A188158 -id:A188158 - cp's OEIS frontend

A286328 Least integer k such that the area of the triangle (prime(n), k, k+1) is an integer.

4, 3, 24, 60, 14, 9, 180, 264, 20, 480, 19, 84, 924, 1104, 51, 1740, 155, 2244, 2520, 2664, 3120, 3444, 99, 51, 51, 5304, 5724, 65, 399, 8064, 8580, 9384, 9660, 221, 11400, 12324, 13284, 13944, 14964, 16020, 819, 18240, 194, 99, 19800, 22260, 24864, 25764, 26220

Offset: 2

Michel Lagneau, May 07 2017

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 corresponding areas are 6, 6, 84, 330, 84, 36, 1710, 3036, 210,...
The following table gives the first values of n, the sides (prime(n), k, k+1) and the area A of each triangle.
+-----+---------+------+------+-------+
|  n  | prime(n)|   k  |  k+1 |    A  |
+-----+---------+------+------+-------+
|  2  |    3    |   4  |   5  |    6  |
|  3  |    5    |   3  |   4  |    6  |
|  4  |    7    |  24  |  25  |   84  |
|  5  |   11    |  60  |  61  |  330  |
|  6  |   13    |  14  |  15  |   84  |
|  7  |   17    |   9  |  10  |   36  |
|  8  |   19    | 180  |  181 | 1710  |
|  9  |   23    | 264  |  265 | 3036  |
| 10  |   29    |  20  |   21 |  210  |
.......................................
We observe triangles of sides (prime(m), prime(m)+1, prime(m)+2) = (3, 4, 5), (13, 14, 15), (193, 194, 195), (37633, 37634, 37635), ... with the corresponding areas 6, 84, 16296, 613283664, ... (subsequence of A011945).
We observe Pythagorean triangles for n = 2, 3, 4, 5, 8, 9, 10, ....
In this case, if prime(n) < k, the numbers k of the sequence such that prime(n) = sqrt(2k+1) are given by the numbers {4, 24, 60, 180, 264, ...}, subsequence of {A084921} = {4, 12, 24, 60, 84, 144, 180, 264, ...}. If prime(n) > k, the numbers k of the sequence such that prime(n) = sqrt(2k^2+2k+1) are given by the numbers 3, 20, 4059, 23660, ....
From Chai Wah Wu, May 15 2017: (Start)
Assumes triangle has positive area.
Let p = prime(n). Then
(p+1)/2 <= a(n) <= (p^2-1)/2.
a(n) = (p+1)/2 if n > 1 is a term in A062325, i.e. p is of the form m^2+1 (A002496); otherwise, a(n) > (p+1)/2.
a(n) is the smallest k >= (p+1)/2 such that sum_{i=(p+1)/2}^{k} i*(p^2-1)/2 is a square.
These statements follow from the fact that the area of a triangle with sides of length p, k and k+1 is equal to (p^2-1)*((2k+1)^2-p^2)/16.
(End)

			a(4) = 24 because the area of the triangle (prime(4), 24, 25) = (7, 24, 25) = sqrt(28*(28-7)*(28-24)*(28-25)) = 84, where the semiperimeter 28 = (7+24+25)/2.

Chai Wah Wu, Table of n, a(n) for n = 2..1000

Cf. A002496, A011945, A062325, A084921, A188158.

nn:=10^7:
for n from 2 to 50 do:
a:=ithprime(n):ii:=0:
for k from 1 to nn while(ii=0) do:
p:=(a+2*k+1)/2:q:=p*(p-a)*(p-k)*(p-k-1):
if q>0 and floor(sqrt(q))=sqrt(q) then
       ii:=1: printf(`%d, `,k):
      else
      fi:
     od:
    od:

