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Decimal expansion of (circumradius)/(inradius) of side-golden right triangle.
4

%I #24 Sep 08 2022 08:45:56

%S 2,6,5,6,8,7,5,7,5,7,3,3,7,5,2,1,5,4,9,4,8,9,7,3,2,1,2,2,3,8,4,0,9,3,

%T 0,2,9,7,2,3,6,6,0,2,5,1,5,7,4,6,5,9,0,7,5,6,5,5,0,2,6,7,4,7,8,9,2,6,

%U 9,2,1,0,7,0,6,6,4,4,7,9,0,8,9,3,4,5,0,4,0,6,5,0,2,2,9,4,3,8,5,5,1,2,0,7,0,6,9,3,7,2,2,9,5,4,2,5,5,5,3,2,7,4,5,2,6,3,0,3,8,1

%N Decimal expansion of (circumradius)/(inradius) of side-golden right triangle.

%C This ratio is invariant of the size of the side-golden right triangle ABC. The shape of ABC is given by sidelengths a,b,c, where a=r*b, and c=sqrt(a^2+b^2), where r=(golden ratio)=(1+sqrt(5))/2. This is the unique right triangle matching the continued fraction [1,1,1,...] of r; i.e, under the side-partitioning procedure described in the 2007 reference, there is exactly 1 removable subtriangle at each stage. (This is analogous to the removal of 1 square at each stage of the partitioning of the golden rectangle as a collection of squares.)

%C Largest root of 4*x^4 - 20*x^2 - 20*x - 5. - _Charles R Greathouse IV_, May 07 2011

%H G. C. Greubel, <a href="/A188594/b188594.txt">Table of n, a(n) for n = 1..5000</a>

%H Clark Kimberling, <a href="http://www.heldermann.de/JGG/JGG11/JGG112/jgg11014.htm">Two kinds of golden triangles, generalized to match continued fractions</a>, Journal for Geometry and Graphics, 11 (2007) 165-171.

%F (circumradius)/(inradius)=abc(a+b+c)/(8*area^2), where area=area(ABC).

%F Equals (sqrt(5) + phi*sqrt(2 + phi))/2, where phi = A001622 is the golden ratio. - _G. C. Greubel_, Nov 23 2017

%e 2.656875757337521549489732...

%t r=(1+5^(1/2))/2; b=1; a=r*b; c=(a^2+b^2)^(1/2);

%t area = (1/4)((a+b+c)(b+c-a)(c+a-b)(a+b-c))^(1/2);

%t RealDigits[N[a*b*c*(a+b+c)/(8*area^2), 130]][[1]]

%t RealDigits[(Sqrt[5] + GoldenRatio*Sqrt[2 + GoldenRatio])/(2),10,50][[1]] (* _G. C. Greubel_, Nov 23 2017 *)

%o (PARI) {phi = (1 + sqrt(5))/2}; (sqrt(5) + phi*sqrt(2 + phi))/2 \\ _G. C. Greubel_, Nov 23 2017

%o (Magma) phi := (1+Sqrt(5))/2; [(Sqrt(5) + phi*Sqrt(2 + phi))/2]; // _G. C. Greubel_, Nov 23 2017

%Y Cf. A001622, A152149, A188595, A188614.

%K nonn,easy,cons

%O 1,1

%A _Clark Kimberling_, Apr 05 2011