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A227895 Integer areas of integer-sided triangles where at least one median is of prime length.

%I A227895 #26 Apr 04 2025 22:28:19
%S A227895 12,24,60,120,168,240,420,660,720,840,1092,1320,1680,2448,2520,2640,
%T A227895 3360,3420,3960,5280,5460,6072,6240,6840,9360,10920,12240,14280,15600,
%U A227895 15960,16320,17160,18480,21840,22440,24480,26520,27720,31920,35880,38760,43680
%N A227895 Integer areas of integer-sided triangles where at least one median is of prime length.
%C A227895 Subset of A181924.
%C A227895 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 A227895 4x^2 = 2a^2 + 2b^2 - c^2
%C A227895 4y^2 = 2a^2 + 2c^2 - b^2
%C A227895 4z^2 = 2b^2 + 2c^2 - a^2
%C A227895 A^2 = s(s-a)(s-b)(s-c)
%C A227895 where s = (a+b+c)/2 is the semiperimeter and  x, y, z the medians.
%C A227895 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:
%C A227895 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 A227895 where a, b and c are the lengths of the edges (see Guy in the reference).
%C A227895 Properties of this sequence: There exist three class of triangles (a, b, c):
%C A227895 (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;
%C A227895 (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;
%C A227895 (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 - c^2)/4) = sqrt((2*10^2+2*24^2-26^2)/4) = 13.
%C A227895 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.
%C A227895 +-----+---------------+---------------+----+----+----+-----+
%C A227895 |   A |    m1         |   m2          | m3 |  a |  b |  c  |
%C A227895 +-----+---------------+---------------+----+----+----+-----+
%C A227895 |  12 | 3*sqrt(17)/2  | 3*sqrt(17)/2  |  3 |  5 |  5 |   8 |
%C A227895 |  24 |   sqrt(73)    |  2*sqrt(13)   |  5 |  6 |  8 |  10 |
%C A227895 |  60 | sqrt(1321)/2  | sqrt(1321)/2  |  5 | 13 | 13 |  24 |
%C A227895 | 120 |   sqrt(601)   |  2*sqrt(61)   | 13 | 10 | 24 |  26 |
%C A227895 | 168 | sqrt(5233)/2  | sqrt(5233)/2  |  7 | 25 | 25 |  48 |
%C A227895 | 240 |  2*sqrt(241)  |   sqrt(481)   | 17 | 16 | 30 |  34 |
%C A227895 | 420 | sqrt(8689)/2  | sqrt(6001)/2  | 17 | 25 | 39 |  56 |
%C A227895 | 660 | sqrt(32521)/2 | sqrt(32521)/2 | 11 | 61 | 61 | 120 |
%C A227895 | 720 |  sqrt(6481)   |  2*sqrt(481)  | 41 | 18 | 80 |  82 |
%C A227895 +-----+---------------+---------------+----+----+----+-----+
%H A227895 Andrew Bremner and 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 A227895 Ralph H. Buchholz, <a href="http://citeseerx.ist.psu.edu/pdf/080bc62243d3bf47143e4058674b17214b53d4e9">On Triangles with rational altitudes, angle bisectors or medians</a>, Newcastle University (1989), 21-22.
%H A227895 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 A227895 Eric W. Weisstein, <a href="https://mathworld.wolfram.com/HeronianTriangle.html">MathWorld: HeronianTriangle</a>
%e A227895 1680 is in the sequence because the triangle (a,b,c) = (52, 102, 146) => A = 1680 and m1 = 4*sqrt(949), m2 = 35 and m3 = 97 is a prime number.
%t A227895 nn=800;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]]&&(PrimeQ[(Sqrt[m1])]||PrimeQ[(Sqrt[m2])]||PrimeQ[(Sqrt[m3])]),AppendTo[lst,Sqrt[area2]]]],{a,nn},{b,a},{c,b}];Union[lst]
%Y A227895 Cf. A181924.
%K A227895 nonn
%O A227895 1,1
%A A227895 _Michel Lagneau_, Oct 14 2013