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A181924 Area A of the triangles such that A, the sides, and one of the medians are integers. 4
12, 24, 48, 60, 96, 108, 120, 168, 192, 216, 240, 300, 336, 360, 384, 420, 432, 480, 540, 588, 600, 660, 672, 720, 768, 840, 864, 960, 972, 1008, 1080, 1092, 1176, 1200, 1260, 1320, 1344, 1440, 1452, 1500, 1512, 1536, 1680, 1728, 1848, 1920, 1944, 1980 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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).

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.

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;

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

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

REFERENCES

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

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

LINKS

Ray Chandler, Table of n, a(n) for n = 1..144

Andrew Bremner and Richard K. Guy, A Dozen Difficult Diophantine Dilemmas, American Mathematical Monthly 95(1988) 31-36.

Eric W. Weisstein, MathWorld: HeronianTriangle

EXAMPLE

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.

MAPLE

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):

MATHEMATICA

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]

CROSSREFS

Sequence in context: A190566 A213739 A102067 * A270257 A180617 A081808

Adjacent sequences:  A181921 A181922 A181923 * A181925 A181926 A181927

KEYWORD

nonn

AUTHOR

Michel Lagneau, Apr 02 2012

STATUS

approved

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Last modified February 18 03:33 EST 2020. Contains 332006 sequences. (Running on oeis4.)