A188158 -id:A188158 - cp's OEIS frontend

A385872 Areas of nondegenerate triangles with perimeter A385737(n) whose side lengths are triangular numbers.

1452, 1176, 2376, 3780, 8316, 10626, 14742, 28500, 12558, 32340, 25200, 94500, 18792, 130680, 89250, 158760, 130680, 155250, 53508, 93636, 122958, 208278, 893970, 1199772, 2183328, 1130976, 2058210, 1414098, 3160080, 4000752, 3898800, 324324, 4900500, 1845120, 7427970

Offset: 1

Felix Huber, Jul 18 2025

nonn

45189144 is the smallest integer area of a right triangle whose sides are triangular numbers. This area corresponds to the triangle [8778, 10296, 13530].
From David A. Corneth, Jul 18 2025: (Start) If sidelengths are u, v, w where 0 < u < v < w < u + v then the area can be written as A = ((u + v + w) * (u + v - w) * (u - v + w) * (-u + v + w)) / 16 = k^2. If A is a square then 16*A is a square (possible extraneous resulting from this can be removed at the end).
We may rewrite 16*A as ((u + v)^2 - w^2) * (w^2 - (v - u)^2) = k^2
Since their product is a square we may write
((u + v)^2 - w^2) * t^2 = (w^2 - (v - u)^2). where t > 1 is a rational. When u, v and t are chosen we can solve for w.
w^2 = (t^2*(u-v)^2 + (u+v)^2) / (t^2 + 1). (End)

			a(1) = 1452 is the area of the triangle [55, 55, 66] with perimeter A385737(1) = 176, where 55 and 66 are triangular numbers.
a(2) = 1176 is the area of the triangle [28, 91, 105] with perimeter A385737(2) = 224, where 28, 91 and 105 are triangular numbers.
From _David A. Corneth_, Jul 18 2025: (Start)
From (u, v) = (28, 91) we get
((u + v)^2 - w^2) * t^2 = (w^2 - (v - u)^2)
(119 - w^2) * t^2 = (w^2 - 63^2). Testing t = 2/3 gives the desired w. (End)

Felix Huber, Table of n, a(n) for n = 1..315
Felix Huber, Maple program
Eric Weisstein's World of Mathematics, Triangular Number

Cf. A000217, A051516, A188158, A385736, A385737.

A385872:=proc(P) # To get all integer areas of triangles with perimeters <= P.
    local p,x,y,z,u,v,w,s,i;
    p:=[];
    for z to floor((sqrt(24*P+9)-3)/6) do
        for x from z to floor((sqrt(4*P-3)-1)/2) do
            for y from max(z,floor((sqrt(1+4*(x^2+x-z^2-z))-1)/2)+1) to min(x,floor((sqrt(1+4*(2*P-x^2-x-z^2-z))-1)/2)) do
            	u:=z*(z+1)/2;
            	v:=y*(y+1)/2;
            	w:=x*(x+1)/2;
            	s:=(u+v+w)/2;
            	if issqr(s*(s-u)*(s-v)*(s-w)) then
               	    p:=[op(p),[u+v+w,sqrt(s*(s-u)*(s-v)*(s-w))]]
               	fi
            od
        od
    od;
    return seq(sort(p)[i,2],i=1..nops(p))
end proc;
A385872(16236);

A181928 Area A of the triangles such that A, the sides and two medians are integers.

