cp's OEIS Frontend

A181928 Area A of the triangles such that A, the sides and two medians are integers.

%I A181928 #59 Feb 16 2025 08:33:13
%S A181928 1680,6720,15120,26880,42000,60480,82320,107520,136080,168000,203280,
%T A181928 221760,241920,283920,329280,378000,430080,485520,544320,606480,
%U A181928 672000,740880,813120,887040,888720,967680,1050000,1135680,1224720
%N A181928 Area A of the triangles such that A, the sides and two medians are integers.
%C A181928 The first six primitives triangles (with areas {1680, 221760, 8168160, 95726400, 302793120, 569336866560}) have been discovered by Ralph H. Buchholz and are listed in a table of the chapter 4 of his thesis (see Links).
%C A181928 Later on, Buchholz & Rathbun identified an infinite family of Heronian triangles with 2 integer medians (comprising 4 of the 6 triangles above). The next two primitive triangles in such family have areas 8548588738240320 and 17293367819066194215360. - _Giovanni Resta_, Apr 05 2017
%C A181928 The areas of non-primitive triangles are of the form {1680*k^2}, {221760*k^2}, {8168160*k^2}, {95726400*k^2}, {302793120*k^2}, ...
%C A181928 Using Heron's formula for the area A of a triangle with sides (a, b, c), the existence of a triangle with three rational medians and integer (or rational) area implies a solution of the Diophantine system:
%C A181928 4x^2 = 2a^2 + 2b^2 - c^2
%C A181928 4y^2 = 2a^2 + 2c^2 - b^2
%C A181928 4z^2 = 2b^2 + 2c^2 - a^2
%C A181928 A^2 = s(s-a)(s-b)(s-c)
%C A181928 where s = (a+b+c)/2 is the semiperimeter and x, y, z the medians.
%C A181928 There is no solution known to this system at this time. The problem is similar to the more famous unsolved problem of finding a box with edges, faces diagonals and body diagonals all rational. Such a box also involves seven quantities which must satisfy a system of four Diophantine equations:
%C A181928 d^2 = a^2 + b^2; e^2 = a^2 + c^2; f^2 = b^2 + c^2; g^2 = a^2 + b^2 + c^2
%C A181928 where a, b and c are the lengths of the edges (see Guy in the reference).
%C A181928 Theorems (from Ralph H. Buchholz)
%C A181928 (i) Any triangle with two integer medians has an even semiperimeter.
%C A181928 (ii) If a Heron triangle has two integer medians then its area is divisible by 120.
%C A181928 It seems that, for any n, a(n) == 0 (mod 1680). The reverse is not always true: e.g., as mentioned by _Giovanni Resta_, the triangle with sides (56*k, 61*k, 75*k) has area of the form 1680 * k^2, but it cannot be a term of a(n). - _Sergey Pavlov_, Mar 31 2017
%D A181928 Ralph H. Buchholz, On triangles with rational altitudes, angles bisectors or medians, PHD Thesis, University of Newcastle, Nov 1989.
%H A181928 Ralph H. Buchholz, <a href="http://www.scribd.com/doc/78239615/39/Two-Median-Heron-Triangles">Triangles with two integer medians</a>
%H A181928 Ralph H. Buchholz and Randall L. Rathbun, <a href="http://www.jstor.org/stable/2974977">An infinite set of Heron triangles with two rational medians</a>, The American Mathematical Monthly, Vol. 104, No. 2 (Feb., 1997), pp. 107-115.
%H A181928 Richard K. Guy, <a href="http://www.jstor.org/stable/2323442">A Dozen Difficult Diophantine Dilemmas</a>, American Mathematical Monthly 95(1988) 31-36.
%H A181928 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/HeronianTriangle.html">Heronian Triangle.</a>
%e A181928 1680 is in the sequence because the corresponding triangle (52, 102, 146) contains two integer medians 35 and 97;
%e A181928 221760 is in the sequence because the corresponding triangle (582, 1252, 1750) contains two integer medians 433 and 1144.
%p A181928 with(numtheory):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):if s>0 then s1:=sqrt(s): m11:=sqrt((2*b^2+2*c^2-a^2)/4): m22:=sqrt((2*c^2+2*a^2-b^2)/4): m33:=sqrt((2*a^2+2*b^2-c^2)/4):if (s1=floor(s1) and m11=floor(m11) and m22=floor(m22)) or (s1=floor(s1) and m11=floor(m11) and m33=floor(m33))  or (s1=floor(s1) and m22=floor(m22) and m33=floor(m33))  then print(s1):print(a):print(b):print(c):print(m11):print(m22):print(m33):else fi:fi:od:od:od:
%t A181928 nn=600; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s (s-a) (s-b) (s-c); m1=(2*b^2+2*c^2-a^2)/4; m2=(2*c^2+2*a^2-b^2)/4; m3=(2*a^2+2*b^2-c^2)/4; If[0 < area2 && (IntegerQ[Sqrt[area2]] && IntegerQ[(Sqrt[m1])] && IntegerQ[Sqrt[m2]]) || (IntegerQ[Sqrt[area2]] && IntegerQ[Sqrt[m1]] && IntegerQ[Sqrt[m3]]) || (IntegerQ[Sqrt[area2]] && IntegerQ[Sqrt[m2]] && IntegerQ[Sqrt[m3]]), AppendTo[lst,Sqrt[area2]]]], {a,nn}, {b,a}, {c,b}]; Union[lst]
%Y A181928 Cf. A188158, A181924.
%K A181928 nonn
%O A181928 1,1
%A A181928 _Michel Lagneau_, Apr 02 2012