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 A192408 Decimal expansion of the solution to x = sin( Pi/6 - x*sqrt(1 - x^2) ). 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 (list; constant; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 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 independant of excentricity. 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... REFERENCES Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 487. LINKS Jean-François Alcover, L. A. Graham, Ingenious Mathematical Problems and Methods, Dover, 1959, p. 6. Eric Weisstein, Ellipse, MathWorld EXAMPLE 0.26493208460277686243411649476257106865019006604136445287874489329209025087... MATHEMATICA RealDigits[ x /. FindRoot[x == Sin[Pi/6 - x*Sqrt[1 - x^2]], {x, 1/4}, WorkingPrecision -> 100]][[1]] PROG (PARI) solve(x=.2, .3, sin(Pi/6-x*sqrt(1-x^2))-x) \\ Charles R Greathouse IV, Jun 30 2011 CROSSREFS Sequence in context: A324651 A342808 A201895 * A074208 A333775 A334205 Adjacent sequences:  A192405 A192406 A192407 * A192409 A192410 A192411 KEYWORD nonn,cons AUTHOR Jean-François Alcover, Jun 30 2011 STATUS approved

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Last modified May 17 05:13 EDT 2021. Contains 343965 sequences. (Running on oeis4.)