Do[kk=0;Do[s=(Prime[n]+2k+1)/2;If[IntegerQ[s],area2=s(s-Prime[n])(s-k)(s-k-1);If[area2>0&&kk==0&&IntegerQ[Sqrt[area2]],Print[n," ",k];kk=1]],{k,1,3*10^4}],{n,2,10}] (* or *)
a[n_] := Block[{p = Prime@n, k}, k = (p + 1)/2; While[! IntegerQ@ Sqrt[(4 k^2 - p^2 + 4 k + 1) (p^2 - 1)/16], k++]; k]; a /@ Range[2, 50] (* Giovanni Resta, May 07 2017 *)

from _future_ import division
from sympy import prime
from gmpy2 import is_square
def A286328(n): # assumes n >= 2
    p, area = prime(n), 0
    k, q, kq = (p + 1)//2, (p**2 - 1)//2, (p - 1)*(p + 1)**2//4
    while True:
        area += kq
        if is_square(area):
            return k
        k += 1
        kq += q # Chai Wah Wu, May 15 2017

A295554 a(n) is the number of distinct integer-sided triangles inscribed in a circle of radius A009003(n) whose inradius are integers.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 5, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 12, 1, 1, 1, 1, 1, 12, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 5, 12, 1, 1, 5, 1, 1

Offset: 1

Michel Lagneau, Feb 03 2018

nonn

For n <= 200, the number of distinct integer-sided triangles inscribed in a circle of radius A009003(n) whose inradius are integers belongs to the set E = {1, 5, 10, 12, 38} where a(168) = 38 (see the table given in reference). Is the set E infinite when n is infinite?
a(m) > 1 for m = 7, 18, 26, 31, 35, ... and {A009003(m)} = {25, 50, 65, 75, 85, ...} = {A009177}.
We observe geometric properties:
If a(n) = 1, the unique triangle is a right triangle.
If a(n) = 5, we find two right triangles, two isosceles triangles and another triangle (neither isosceles nor right triangle).
If a(n) = 10, we find three right triangles, two isosceles triangles and five other triangles.
If a(n) = 12, we find four right triangles and eight other triangles.
The area A of a triangle whose sides have lengths u, v, and w is given by Heron's formula: A = sqrt(s*(s-u)*(s-v)*(s-w)), where s = (u+v+w)/2.
The inradius r is given by r = A/s and the circumradius is given by R = u*v*w/4A.

			a(7) = 5 because there exists 5 distinct triangles of integer circumradius R = A009003(7)= 25 with the corresponding integer inradius {4, 6, 8, 10, 12}.

Felix Huber, Illustration of the term a(7)
Michel Lagneau, Triangles
Eric Weisstein's World of Mathematics, Circumradius
Eric Weisstein's World of Mathematics, Inradius

Cf. A009003, A188158, A208984, A210207.

A009003=Select[Range[200], Length[PowersRepresentations[#^2, 2, 2]] > 1 &];lst= {};Do[R=Part[A009003,n];it=0;Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s (s-a) (s-b) (s-c);If[area2>0&&IntegerQ[Sqrt[area2]]&&R==a*b*c/(4*Sqrt[area2])&&IntegerQ[Sqrt[area2]/s],it=it+1]],{a,2*R},{b,a},{c,b}];AppendTo[lst,it],{n,1,30}];lst

A309312 Areas of Heronian triangles that uniquely define the lengths of the triangle's sides.

Original entry on oeis.org

6, 30, 42, 54, 66, 72, 114, 132, 150, 198, 204, 234, 294, 306, 378, 390, 408, 468, 486, 522, 570, 594, 648, 684, 726, 744, 750, 798, 816, 876, 984, 1014, 1026, 1110, 1170, 1218, 1254, 1290, 1332, 1350, 1368, 1428, 1494, 1518, 1548, 1554, 1590, 1734, 1770, 1782, 1806, 1938, 1950, 1974

Offset: 1

Jacob Vecht, Jul 22 2019

These are not necessarily primitive Heronian triangles. The corresponding sequence for primitive Heronian triangles is neither a subset nor a superset of this sequence.

