. Sloane, Aug 16 2019
STATUS
approved