%I
%S 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,
%T 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,
%U 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
%N Decimal expansion of the solution to x = sin( Pi/6  x*sqrt(1  x^2) ).
%C Trisecting an ellipse area.
%C 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).
%C Setting s = a * t, the equation in t becomes t = sin( Pi/6  t*sqrt(1  t^2) ), which is noticeably independant of excentricity.
%C In the case of a unit radius circle, total cut length is 4*sqrt(1t^2) = 3.857068297..., which is quite larger than cutting along 3 radii.
%C 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...
%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 487.
%H JeanFrançois Alcover, <a href="/A192408/a192408.gif">L. A. Graham, Ingenious Mathematical Problems and Methods, Dover, 1959, p. 6.</a>
%H Eric Weisstein, <a href="http://mathworld.wolfram.com/Ellipse.html">Ellipse</a>, MathWorld
%e 0.26493208460277686243411649476257106865019006604136445287874489329209025087...
%t RealDigits[ x /. FindRoot[x == Sin[Pi/6  x*Sqrt[1  x^2]], {x, 1/4}, WorkingPrecision > 100]][[1]]
%o (PARI) solve(x=.2,.3,sin(Pi/6x*sqrt(1x^2))x) \\ _Charles R Greathouse IV_, Jun 30 2011
%K nonn,cons
%O 0,1
%A _JeanFrançois Alcover_, Jun 30 2011
