OFFSET
1,1
COMMENTS
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).
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
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}, ...
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:
4x^2 = 2a^2 + 2b^2 - c^2
4y^2 = 2a^2 + 2c^2 - b^2
4z^2 = 2b^2 + 2c^2 - a^2
A^2 = s(s-a)(s-b)(s-c)
where s = (a+b+c)/2 is the semiperimeter and x, y, z the medians.
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:
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
where a, b and c are the lengths of the edges (see Guy in the reference).
Theorems (from Ralph H. Buchholz)
(i) Any triangle with two integer medians has an even semiperimeter.
(ii) If a Heron triangle has two integer medians then its area is divisible by 120.
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
REFERENCES
Ralph H. Buchholz, On triangles with rational altitudes, angles bisectors or medians, PHD Thesis, University of Newcastle, Nov 1989.
LINKS
Ralph H. Buchholz, Triangles with two integer medians
Ralph H. Buchholz and Randall L. Rathbun, An infinite set of Heron triangles with two rational medians, The American Mathematical Monthly, Vol. 104, No. 2 (Feb., 1997), pp. 107-115.
Richard K. Guy, A Dozen Difficult Diophantine Dilemmas, American Mathematical Monthly 95(1988) 31-36.
Eric Weisstein's World of Mathematics, Heronian Triangle.
EXAMPLE
1680 is in the sequence because the corresponding triangle (52, 102, 146) contains two integer medians 35 and 97;
221760 is in the sequence because the corresponding triangle (582, 1252, 1750) contains two integer medians 433 and 1144.
MAPLE
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:
MATHEMATICA
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]
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Apr 02 2012
STATUS
approved