Original entry on oeis.org

1680, 6720, 15120, 26880, 42000, 60480, 82320, 107520, 136080, 168000, 203280, 221760, 241920, 283920, 329280, 378000, 430080, 485520, 544320, 606480, 672000, 740880, 813120, 887040, 888720, 967680, 1050000, 1135680, 1224720

Offset: 1

Michel Lagneau, Apr 02 2012

nonn

The first six primitives triangles (with areas {1680, 221760, 8168160, 95726400, 302793120, 569336866560}) have been discovered by Ralph H. Buchholz and are listed in a table of the chapter 4 of his thesis (see Links).
Later on, Buchholz & Rathbun identified an infinite family of Heronian triangles with 2 integer medians (comprising 4 of the 6 triangles above). The next two primitive triangles in such family have areas 8548588738240320 and 17293367819066194215360. - Giovanni Resta, Apr 05 2017
The areas of non-primitive triangles are of the form {1680*k^2}, {221760*k^2}, {8168160*k^2}, {95726400*k^2}, {302793120*k^2}, ...
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, faces 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).
Theorems (from Ralph H. Buchholz)
(i) Any triangle with two integer medians has an even semiperimeter.
(ii) If a Heron triangle has two integer medians then its area is divisible by 120.
It seems that, for any n, a(n) == 0 (mod 1680). The reverse is not always true: e.g., as mentioned by Giovanni Resta, the triangle with sides (56*k, 61*k, 75*k) has area of the form 1680 * k^2, but it cannot be a term of a(n). - Sergey Pavlov, Mar 31 2017

			1680 is in the sequence because the corresponding triangle (52, 102, 146) contains two integer medians 35 and 97;
221760 is in the sequence because the corresponding triangle (582, 1252, 1750) contains two integer medians 433 and 1144.

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

Ralph H. Buchholz, Triangles with two integer medians
Ralph H. Buchholz and Randall L. Rathbun, An infinite set of Heron triangles with two rational medians, The American Mathematical Monthly, Vol. 104, No. 2 (Feb., 1997), pp. 107-115.
Richard K. Guy, A Dozen Difficult Diophantine Dilemmas, American Mathematical Monthly 95(1988) 31-36.
Eric Weisstein's World of Mathematics, Heronian Triangle.

Cf. A188158, A181924.

with(numtheory):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) and m22=floor(m22)) or (s1=floor(s1) and m11=floor(m11) and m33=floor(m33))  or (s1=floor(s1) and m22=floor(m22) and m33=floor(m33))  then print(s1):print(a):print(b):print(c):print(m11):print(m22):print(m33):else fi:fi:od:od:od:

nn=600; 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[area2]] && IntegerQ[Sqrt[m1]] && IntegerQ[Sqrt[m3]]) || (IntegerQ[Sqrt[area2]] && IntegerQ[Sqrt[m2]] && IntegerQ[Sqrt[m3]]), AppendTo[lst,Sqrt[area2]]]], {a,nn}, {b,a}, {c,b}]; Union[lst]

A213158 Positive integers of the form (x+y+z)xy*z (x,y,z positive integers).

Original entry on oeis.org

3, 8, 15, 20, 24, 35, 36, 48, 56, 63, 80, 84, 96, 99, 108, 120, 128, 135, 140, 143, 144, 168, 176, 180, 195, 200, 216, 224, 231, 240, 243, 255, 260, 264, 275, 288, 300, 308, 320, 323, 336, 351, 360, 384, 396, 399, 416, 420, 440, 455, 468, 476, 483, 495, 504

Offset: 1

Jean-François Alcover, Jun 06 2012

nonn

Square terms are 36, 144, 576,... and the corresponding square roots are 6, 12, 24,... i.e. sequence A188158 (integer areas of primitive integer triangles).

Positive integers of the form (a^2-b^2)*(b^2-c^2) with integers a>b>c>=0. - Michael Somos, May 18 2013

			a(21)=144 for x=1, y=4 and z=4 then the triangle sides are x+y = 5, z+x = 5 and y+z = 8, hence half-perimeter = p = x+y+z = 9 and Heron's formula is checked: area = sqrt(p*(p-5)*(p-5)*(p-8)) = sqrt(144) = 12.
36 = (4^2-2^2) * (2^2-1^2). 63 = (5^2-2^2) * (2^2-1^2) = (5^2-4^2) * (4^2-2^2)= (8^2-1^2) * (1^2-0^2). - _Michael Somos_, May 19 2013

R. D. Carmichael, Diophantine Analysis, Wiley, 1915, p. 9.

Cf. A188158.

nmax = 25; mx = nmax (nmax + 2); Union[Reap[Do[a = (x + y + z)*x*y*z; If[a <= mx, Sow[a]], {x, 1, nmax}, {y, x, nmax}, {z, y, nmax}]][[2, 1]]]

A218243 Triangle numbers: m = abc such that the integers a,b,c are the sides of a triangle with integer area.