			A triangle with integer sides and an area of 42 square units necessarily has sides 7,15,20, so 42 is a term. 12 is not a term because an area of 12 could have sides 6,5,5 or 8,5,5.

Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 500 terms from Jacob Vecht)
Wikipedia, Heronian triangle

Subsequence of A188158.

Corrected 37th term "1248" to "1254".

A383413 Area A of triangles such that the sides are distinct integers and A is an integer.

Original entry on oeis.org

6, 24, 30, 36, 42, 54, 60, 66, 72, 84, 90, 96, 114, 120, 126, 132, 144, 150, 156, 168, 180, 198, 204, 210, 216, 234, 240, 252, 264, 270, 288, 294, 300, 306, 324, 330, 336, 360, 378, 384, 390, 396, 408, 420, 456, 462, 468, 480, 486, 504, 510, 522, 528, 540, 546, 576, 594

Offset: 1

Karl-Heinz Hofmann, Apr 26 2025

nonn

All terms are multiples of 6.

Subsequence of A188158.

			72 is in the sequence because the triangle with sides {a=5, b=29, c=30} has an area of exactly 72 and all sides are distinct.
12 is not in the sequence because this area is only possible with the isosceles triangles {a=5, b=5, c=6} and {a=5, b=5, c=8} with a and b not distinct.

Wikipedia, Heronian triangle

Cf. A188158, A316853.

nn = 450; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[a!=b!=c&&0 < area2 <= nn^2 && IntegerQ[Sqrt[area2]], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a-1}, {c, b-1}]; Union[lst]; lst (* James C. McMahon, May 10 2025 *)

A385819 Numbers k such that there are least five primitive Heron triangles having the same area and perimeter k.

Original entry on oeis.org

2842, 3542, 5642, 5750, 6314, 7238, 7546, 9790, 15470, 15778, 17710, 20026, 21658, 21970, 22610, 26962

Offset: 1

Zhining Yang, Jul 09 2025

			3542 is a term because there exists 5 primitive Heron triangles: {{421,1518,1603}, {511,1375,1656}, {583,1288,1671},{759,1096,1687}, {851,1001,1690}} with same perimeter 3542 and same area 318780.
20026 is a term because there exists 6 primitive Heron triangles: {{2108,8493,9425}, {2173,8398,9455}, {2261,8277,9488}, {2418,8075,9533}, {4123,6205,9698}, {4588,5729,9709}} with same perimeter 20026 and same area 8410920.

BBS Math Blog, Primitive Heron triangles with equal perimeter and area (in Chinese).

Cf. A051516, A070138, A188158

sol = Association[];
For[n = 2, n <= 6000, n += 2,
For[z = Ceiling[n/3], z < Floor[n/2], z++,
For[x = 1, x < Floor[n/3], x++, y = n - x - z;
   If[x + y > z > y > x && GCD[x, y, z] == 1, p = (x + y + z)/2;
    A = Sqrt[p (p - x) (p - y) (p - z)];
    If[IntegerQ[A], d = ToString@n <> "->" <> ToString@A; t = {x, y, z};
     If[KeyExistsQ[sol, d], AppendTo[sol[d], t], sol[d] = {t}]]]]]];
Select[sol, Length@# > 4 &]

A228097 Integer areas of excentral triangles of integer-sided triangles.

Original entry on oeis.org

30, 50, 75, 120, 195, 200, 260, 270, 300, 340, 450, 480, 510, 525, 585, 675, 700, 750, 765, 780, 800, 845, 1014, 1015, 1040, 1080, 1156, 1200, 1250, 1360, 1365, 1470, 1530, 1554, 1740, 1755, 1800, 1845, 1875, 1920, 2040, 2100, 2210, 2220, 2275, 2340, 2430

