A181924 - OEIS

%I #20 Feb 25 2023 15:17:58

%S 12,24,48,60,96,108,120,168,192,216,240,300,336,360,384,420,432,480,

%T 540,588,600,660,672,720,768,840,864,960,972,1008,1080,1092,1176,1200,

%U 1260,1320,1344,1440,1452,1500,1512,1536,1680,1728,1848,1920,1944,1980

%N Area A of the triangles such that A, the sides, and one of the medians are integers.

%C 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    4x^2 = 2a^2 + 2b^2 - c^2

%C    4y^2 = 2a^2 + 2c^2 - b^2

%C    4z^2 = 2b^2 + 2c^2 - a^2

%C    A^2 = s(s-a)(s-b)(s-c)

%C where s = (a+b+c)/2 is the semiperimeter and x, y, z the medians.

%C 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 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 where a, b and c are the lengths of the edges (see Guy in the reference).

%C But there exists Heron triangles with two integer medians, for example the triangle (a,b,c) = (52, 102, 146) => A = 1680 and m1 = 4*sqrt(949), m2 = 97 and m3 = 35.

%C Properties of this sequence: There exist three class of triangles (a, b, c):

%C (i) A class of isosceles triangles where a = b < c => the median m = 2*A/c;

%C (ii) A class of Pythagorean where a^2 + b^2 = c^2, and it is easy to check that the median m = c/2.

%C (iii) A class of non-isosceles and non-Pythagorean triangles (a,b,c) having one or two integer medians.

%D Ralph H. Buchholz and Randall L. Rathbun, An infinite set of Heron triangles with two rational medians, Newcastle University, Newcastle, Jan 1997.

%D Ralph H. Buchholz, On triangles with rational altitudes, angles bisectors or medians, PHD Thesis, University of Newcastle, Nov 1989.

%H Ray Chandler, <a href="/A181924/b181924.txt">Table of n, a(n) for n = 1..144</a>

%H 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 Ralph H. Buchholz, <a href="https://doi.org/10.1017/S0004972700030446">On triangles with rational altitudes, angles bisectors or medians</a>, Bulletin of the Australian Mathematical Society , Volume 45 , Issue 3 , June 1992 , pp. 525-526.

%H Ralph H. Buchholz and Randall L. Rathbun, <a href="https://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 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HeronianTriangle.html">Heronian Triangle</a>

%e 336 is in the sequence, because for the sides (14,48,50), A = sqrt(56*(56-14)*(56-48)*(56-50)) = sqrt(112896) = 336, and m = sqrt(2a^2 + b^2 - c^2)/2 = sqrt(2*14^2 + 2*48^2 - 50^2)/2 = 25.

%p with(numtheory):T:=array(1..1000):k:=0: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) or m22=floor(m22) or m33=floor(m33))  then k:=k+1:T[k]:=s1:else fi:fi:od:od:od: L := [seq(T[i],i=1..k)]:L1:=convert(T,set):A:=sort(L1, `<`): print(A):

%t nn = 300; 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[m3])]), AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

%K nonn

%O 1,1

%A _Michel Lagneau_, Apr 02 2012