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 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...
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.
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
(PARI) sin(solve(x=0, 1, sin(x)+x-Pi/3)/2) \\ Gleb Koloskov, Aug 25 2021
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Jean-François Alcover, Jun 30 2011
STATUS
approved