Offset: 1

Michel Lagneau, Oct 26 2013

nonn

The excentral triangle, also called the tritangent triangle, of a triangle ABC is the triangle IJK with vertices corresponding to the excenters of ABC.
The excentral triangle has side lengths:
  a' = a*csc(A/2) where csc(z)=1/sin(z);
  b' = b*csc(B/2);
  c' = c*csc(C/2);
and area:
  A' = 4*A*a*b*c/((a+b-c)*(a-b+c)*(-a+b+c)).
Property of this sequence:
The areas of the original triangles are integers. The primitive triangles with areas a(n) are 30, 50, 75, 195, ...
The non-primitive triangles with areas 4*a(n) are in the sequence.
The following table gives the first values (A', A, a, b, c) where A' is the area of the excentral triangles, A is the area of the reference triangles ABC, a, b, c the integer sides of the original triangles ABC.
   ----------------------
   | A'|  A |  a|  b|  c|
   ----------------------
   | 30|   6|  3|  4|  5|
   | 50|  12|  5|  5|  6|
   | 75|  12|  5|  5|  8|
   |120|  24|  6|  8| 10|
   |195|  30|  5| 12| 13|
   |200|  48| 10| 10| 12|
   |260|  24|  4| 13| 15|
   |270|  54|  9| 12| 15|
   |300|  48| 10| 10| 16|
   |340|  60|  8| 15| 17|
   ......................

			30 is in the sequence because the area A' = 4*A*a*b*c/((a+b-c)*(a-b+c)*(-a+b+c)) of the excentral triangle corresponding to the initial triangle (3,4,5) is A' = 4*6*3*4*5/((3+4-5)*(3-4+5)*(-3+4+5)) = 30, where A = 6 obtained by Heron's formula A = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(6*(6-3)*(6-4)*(6-5)) = 6, and where s = 6 is the semiperimeter.
The sides of the excentral triangle are:
  a' = 3*csc(1/2*arcsin(3/5)) = 9.48683298...
  b' = 4*csc(1/2*arcsin(4/5)) = 8.94427191...
  c' = 5*sqrt(2) = 7.07106781...

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

Eric Weisstein's World of Mathematics, Excentral Triangles

Cf. A188158, A227879.

nn = 500; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 && IntegerQ[4*Sqrt[area2]*a*b*c/((a + b - c)*(a - b + c)*(-a + b + c))], AppendTo[lst, 4*Sqrt[area2]*a*b*c/((a + b - c)*(a - b + c)*(-a + b + c))]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

A228435 Integer areas of the mixtilinear triangles of integer-sided triangles.

Original entry on oeis.org

6, 14, 24, 54, 56, 84, 96, 112, 124, 126, 130, 150, 154, 156, 210, 216, 224, 266, 294, 336, 342, 344, 350, 366, 384, 436, 448, 456, 486, 496, 504, 520, 531, 546, 570, 600, 616, 624, 630, 660, 686, 726, 756, 814, 840, 864, 896, 924, 1008, 1014, 1064, 1116, 1134

