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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, face 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).
Properties of this sequence: There exist three class of triangles (a, b, c):
(i) A class of isosceles triangles where a = b < c => the median m = 2*A/c. Example: for a(1) = 12, m = 2*12/8 = 3;
(ii) A class of Pythagorean where a^2 + b^2 = c^2, and it is easy to check that the median m = c/2. Example: for a(2) = 24, 6^2 + 8^2 = 10^2 and m = 10/2 = 5;
(iii) A class of non-isosceles and non-Pythagorean triangles (a,b,c) having one or two integer medians. Example: for a(4) = 120, (a,b,c) = (10, 24, 26)and m = sqrt((2a^2 + 2b^2 - a^2)/4) = sqrt((2*10^2+2*24^2-26^2)/4) = 13.
The following table gives the first values (A, m1, m2, m3, a,b,c) where A is the area, m1, m2, m3 are the medians and a, b, c the integer sides of the triangles.
+-----+---------------+---------------+----+----+----+-----+
| A | m1 | m2 | m3 | a | b | c |
+-----+---------------+---------------+----+----+----+-----+
| 12 | 3*sqrt(17)/2 | 3*sqrt(17)/2 | 3 | 5 | 5 | 8 |
| 24 | sqrt(73) | 2*sqrt(13) | 5 | 6 | 8 | 10 |
| 60 | sqrt(1321)/2 | sqrt(1321)/2 | 5 | 13 | 13 | 24 |
| 120 | sqrt(601) | 2*sqrt(61) | 13 | 10 | 24 | 26 |
| 168 | sqrt(5233)/2 | sqrt(5233)/2 | 7 | 25 | 25 | 48 |
| 240 | 2*sqrt(241) | sqrt(481) | 17 | 16 | 30 | 34 |
| 420 | sqrt(8689)/2 | sqrt(6001)/2 | 17 | 25 | 39 | 56 |
| 660 | sqrt(32521)/2 | sqrt(32521)/2 | 11 | 61 | 61 | 120 |
| 720 | sqrt(6481) | 2*sqrt(481) | 41 | 18 | 80 | 82 |
+-----+---------------+---------------+----+----+----+-----+
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