Original entry on oeis.org

60, 150, 200, 480, 780, 1200, 1530, 1600, 1620, 1690, 1950, 2040, 2100, 2730, 2860, 3570, 3840, 4050, 4056, 4200, 4350, 4624, 5100, 5400, 5460, 6240, 7500, 8120, 8250, 8670, 8750, 9600, 10812, 11050, 11900, 12180, 12240, 12800, 12960, 13260, 13520, 13650

Offset: 1

Michel Lagneau, Oct 24 2012

nonn

A triangle number m is an integer with at least one decomposition m = a*b*c such that the area of the triangle of sides (a,b,c) is an integer. Because this property is not always unique, we introduce the notion of "triangle order" for each triangle number m, denoted by TO(m). For example, TO(60) = 1 because the decomposition 60 = 3*4*5 is unique with the triangle (3,4,5) whose area A is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2 => A = sqrt(6*(6-3)*(6-4)*(6-5)) = 6, but TO(780) = 2 because 780 = 4*13*15 = 5*12*13 and the area of the triangle (4,13,15) is sqrt(16*(16-4)*(16-13)*(16-15)) = 24 and the area of the triangle (5,12,13) is sqrt(15*(15-5)*(15-12)*(15-13)) = 30.
Given an area A of A188158, there exists either a unique triangle number (for example for A = 6 => m = 60 = 3*4*5), or several triangle numbers (for example for A=60 => m1 = 4350 = 6*25*29, m2 = 2040 = 8*15*17, m3 = 1690 = 13*13*10).
The number of ways to write m = a*b*c with 1<=a<=b<=c<=m is given by A034836, thus: TO(m) <= A034836(m).
If n is in this sequence, so is nk^3 for any k > 0. Thus this sequence is infinite. - Charles R Greathouse IV, Oct 24 2012
In view of the preceding comment, one might call "primitive" the elements of the sequence for which there is no k>1 such that n/k^3 is again a term of the sequence. These elements 60, 150, 200, 780, 1530, 1690, 1950,... are listed in A218392. - M. F. Hasler, Oct 27 2012

			60 is in the sequence because 60 = 3*4*5 and the corresponding area is sqrt(6*(6-3)*(6-4)*(6-5)) = 6 = A188158(1).

Charles R Greathouse IV, Table of n, a(n) for n = 1..2000
Eric Weisstein's World of Mathematics, Triangle

Subsequence of A139270.

Cf. A034836, A188158, A083875.

nn = 500; 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, a*b*c]]], {a, nn}, {b, a}, {c, b}]; Union[lst] (* Program from T. D. Noe, adapted for this sequence - see A188158 *)

Heron(a,b,c)=a*=a;b*=b;c*=c;((a+b+c)^2-2*(a^2+b^2+c^2))
is(n)=fordiv(n,a, if(a^3<=n, next); fordiv(n/a,b, my(c=n/a/b,h); if(a>=b && b>=c && aCharles R Greathouse IV, Oct 24 2012

A181925 Area A of the triangles such that A, the sides, and at least one of the three bisectors are integers.

