OFFSET
0,1
COMMENTS
Given an equilateral triangle of unit length with two cevians drawn from one vertex to the side opposite that divide the equilateral triangle into 3 subtriangles. Adjust these cevians so that the 3 subtriangles all have congruent incircles. Then the real positive root of 48*x^4 + 16*x^3 - 27*x^2 - 18*x - 3 = 0 gives x = (3^(1/3)+3^(2/3))/4 = 0.880583348... as the length of these cevians and the radius of the three congruent incircles is given by A/(s+2*x) where A is the area and s the semiperimeter of the equilateral triangle. Hence the congruent inradius = sqrt(3)/(2*(3+3^(2/3)+3^(1/3))).
A cubic number with denominator 4. - Charles R Greathouse IV, Aug 26 2017
LINKS
FORMULA
Equals (3^(1/3)+3^(2/3))/4.
16*x^3 - 9*x - 3 is the irreducible polynomial. - Michael Somos, Jun 09 2014
EXAMPLE
0.880583348339828124212923783784498743682418648468...
MATHEMATICA
N[Select[x/.Solve[48x^4+16x^3-27x^2-18x-3==0, {x}], Im[#]==0&&Re[#]>0 &], 100]
PROG
(PARI) polrootsreal(16*x^3-9*x-3)[1] \\ Charles R Greathouse IV, Aug 26 2017
CROSSREFS
KEYWORD
AUTHOR
Frank M Jackson, Jun 05 2014
STATUS
approved