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

%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(1-t^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 Jean-Fran├žois 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/6-x*sqrt(1-x^2))-x) \\ _Charles R Greathouse IV_, Jun 30 2011

%K nonn,cons

%O 0,1

%A _Jean-Fran├žois Alcover_, Jun 30 2011

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Last modified June 21 09:22 EDT 2021. Contains 345358 sequences. (Running on oeis4.)