Original entry on oeis.org

12, 48, 60, 108, 120, 168, 192, 240, 300, 360, 420, 432, 480, 540, 588, 660, 672, 768, 960, 972, 1008, 1080, 1092, 1200, 1260, 1344, 1440, 1452, 1500, 1512, 1680, 1728, 1848, 1920, 1980, 2028, 2160, 2352, 2448, 2520, 2640, 2688, 2700, 2772, 2940, 3000

Offset: 1

Michel Lagneau, Apr 02 2012

nonn

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.
The lengths of the bisectors are given by:
b1 = sqrt(bc*(b+c-a)(a+b+c))/(b+c)
b2 = sqrt(ac*(a+c-b)(a+b+c))/(a+c)
b3 = sqrt(ab*(a+b-c)(a+b+c))/(a+b)
Properties of this sequence: There exist three subsets of numbers included in a(n):
Case (i): A subset with a majority of isosceles triangles whose area equals the sum of the areas of two Pythagorean triangles with integer sides => the sequence A118903 is included in this sequence. This sort of triangles contains generally only one integer bisector, but more rarely three integer bisectors (see the examples).
Case (ii): Right triangles (a,b,c) where a^2 + b^2 = c^2.
Case (iii): A class of non-isosceles and non-right triangles (a, b, c) where one, two or three bisectors are integers.

			Case (i): 12 is in the sequence because the area of the isosceles triangle (5, 5, 6) equals 12 and one of the bisectors is an integer (4). But the isosceles triangle (546, 975, 975) whose area equals 255528 contains three integer bisectors: 936, 560, 560.
Case (ii): The right triangle (28, 96, 100) => A = 1344, and the integer median is m = 35.
Case (iii): The triangle (31091676, 46267375, 62553491) => A =  690494511777840, and the three bisectors are 51555075, 38342304 and 22314600.

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

Ralph H. Buchholz, Triangles with two integer medians
Eric Weisstein's World of Mathematics, Angle Bisectors

Cf. A118903, A188158.

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):aa:=b*c*(b+c-a)*(a+b+c): bb:=a*c*(a+c-b)*(a+b+c): cc:=a*b*(a+b-c)*(a+b+c):if s>0 and aa>0 and bb>0 and cc>0 then s1:=sqrt(s): aa1:=sqrt(aa)/(b+c): bb1:=sqrt(bb)/(a+c): cc1:=sqrt(cc)/(a+b):if s1=floor(s1) and (aa1=floor(aa1) or bb1=floor(bb1) or cc1=floor(cc1))  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):

nn=300; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s (s-a) (s-b) (s-c); aa=b*c*(b+c-a)*(a+b+c); bb=a*c*(a+c-b)*(a+b+c); cc=a*b*(a+b-c)*(a+b+c); If[0 < area2 && aa > 0 && bb > 0 && cc > 0 && IntegerQ[Sqrt[area2]] && (IntegerQ[Sqrt[aa]/(b+c)] || IntegerQ[Sqrt[bb]/(a+c)] || IntegerQ[Sqrt[cc]/(a+b)]), AppendTo[lst, Sqrt[area2]]]], {a,nn}, {b,a}, {c,b}]; Union[lst]

A182171 Area A of the triangles such that A, the sides and three perpendicular bisectors are integers.

Original entry on oeis.org

108, 384, 432, 768, 972, 1536, 1728, 2700, 3072, 3456, 3888, 5292, 6144, 6912, 8748, 9600, 10800, 12288, 13068, 13824, 15552, 17280, 18252, 18816, 18900, 19200, 21168, 24300, 24576, 27000, 27648, 31104, 31212, 34560, 34992, 37632

