A188158 Area A of the triangles such that A and the sides are integers.

6, 12, 24, 30, 36, 42, 48, 54, 60, 66, 72, 84, 90, 96, 108, 114, 120, 126, 132, 144, 150, 156, 168, 180, 192, 198, 204, 210, 216, 234, 240, 252, 264, 270, 288, 294, 300, 306, 324, 330, 336, 360, 378, 384, 390, 396, 408, 420, 432, 456, 462, 468, 480, 486, 504, 510, 522, 528

Offset: 1

Author

Michel Lagneau, Mar 22 2011

Keywords

• nonn

Comments

The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2. A given area often corresponds to more than one triangle; for example, a(9) = 60 for the triangles (a,b,c) = (6,25,29), (8,17,15), (13,13,10) and (13,13,24).

If only primitive integer triangles (that is, the lengths of the sides are coprime) are considered, then the possible areas are 6 times the terms in A083875. - T. D. Noe, Mar 23 2011

Examples

			a(3) = 24 because the area of the triangle whose sides are 4, 15, 13 is given by sqrt(p(p-4)(p-15)(p-13)) = 24, where p = (4 + 15 + 13)/2 = 16.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Triangle
• Wikipedia, Heronian triangle

Crossrefs

Cf. A007237, A009112, A024153, A024365, A051516, A051584, A051585, A055592, A055593, A055594, A055595.

Programs

Maple

# storage of areas in T(i)
T:=array(1..4000):nn:=100:k:=1:for a from 1
  to nn do: for b from 1 to nn do: for c from 1 to nn do: p:=(a+b+c)/2 : x:=p*(p-a)*(p-b)*(p-c):   if x>0 then x1:=abs(x):s:=sqrt(x1) :else fi:if s=floor(s) then T[k]:=s:k:=k+1:else
  fi:od:od:od:
# sort of T(i)
for jj from 1 to k-1 do: ii:=jj:for k1 from  ii+1 to k-1 do:if T[ii]>T[k1] then ii:=k1:else fi:od: m:=T[jj]:T[jj]:=T[ii]:T[ii]:=m:od:liste:=convert(T,set):print(liste):
# second program:
isA188158 := proc(A::integer)
    local Asqr, s,a,b,c ;
    Asqr := A^2 ;
    for s in numtheory[divisors](Asqr) do
        if s^2> A then
        for a from 1 to s-1 do
            if modp(Asqr,s-a) = 0 then
                for b from a to s-1 do
                    c := 2*s-a-b ;
                    if s*(s-a)*(s-b)*(s-c) = Asqr then
                        return true ;
                    end if;
                end do:
            end if;
        end do:
        end if;
    end do:
    false ;
end proc:
for n from 3 to 600 do
    if isA188158(n) then
        printf("%d,\n",n) ;
    end if;
end do: # R. J. Mathar, May 02 2018

Mathematica

nn = 528; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 <= nn^2 && IntegerQ[Sqrt[area2]], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst] (* T. D. Noe, Mar 23 2011 *)