Offset: 1

Michel Lagneau, Nov 10 2013

nonn

The mixtilinear triangle is the triangle connecting the centers of the mixtilinear incircles.
A circle that in internally tangent to two sides of a triangle and to the circumcircle is called a mixtilinear incircle. There are three mixtilinear incircles, one corresponding to each angle of the triangle.
The area of the mixtilinear triangle is given by
A' = A*(a^3 - a^2*b - a*b^2 + b^3 - a^2*c + 6*a*b*c - b^2*c - a*c^2 - b*c^2 + c^3)/(a + b + c)^3
where A is the area of the reference triangle of sides (a, b, c).
See the link for the side lengths of the mixtilinear triangles.
Properties of this sequence:
The primitive mixtilinear triangles are 6, 14, 84, 112, 124, 130, ...
The nonprimitive triangles of areas 4*a(n), 9*a(n), ..., p^2*a(n), ... are in the sequence.
It appears that if the triangles are isosceles, one of the sides of the mixtilinear triangles is an integer, for example, for the triangles of areas a(n) = 112, 342, 448, 1008, 1368, 1792, 2800, 3078, 4032, 5472, ... corresponding to a triangle with an integer side of values respectively 8, 24, 16, 24, 48, 32, 40, 72, 48, 96, ...
The following table gives the first values (A', A, a, b, c, a', b', c') where A' is the area of the mixtilinear triangles, A is the area of the initial triangles, a, b, c are the integer sides of the initial triangles, and a', b', c' are the sides of the mixtilinear triangles.
-------------------------------------------------------------------------------
A' |   A| a| b|  c|        a'         |        b'        |          c'
-------------------------------------------------------------------------------
6  |  54| 9|12| 15|  9*sqrt(5)/4      | 4*sqrt(13)/3     |  5*sqrt(37)/12
14 | 126|15|28| 41|135*sqrt(1717)/196 |28*sqrt(757)/27   | 41*sqrt(16045)/5292
24 | 216|18|24| 30|  9*sqrt(5)/2      | 8*sqrt(13)/3     |  5*sqrt(37)/6
56 | 504|30|56| 82|135*sqrt(1717)/98  |56*sqrt(757)/27   | 41*sqrt(16045)/2646
84 | 756|39|42| 45|104*sqrt(445)/147  | 7*sqrt(37)/3     | 45*sqrt(197)/49
96 | 864|36|48| 60|  9*sqrt(5)        |16*sqrt(13)/3     |  5*sqrt(37)/3
112| 972|45|45| 72| 20*sqrt(2)        |20*sqrt(2)        |        8
124|1116|39|62| 85|351*sqrt(8605)/961 |31*sqrt(853)/27   |680*sqrt(79957)/25947
126|1134|45|84|123|405*sqrt(1717)/196 |28*sqrt(757)/9    | 41*sqrt(16045)/1764
...............................................................................

			6 is in the sequence. We use two ways:
First way: from the initial triangle (9, 12, 15) the formula given in the comments gives directly the area of the mixtilinear triangle: A' = 54*(9^3 - 9^2*12 - 9*12^2 + 12^3 - 9^2*15 + 6*9*12*15 - 12^2*15 - 9*15^2 - 12*15^2 + 15^3)/(9 + 12 + 15)^3 = 6, where the area of the initial triangle A = 54 is obtained by Heron's formula A = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(18*(18-9)*(18-12)*(18-15)) = sqrt(2916) = 54, where s=18 is the semiperimeter.
Second way: by calculation of the sides a', b', c' and by use of Heron's formula. With the formulas given in the link, we find
a’ = 9*sqrt(5)/4;
b’ = 4*sqrt(13)/3;
c’ = 5*sqrt(37)/12.
Now, we use Heron's formula with (a',b',c'). We find A' = sqrt(s1*(s1-a')*(s1-b')*(s1-c')) with:
s1 = (a' + b' + c')/2 = (9*sqrt(5)/4 + 4*sqrt(13)/3 + 5*sqrt(37)/12)/2.
We find A' = 6.

Eric Weisstein's World of Mathematics, Mixtilinear triangles

Cf. A188158.

nn = 500; lst = {}; Do[s = (a + b + c)/2; t = a^3 - a^2*b - a*b^2 + b^3 - a^2*c + 6*a*b*c - b^2*c - a*c^2 - b*c^2 + c^3; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 && IntegerQ[Sqrt[area2] * t/(a + b + c)^3], AppendTo[lst, Sqrt[area2] * t/(a + b + c)^3]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

A229926 Integer areas of the integer-sided triangles T(n) defined by the property: a(0) = 6 ; for n > 0, a(n) is the area A where the smallest side of T(n) is the greatest side of T(n-1).