Offset: 1

Michel Lagneau, Apr 16 2012

nonn

Let a triangle with the angles (A, B, C) and the sides opposite the angles labeled (a, b, c). The length of the perpendicular bisectors is given by (x, y, z) where:
x is the perpendicular bisector passing through the midpoint of the segment BC = a;
y is the perpendicular bisector passing through the midpoint of the segment AC = b;
z is the perpendicular bisector passing through the midpoint of the segment AB = c.
We obtain the relations:
x = (a/2)*tg B if x intersects AB or (a/2)* tg C if x intersects AC;
y = (b/2)* tg A if y intersects AB or (b/2)* tg C if y intersects BC;
z = (c/2)*tg A if z intersects AC or (c/2) *tg B if z intersects BC.
The area A of the 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.
Finally, we obtain:
x = (a/2) * min {tg B, tg C }; y = (b/2) * min {tg A, tg C }; z = (c/2) * min {tg A, tg B } with tg A = 4*A/(b^2+c^2-a^2) ; tg B = 4*A/(c^2+a^2-b^2) ; tg C = 4*A/(a^2+b^2-c^2).
Properties of this sequence:
The numbers of the form 108*n^2, 384*n^2, 768*n^2, 17280*n^2, 18900*n^2 are in the sequence because the area of the primitive triangles (15, 15, 18), (24, 32, 40), (40, 40, 64), (120, 288, 312), (150, 255, 315) are 108, 384, 768 , 17280 and 18900 respectively.
There exists three class of numbers included into a(n) :
Case (i) : a subset of isosceles triangles;
Case (ii) : a subset of right triangles;
Case (iii) : other (neither isosceles nor right triangle).

			Primitive solutions follow:
Area,  ( a,   b,   c),  ( x,   y,   z), Case
  108,  (15,  15,  18),  (10,  10,  12), Isosceles,
  384,  (24,  32,  40),  (16,  12,  15), Right,
  768,  (40,  40,  64),  (15,  15,  24), Isosceles,
17280, (120, 288, 312), (144,  60,  65), Right,
18900, (150, 255, 315), (100,  68,  84), Other,
27000, (255, 255, 450),  (68,  68, 120), Isosceles,
34560, (312, 312, 576),  (65,  65, 120), Isosceles,
53760, (272, 400, 448), (255, 150, 168), Other,
54000, (240, 450, 510), (225, 120, 136), Right,
91476, (429, 462, 495), (364, 308, 330), Other,
95256, (252, 819, 945), (168, 104, 120), Other,
96768, (336, 720, 960), (126, 105, 140), Other.

Ray Chandler, Table of n, a(n) for n = 1..65
Eric W. Weisstein, MathWorld: Triangle

Cf. A188158, A210643, A181924, A181925, A181928.

zz:=evalf(1/10^6):k:=0:nn:=350:
for a from 15 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):
             u:= a^2+b^2-c^2:v:= b^2+c^2-a^2 :w:=c^2+a^2-b^2:
             if s>0 then s1:=sqrt(s):else fi:
             if u>0 then u0:=u:else u0:=zz:fi:
             if v>0 then v0:=v:else vo:=zz:fi:
             if w>0 then w0:=w:else w0:=zz:fi:
a0:= evalf(2*a*s1/w0):a1:=evalf(2*a*s1/u0): b0:= evalf(2*b*s1/u0):b1:=evalf(2*b*s1/v0): c0:= evalf(2*c*s1/w0):c1:=evalf(2*c*s1/v0):
             if a0

x = Min{2*a*A/(c^2+a^2-b^2) , 2*a*A/(a^2+b^2-c^2)};
y = Min{2*b*A/(a^2+b^2-c^2)  , 2*b*A/(b^2+c^2-a^2)};
z = Min{2*c*A/(c^2+a^2-b^2)  , 2*c*A/(b^2+c^2-a^2)}.

More terms from Ray Chandler, Apr 24 2013

A209432 Area A of the triangles such that A and the sides are integers and there exists at least one square inscribed in the triangle whose sides are also integers.

Original entry on oeis.org

24, 96, 216, 294, 300, 324, 384, 600, 810, 864, 1176, 1200, 1296, 1452, 1536, 1920, 1944, 2400, 2520, 2646, 2700, 2904, 2916, 3240, 3456, 4056, 4320, 4704, 4800, 4950, 5184, 5400, 5808, 6144, 6300, 6936, 7260, 7290, 7350, 7500, 7680, 7776, 8064, 8100, 8214

