A192408 Decimal expansion of the solution to x = sin( Pi/6 - x*sqrt(1 - x^2) ).

2, 6, 4, 9, 3, 2, 0, 8, 4, 6, 0, 2, 7, 7, 6, 8, 6, 2, 4, 3, 4, 1, 1, 6, 4, 9, 4, 7, 6, 2, 5, 7, 1, 0, 6, 8, 6, 5, 0, 1, 9, 0, 0, 6, 6, 0, 4, 1, 3, 6, 4, 4, 5, 2, 8, 7, 8, 7, 4, 4, 8, 9, 3, 2, 9, 2, 0, 9, 0, 2, 5, 0, 8, 7, 0, 6, 8, 8, 6, 3, 8, 9, 7, 2, 7, 3, 4, 9, 8, 5, 2, 3, 3, 7, 4, 6, 1, 8, 4, 4

Offset: 0

Author

Jean-François Alcover, Jun 30 2011

Keywords

• nonn
• cons

Comments

Trisecting an ellipse area.

Given the ellipse x^2/a^2 + y^2/b^2 = 1, one way to trisect its area is to use the symmetric lines x = s and x = -s, s being the unique real solution to s = a*sin(Pi/6 - (s*sqrt(a^2 - s^2))/a^2).

Setting s = a * t, the equation in t becomes t = sin( Pi/6 - t*sqrt(1 - t^2) ), which is noticeably independent of eccentricity.

In the case of a unit radius circle, total cut length is 4*sqrt(1-t^2) = 3.857068297..., which is quite larger than cutting along 3 radii.

This constant is also the solution to an elementary problem involving two overlapping circles, known as "Mrs. Miniver's problem" (cf. S. R. Finch, p. 487). The distance between the centers of the two circles is 2*x = 0.5298641692...

Examples

			0.26493208460277686243411649476257106865019006604136445287874489329209025087...
		

References

• Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 487.

Links

• L. A. Graham, Mrs Miniver's problem, Ingenious Mathematical Problems and Methods, Dover, 1959, p. 6.
• Eric Weisstein's World of Mathematics, Ellipse.

Programs

Mathematica

RealDigits[ x /. FindRoot[x == Sin[Pi/6 - x*Sqrt[1 - x^2]], {x, 1/4}, WorkingPrecision -> 100]][[1]]

PARI

solve(x=.2,.3,sin(Pi/6-x*sqrt(1-x^2))-x) \\ Charles R Greathouse IV, Jun 30 2011

PARI

sin(solve(x=0,1,sin(x)+x-Pi/3)/2) \\ Gleb Koloskov, Aug 25 2021