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%I #13 Sep 06 2021 04:23:27
%S 1,2,3,4,5,6,7,8,9,10,11,12,15,16,18,20,22,24,25,27,28,29,30,32,33,34,
%T 35,36,38,40,42,44,45,48,49,50,54,55,56,58,60,62,63,64,66,68,70,72,75,
%U 76,77,78,80,81,84,88,90,95,96,98,99,100,108,110,112,114,116
%N Integer area A of triangles having their side lengths in the commutative ring Z[phi] where phi is the golden ratio.
%C Generalized integer areas triangles in the ring Z[phi] = {a + b*phi| a,b in Z}. Z[phi] is a ring because if x = a + b*phi and y = c + d*phi are in the ring, the sum x+y = a+c + (b+d)*phi is in the ring, and the product x*y = (a*c + b*d) + (a*d + b*c + b*d)*phi is also in the ring.
%C This sequence is tested with a and b in the range [-40, ..., +40]. For the values of areas > 150 it is necessary to expand the range of variation, but nevertheless the calculations become very long.
%C The sequence A188158 is included in this sequence. The numbers 5*a(n) are in the sequence because if the integer area of the integer-sided triangle (a, b, c) is A, the area of the triangle of sides (a*sqrt(5), b*sqrt(5), c*sqrt(5)) is 5*A, where sqrt(5)= -1 + 2*phi.
%C The primitive areas are p = 1, 2, 3, 6, 7, 11, 22, ... and the areas p^2*a(n) are also in the sequence.
%C The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.
%C For the same area, the number of triangles is not unique, for example the area of the triangles (1,5,-2+4*phi), (2,2*phi-1, 2*phi-1),(3,3-phi,2+phi) and (4,2*phi-1, 2*phi-1) is A = 2.
%C It is possible to obtain rational values and also values in the ring Z[phi] for the circumradius (see the table below).
%C The following table gives the first values (A, a, b, c, R) where A is the integer area, a,b,c are the sides in Z[phi] and R = a*b*c/(4*A) are the values of circumradius.
%C ----------------------------------------------------------------
%C | A | a | b | c | R |
%C ----------------------------------------------------------------
%C | 1 | 1 | 2 | -1 + 2*phi | sqrt(5)/2 |
%C | 2 | 1 | 5 | -2 + 4*phi | 5*sqrt(5)/4 |
%C | 3 | 3 | -1 + 2*phi | -2 + 4*phi | 5/2 |
%C | 4 | 2 | 4 | -2 + 4*phi | sqrt(5) = -1 + 2*phi |
%C | 5 | 2 | 13 | -5 + 10*phi | 13*sqrt(5)/2 |
%C | 6 | 3 | 4 | 5 | 5/2 |
%C | 7 | 7 | -2 + 4*phi | -5 + 10*phi | 25/2 |
%C | 8 | 5 | 13 | -8 + 16*phi | 65*sqrt(5)/4 |
%C | 10 | 5 | 5 | -2 + 4*phi | 5*sqrt(5)/4 |
%C | 11 | 2 | 11 | -5 + 10*phi | 5*sqrt(5)/2 |
%C | 12 | 4 | 10 | -6 + 12*phi | 5*sqrt(5) = -5 + 10*phi|
%C | 15 | 5 | 10 | -3 + 6*phi | 5*sqrt(5)/2 |
%C | 16 | 4 | 8 | -4 + 8*phi | 2*sqrt(5) = -2 + 4*phi |
%C | 18 | 3 | 15 | -6 + 12*phi | 15*sqrt(5)/4 |
%C | 20 | 10 | -2 + 4*phi | -4 + 8*phi | 15*sqrt(5)/4 |
%C ----------------------------------------------------------------
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Ring.html">Ring</a>
%t err=1/10^10;nn=40;q=(Sqrt[5]+1)/2;lst={};lst1={};Do[If[u+q*v>0,lst=Union[lst,{u+q*v}]],{u,nn,-nn,-1},{v,nn,-nn,-1}];n1=Length[lst];Do[a=Part[lst,i];b=Part[lst,j];c=Part[lst,k];s=(a+b+c)/2;area2=s*(s-a)*(s-b)*(s-c);If[a*b*c!=0&&N[area2]>0&&Abs[N[Sqrt[area2]]-Round[N[Sqrt[area2]]]]<err,AppendTo[lst1,Round[Sqrt[N[area2]]]]],{i,1,n1},{j,i,n1},{k,j,n1}];Union[lst1]
%Y Cf. A188158.
%K nonn
%O 1,2
%A _Michel Lagneau_, Feb 25 2014
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