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A181928
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Area A of the triangles such that A, the sides and two medians are integers.
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2
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1680, 6720, 15120, 26880, 42000, 60480, 82320, 107520, 136080, 168000, 203280, 221760, 241920, 283920, 329280, 378000, 430080, 485520, 544320, 606480, 672000, 740880, 813120, 887040, 888720, 967680, 1050000, 1135680, 1224720
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OFFSET
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1,1
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COMMENTS
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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
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REFERENCES
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Ralph H. Buchholz, On triangles with rational altitudes, angles bisectors or medians, PHD Thesis, University of Newcastle, Nov 1989.
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LINKS
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EXAMPLE
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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.
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MAPLE
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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:
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MATHEMATICA
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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]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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