Offset: 1

Michel Lagneau, Mar 09 2012

nonn

Every triangle has three inscribed squares (squares in its interior such that all four of the square's vertices lie on sides of the triangle, so two of them lie on the same side and hence one side of the square coincides with part of a side of the triangle). However, in the case of a right triangle, two of the squares coincide and have a vertex at the triangle's right angle, so a right triangle has only two distinct inscribed squares. Within a given triangle, a longer common side is associated with a smaller inscribed square. If an inscribed square has sides of length x and the triangle has a side of length a, part of which side coincides with a side of the square, then x, a, and the triangle's area A are related according to x = 2Aa/(a^2+2A).
Property of this sequence: the numbers of the form 24*k^2 are in the sequence.
Theorem: Consider a triangle whose area A and sides (a, b, c) are integers such that there exists at least one square inscribed in this triangle whose sides x are also integers. Then, if the smallest side a = min {a, b, c} of this triangle is of the form a = 4k, k integer, then x = 3k and A = 24k^2.
Proof: Let k be an integer, and let the sides of a triangle be a = 4k, b = 13k, c = 15k. Then s = (a+b+c)/2 = 16k and A = sqrt(s(s-a)(s-b)(s-c)) = 24k^2. With x = 2Aa/(a^2+2A), we find x = 3k.

			294 is in the sequence because for (a, b, c) = (21, 28, 35) => x1 = 2*21*294/(2*294+21^2) = 12348/1029 = 12 is the integer value of the side of the square inscribed in the triangle (21, 28, 35) whose area equals 294 and whose side coincides with the side [21] of this triangle. But we also have a second square with the side x2 = 2*28*294/(2*294+28^2) = 16464/1372 = 12 whose side coincides with the side [28] of the same triangle.

Herbert Bailey and Duane Detemple, Squares inscribed in angles and triangles, Mathematics Magazine 71(4), 1998, 278-284.
Eric W. Weisstein, MathWorld: Triangle Square Inscribing

Cf. A120062, A188158, A208984.

with(numtheory):T:=array(1..1500):k:=0:nn:=500: 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 : x:=p*(p-a)*(p-b)*(p-c): if x>0 then s:=sqrt(x) :if s=floor(s) and (irem(2*a*s,2*s+a^2) = 0 or irem(2*b*s,2*s+b^2) = 0 or irem(2*c*s,2*s+c^2) = 0) then k:=k+1:T[k]:= s: else fi:fi:od:od:od: L := [seq(T[i],i=1..k)]:L1:=convert(T,set):A:=sort(L1, `<`): print(A):

nn=500;lst={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s (s-a) (s-b) (s-c);If[0

A = sqrt(p*(p-a)*(p-b)*(p-c)) with p = (a+b+c)/2 (Heron's formula);

Sides of the three squares: x1 = 2*A*a/(a^2+2*A); x2 = 2*A*b/(b^2+2*A); x3 = 2*A*c/(c^2+2*A).

A210645 Area A of the triangles such that A, the sides and one of the altitudes are four consecutive integers of an arithmetic progression d.

Original entry on oeis.org

84, 336, 756, 1344, 2100, 3024, 4116, 5376, 6804, 8400, 10164, 12096, 14196, 16464, 18900, 21504, 24276, 27216, 30324, 33600, 37044, 40656, 44436, 48384, 52500, 56784, 61236, 65856, 70644, 75600, 80724, 86016, 91476, 97104, 102900, 108864, 114996, 121296

Offset: 1

Michel Lagneau, Mar 26 2012

nonn

See the first link for the comments. We use Heron's Formula for a triangle: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.The sides and one of the altitudes are of the form q+d, q, q-d, q-2d.

			84 is in the sequence because (a, b, c, h) = (15, 14, 13, 12) => A = sqrt(21*(21-15)*(21-14)*(21-13)) = sqrt(7056) = 84 but A = (1/2)*h*b = (1/2)*12*14 = 84.

