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%I #20 Nov 09 2024 08:07:10
%S 8,8,0,5,8,3,3,4,8,3,3,9,8,2,8,1,2,4,2,1,2,9,2,3,7,8,3,7,8,4,4,9,8,7,
%T 4,3,6,8,2,4,1,8,6,4,8,4,6,8,1,5,3,1,7,1,8,1,1,0,0,1,8,1,8,6,8,5,4,4,
%U 8,4,7,7,0,5,6,8,1,6,5,2,8,3,6,5,2
%N Decimal expansion of the real positive root of 48*x^4 + 16*x^3 - 27*x^2 - 18*x - 3 = 0.
%C 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))).
%C A cubic number with denominator 4. - _Charles R Greathouse IV_, Aug 26 2017
%H Don McConnell, <a href="http://math.stackexchange.com/questions/781810">Three Congruent Incircles of a divided Equilateral triangle</a>
%H Paul Yiu, <a href="http://www.math-cs.ucmo.edu/~mjms/2003.1/pyiu.pdf">The Congruent-incircle Cevians of a triangle</a>
%H <a href="/index/Al#algebraic_03">Index entries for algebraic numbers, degree 3</a>
%F Equals (3^(1/3)+3^(2/3))/4.
%F 16*x^3 - 9*x - 3 is the irreducible polynomial. - _Michael Somos_, Jun 09 2014
%e 0.880583348339828124212923783784498743682418648468...
%t N[Select[x/.Solve[48x^4+16x^3-27x^2-18x-3==0, {x}], Im[#]==0&&Re[#]>0 &], 100]
%o (PARI) polrootsreal(16*x^3-9*x-3)[1] \\ _Charles R Greathouse IV_, Aug 26 2017
%K nonn,cons,easy
%O 0,1
%A _Frank M Jackson_, Jun 05 2014