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, 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

Author

Zhao Hui Du, Mar 30 2019

Keywords

• nonn
• cons

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.

Examples

			3.945702967267185713842899552111799188874835401074741524...
		

Links

• Zhao Hui Du, Picture showing the minimum cut length
• J. Hu, A Chinese BBS discussing the problem

Crossrefs

Cf. A307234, A207235, A307237.

Programs

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 *)

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))

Extensions

Terms a(32) onward added by G. C. Greubel, Jul 02 2019

Edited by N. J. A. Sloane, Aug 16 2019