Eric Weisstein, Altitude

Cf. A188158, A210643.

with(numtheory):T:=array(1..1000):k:=0:nn:=800:for q from 1 to nn do: for d from 1 to nn do: a:=q+d:b:=q:c:=q-d:h1:=q-2*d:p:=(a+b+c)/2:s:=p*(p-a)*(p-b)*(p-c):if s>0 then s1:=sqrt(s): h11:=2*s1/a: h22:=2*s1/b:h33:=2*s1/c:if s1=floor(s1) and (h1=h11 or h1=h22 or h1=h33)  then k:=k+1:T[k]:=s1:else fi:fi:od:od: L := [seq(T[i],i=1..k)]:L1:=convert(T,set):A:=sort(L1, `<`): print(A):

Conjecture: a(n) = 84*n^2. a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). G.f.: 84*x*(1+x)/(1-x)^3. - Colin Barker, Apr 19 2012

A227879 Integer areas of incentral triangles of integer-sided triangles.

Original entry on oeis.org

70, 280, 360, 480, 630, 1120, 1312, 1440, 1750, 1768, 1920, 2132, 2520, 3240, 3430, 4320, 4480, 5248, 5670, 5760, 7000, 7038, 7072, 7680, 7800, 8470, 8528, 9000, 9240, 10080, 11808, 11830, 12000, 12960, 13720, 13950, 14744, 15750, 15912, 17280, 17640, 17920

Offset: 1

Michel Lagneau, Oct 25 2013

nonn

The incentral triangle IJK is the Cevian triangle of a triangle ABC with respect to its incenter. It is therefore also the triangle whose vertices are determined by the intersections of the reference triangle's angle bisectors with the respective opposite sides.
The area is given by:
  A' = 2*A*a*b*c/((a+b)*(b+c)*(c+a)) where A is the area of the original triangle.
The side lengths are:
  a' = a*b*c*sqrt(3+2*(-cos A + cos B + cos C))/((a+b)*(a+c))
  b' = a*b*c*sqrt(3+2*(cos A - cos B + cos C))/((b+c)*(b+a))
  c' = a*b*c*sqrt(3+2*(cos A + cos B - cos C))/((c+a)*(c+b))
Properties of this sequence:
The areas of the original triangles are integers. The primitive triangles with areas a(n) are 70, 360, 480, 630, 1312, ...
The nonprimitive triangles with areas 4*a(n) are in the sequence.
It appears that if the original triangle is isosceles, a side of the corresponding incenter triangle is an integer.
The following table gives the first values (A', A, a, b, c, t1, t2, t3) where A' is the area of the incentral triangles, A is the area of the reference triangles ABC, a, b, c the integer sides of the original triangles ABC and t1, t2, t3 are the sides of the incentral triangles.
------------------------------------------------------------------------
  A'|  A | a |  b | c |      t1        |      t2        |    t3
------------------------------------------------------------------------
70  | 294| 21|  28| 35|3*sqrt(65)/2    |4*sqrt(85)/3    |7*sqrt(145)/6
280 |1176| 42|  56| 70|3*sqrt(65)      |8*sqrt(85)/2    |7*sqrt(145)/3
360 |1452| 55|  55| 66|3*sqrt(89)      |3*sqrt(89)      |    30
480 |2028| 65|  65|104|4*sqrt(61)      |4*sqrt(61)      |    40
630 |2646| 63|  84|105|9*sqrt(65)/2    |4*sqrt(85)      |7*sqrt(145)/2
1120|4704| 84| 112|140|6*sqrt(65)      |16*sqrt(85)/3   |14*sqrt(145)/3
1312|8820| 63| 280|287|36*sqrt(2501)/35|40*sqrt(7585)/63|28*sqrt(9061)/45
1440|5808|110| 110|132|6*sqrt(89)      |6*sqrt(89)      |    60
1750|7350|105| 140|175|15*sqrt(65)/2   |20*sqrt(85)/3   |35*sqrt(145)/6
1768|8670| 85| 204|221|50*sqrt(13)/3   |12*sqrt(689)/5  |34*sqrt(949)/15
1920|8112|130| 130|208|8*sqrt(61)      |8*sqrt(61)      |    80
.......................................................