Original entry on oeis.org

6, 12, 24, 48, 96, 192, 384, 768, 1080, 1080, 3888, 4320, 15360, 69120, 69120, 248832, 349920, 349920, 1259712, 342144, 7226112, 10782720, 17031168, 18095616, 19226592, 21660210, 30270240, 44706816, 81544320, 128798208

Offset: 0

Michel Lagneau, Oct 03 2013

nonn

Subsequence of A188158.
The sequence of the common sides is {5, 6, 10, 12, 20, 24, 40, 48, 51, 90, 108, 208, 384, 408, 720, 864, 918, 1620, 1944, 3880, 4656, 6240, 6336, ...}
a(n) = 6*2^n for n = 0, 1, 2,..., 7, and then this property disappears.
The area is given by Heron's formula A = sqrt(s(s-a)(s-b)(s-c)) where the semiperimeter s = (a + b + c)/2.
The following table gives the first values (n, A, a, b, c) where a <= b <= c are the integer sides of the triangles.
+----+------+-----+-----+-----+
|  n |    A |   a |   b |   c |
+----+------+-----+-----+-----+
|  0 |    6 |   3 |   4 |   5 |
|  1 |   12 |   5 |   5 |   6 |
|  2 |   24 |   6 |   8 |  10 |
|  3 |   48 |  10 |  10 |  12 |
|  4 |   96 |  12 |  16 |  20 |
|  5 |  192 |  20 |  20 |  24 |
|  6 |  384 |  24 |  32 |  40 |
|  7 |  768 |  40 |  40 |  48 |
|  8 | 1080 |  48 |  51 |  51 |
|  9 | 1080 |  51 |  51 |  90 |
| 10 | 3888 |  90 |  90 | 108 |
| 11 | 4320 | 108 | 116 | 208 |
+----+------+-----+-----+-----+

Cf. A188158.

with(numtheory):nn:=15000:a:=5: printf ( "%d %d %d %d %d \n",1,6,3,4,a):
    for n from 2 to 40 do:
       ii:=0:
      for b from a to nn while(ii=0) do:
        for c from b to nn while(ii=0)  do:
          p:=(a+b+c)/2 : x:=p*(p-a)*(p-b)*(p-c):
          if x>0
          then
          x0:= sqrt(x):
          else
          fi:
           if x0=floor(x0)
           then
           ii:=1:printf ( "%d %d %d %d %d \n",n,x0,a,b,c):
           a:=max(b,c):
           else
           fi:
          od:
        od:
       od:

A230361 Integer areas of the tangential triangles corresponding to the integer-sided triangles with integer areas.

Original entry on oeis.org

23660, 26250, 53235, 94640, 105000, 147875, 212940, 222530, 236250, 378560, 390390, 420000, 479115, 591500, 656250, 788970, 851760, 945000, 1286250, 1330875, 1561560, 1680000, 1916460, 2126250, 2608515, 2625000, 3176250, 3407040, 3513510

Offset: 1

Michel Lagneau, Oct 16 2013

nonn

The tangential triangle is the triangle (A', B', C') of side lengths (a', b', c') formed by the lines tangent to the circumcircle of a given triangle (A, B, C) of side lengths (a, b, c) at its vertices.
The area of the tangential triangle is given by Area = (1/2)*Area(A,B,C)*|sec A * sec B * sec C|
The side lengths of the tangential triangle are:
a' = 2*a^3*b*c/|a^4 - (b^2 - c^2)^2|
b' = 2*a*b^3*c/|b^4 - (c^2 - a^2)^2|
c' = 2*a*b*c^3/|c^4 - (a^2 - b^2)^2|
In the general case, these sides lengths are rational numbers (see the examples in the table below). But it is possible to find integer sides, for example the triangle (210, 210, 252) generates a tangential triangle (625, 625, 350).
It is possible to find the same tangential triangle with two distinct triangles of side lengths (a1,b1,c1) and (a2,b2,c2); for example, the triangles (105, 105, 175) and (140, 140, 224) generate the same tangential triangle (625/2, 625/2, 175).
The following table gives the first values (S', S, a, b, c, a', b', c') where S' is the area of the tangential triangle (A', B', C'), S is the area of the initial triangle (A, B, C), a, b, c the integer sides of the triangle (A, B, C) and a', b', c' are the integer sides of the tangential triangle (A', B', C').
**************************************************************
   S'  *   S   *  a  *  b  *  c  *   a'    *    b'   *  c'
**************************************************************
23660  * 10584 *  84 * 273 * 315 *  182    *  2197/6  * 1625/6
26250  *  5292 * 105 * 105 * 126 *  625/2  *   625/2  *  175
26250  *  9408 * 140 * 140 * 224 *  625/2  *   625/2  *  175
26250  * 24192 * 168 * 360 * 480 *  175    *   625/2  *  625/2
53235  *  8064 * 104 * 160 * 168 * 2197/4  *  1625/4  *  273
94640  * 42336 * 168 * 546 * 630 *  364    *  2197/3  * 1625/3
105000 * 21168 * 210 * 210 * 252 *  625    *   625    *  350
105000 * 37632 * 280 * 280 * 448 *  625    *   625    *  350
147875 * 40320 * 200 * 416 * 504 * 8125/12 * 10985/12 *  455
212940 * 32256 * 208 * 320 * 336 * 2197/2  *  1625/2  *  546
.........................................

