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A181925 Area A of the triangles such that A, the sides, and at least one of the three bisectors are integers.

%I A181925 #29 Feb 16 2025 08:33:13
%S A181925 12,48,60,108,120,168,192,240,300,360,420,432,480,540,588,660,672,768,
%T A181925 960,972,1008,1080,1092,1200,1260,1344,1440,1452,1500,1512,1680,1728,
%U A181925 1848,1920,1980,2028,2160,2352,2448,2520,2640,2688,2700,2772,2940,3000
%N A181925 Area A of the triangles such that A, the sides, and at least one of the three bisectors are integers.
%C A181925 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 A181925 The lengths of the bisectors are given by:
%C A181925 b1 = sqrt(bc*(b+c-a)(a+b+c))/(b+c)
%C A181925 b2 = sqrt(ac*(a+c-b)(a+b+c))/(a+c)
%C A181925 b3 = sqrt(ab*(a+b-c)(a+b+c))/(a+b)
%C A181925 Properties of this sequence: There exist three subsets of numbers included in a(n):
%C A181925 Case (i): A subset with a majority of isosceles triangles whose area equals the sum of the areas of two Pythagorean triangles with integer sides => the sequence A118903 is included in this sequence. This sort of triangles contains generally only one integer bisector, but more rarely three integer bisectors (see the examples).
%C A181925 Case (ii): Right triangles (a,b,c) where a^2 + b^2 = c^2.
%C A181925 Case (iii): A class of non-isosceles and non-right triangles (a, b, c) where one, two or three bisectors are integers.
%D A181925 Ralph H. Buchholz, On triangles with rational altitudes, angles bisectors or medians, PhD Thesis, University of Newcastle, Nov 1989.
%H A181925 Ralph H. Buchholz, <a href="http://www.scribd.com/doc/78239615/39/Two-Median-Heron-Triangles">Triangles with two integer medians</a>
%H A181925 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/AngleBisector.html">Angle Bisectors</a>
%e A181925 Case (i): 12 is in the sequence because the area of the isosceles triangle (5, 5, 6) equals 12 and one of the bisectors is an integer (4). But the isosceles triangle (546, 975, 975) whose area equals 255528 contains three integer bisectors: 936, 560, 560.
%e A181925 Case (ii): The right triangle (28, 96, 100) => A = 1344, and the integer median is m = 35.
%e A181925 Case (iii): The triangle (31091676, 46267375, 62553491) => A =  690494511777840, and the three bisectors are 51555075, 38342304 and 22314600.
%p A181925 with(numtheory):T:=array(1..1000):k:=0:nn:=300:for a from 1 to nn do: for b from a to nn do: for c from b to nn do:p:=(a+b+c)/2:s:=p*(p-a)*(p-b)*(p-c):aa:=b*c*(b+c-a)*(a+b+c): bb:=a*c*(a+c-b)*(a+b+c): cc:=a*b*(a+b-c)*(a+b+c):if s>0 and aa>0 and bb>0 and cc>0 then s1:=sqrt(s): aa1:=sqrt(aa)/(b+c): bb1:=sqrt(bb)/(a+c): cc1:=sqrt(cc)/(a+b):if s1=floor(s1) and (aa1=floor(aa1) or bb1=floor(bb1) or cc1=floor(cc1))  then k:=k+1:T[k]:=s1:else fi:fi:od:od:od: L := [seq(T[i],i=1..k)]:L1:=convert(T,set):A:=sort(L1, `<`): print(A):
%t A181925 nn=300; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s (s-a) (s-b) (s-c); aa=b*c*(b+c-a)*(a+b+c); bb=a*c*(a+c-b)*(a+b+c); cc=a*b*(a+b-c)*(a+b+c); If[0 < area2 && aa > 0 && bb > 0 && cc > 0 && IntegerQ[Sqrt[area2]] && (IntegerQ[Sqrt[aa]/(b+c)] || IntegerQ[Sqrt[bb]/(a+c)] || IntegerQ[Sqrt[cc]/(a+b)]), AppendTo[lst, Sqrt[area2]]]], {a,nn}, {b,a}, {c,b}]; Union[lst]
%Y A181925 Cf. A118903, A188158.
%K A181925 nonn
%O A181925 1,1
%A A181925 _Michel Lagneau_, Apr 02 2012