			70 is in the sequence because the formula A' = 2*A*a*b*c/((a+b)*(b+c)*(c+a)) gives with the initial triangle (21,28,35): A' = 2*294*21*28*35/((21+28)*(28+35)*(35+21)) = 70, with the area A = 294 obtained by Heron's formula A = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(42*(42-21)*(42-28)*(42-35)) = 294, where s = 42 is the semiperimeter.

C. Kimberling, Triangle Centers and Central Triangles. Congr. Numer. 129, 1-295, 1998.

Wolfram MathWorld, Incentral Triangles

Cf. A188158.

nn=1000; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s(s-a)(s-b)(s-c); t= 2*Sqrt[area2]*a*b*c/((a+b)*(b+c)*(c+a)); If[0

A228847 Let A(x,y,z) denote the integer area A of the triangle with integer side lengths (x,y,z). a(n) is the smallest area such that there exist at least n distinct triangles (x_i, y_i, z_i) satisfying A(x_1, y_1, z_1) = A(x_2, y_2, z_2) = ... = A(x_n, y_n, z_n).

Original entry on oeis.org

6, 12, 60, 60, 210, 210, 336, 336, 420, 420, 420, 420, 420, 420, 420, 840, 840, 1680, 1680, 1680, 1680, 1680, 1680, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560, 7560

Offset: 1

Michel Lagneau, Sep 05 2013

nonn

The area A of a triangle whose sides have lengths x, y, and z is given by A188158. The area is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.
The sequence A228858 gives the areas such that there exist exactly n triangles (x_i, y_i, z_i) satisfying A(x_1, y_1, z_1) = A(x_2, y_2, z_2) = ... = A(x_n, y_n, z_n).
The Maple program examines all triangles having longest side <= 600.

			a(6) = 210 because there exist 6 triangles:
(3,148,149) => A = sqrt(150*(150-3)*(150-148)*(150-149)) = 210;
(7,65,68) => A = sqrt(70*(70-7)*(70-65)*(70-68)) = 210;
(12,35,37) => A = sqrt(42*(42-12)*(42-35)*(42-37)) = 210;
(17,25,28) => A = sqrt(35*(35-17)*(35-25)*(35-28)) = 210;
(17,28,39) => A = sqrt(42*(42-17)*(42-28)*(42-39)) = 210;
(20,21,29) => A = sqrt(35*(35-20)*(35-21)*(35-29)) = 210.

Eric Weisstein's World of Mathematics, Triangle

Cf. A188158, A228858.

with(numtheory):nn:=600:lst:={}:T:=array(1..30000):k:=0: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 : 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 lst:=lst union {s}:k:=k+1:T[k]:=s:else fi:od:od:od:k1:=nops(lst):for n from 1 to 30 do:jj:=0:for i from 1 to k1 while(jj=0) do:ii:=0:for j from 1 to k while(jj=0) do:if lst[i]=T[j] then ii:=ii+1:if ii=n then jj:=1:printf ( "%d %d \n",n,lst[i]):else fi:fi:od:od:od:

cp's OEIS Frontend

A385872 Areas of nondegenerate triangles with perimeter A385737(n) whose side lengths are triangular numbers.

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A181928 Area A of the triangles such that A, the sides and two medians are integers.

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A213158 Positive integers of the form (x+y+z)*x*y*z (x,y,z positive integers).

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A218243 Triangle numbers: m = a*b*c such that the integers a,b,c are the sides of a triangle with integer area.

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PARI

A181925 Area A of the triangles such that A, the sides, and at least one of the three bisectors are integers.

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A182171 Area A of the triangles such that A, the sides and three perpendicular bisectors are integers.

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A209432 Area A of the triangles such that A and the sides are integers and there exists at least one square inscribed in the triangle whose sides are also integers.

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A213158 Positive integers of the form (x+y+z)xy*z (x,y,z positive integers).

A218243 Triangle numbers: m = abc such that the integers a,b,c are the sides of a triangle with integer area.