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A307238
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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).
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3
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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3.945702967267185713842899552111799188874835401074741524...
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MATHEMATICA
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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 *)
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PROG
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(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))
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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