			26250 is in the sequence because the triangle of integer sides (a, b, c)= (105, 105, 126) generates the tangential triangle (a', b', c') = (625/2, 625/2, 175) where:
a' = 2*a^3*b*c/|a^4 - (b^2 - c^2)^2| = 625/2
b' = 2*a*b^3*c/|b^4 - (c^2 - a^2)^2| = 625/2
c' = 2*a*b*c^3/|c^4 - (a^2 - b^2)^2| = 175
The area of this triangle is given by two possible ways:
First way:
Heron's formula gives Area = sqrt(s'*(s'-a')*(s'-b')*(s'-c')) = sqrt(400*(400-625/2)*(400-625/2)*(400-175)) = 26250 with the semiperimeter s' = (625/2 + 625/2 + 175)/2 = 400.
Second way:
Area of the triangle (a, b, c) = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(168*(168-105)*(168-105)*(168-126)) = 5292 with the semiperimeter s = (105 + 105 + 126)/2 = 168.
Then, we use the formula Area = (1/2)* Area(A,B,C) * |sec A * sec B * sec C| = 2646*5/3*5/3*25/7 = 26250 where:
sec A = 1/cos A = 2*b*c/(b^2+c^2-a^2)= 5/3;
sec B = 1/cos B = 2*a*c/(c^2+a^2-b^2)= 5/3;
sec C = 1/cos C = 2*a*b/(a^2+b^2-c^2)= 25/7.

Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.
Kimberling, C. Triangle Centers and Central Triangles. Congr. Numer. 129, 1-295, 1998.

C. Kimberling, Central Points and Central Lines in the Plane of a Triangle, Math. Mag. 67, 163-187, 1994.
Eric W. Weisstein, MathWorld: Tangential Triangle

Cf. A188158.

nn=1500; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s (s-a) (s-b) (s-c); aa=Abs[((b^2+c^2-a^2)*(c^2+a^2-b^2)*(a^2+b^2-c^2))];If[0 < area2 && aa>0&& IntegerQ[Sqrt[area2]* (4*a^2*b^2*c^2)/aa], AppendTo[lst, Sqrt[area2]* (4*a^2*b^2*c^2)/aa]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

A230479 Integer areas of the integer-sided triangles such that the length of the circumradius is a square.

Original entry on oeis.org

168, 336, 432, 600, 768, 2688, 5376, 6000, 6912, 9600, 12288, 13608, 14280, 20280, 27216, 28560, 30720, 32928, 34560, 34992, 38640, 43008, 46200, 48600, 62208, 69360, 77280, 86016, 96000, 105000, 108000, 110592, 118272, 153600, 196608

Offset: 1

Michel Lagneau, Oct 20 2013

nonn

The primitive areas are 168, 338, 432, 600, 768, 13608, 14280, 20280, 27216, ...
The non-primitive areas 16*a(n) are in the sequence because if R is the circumradius corresponding to a(n), then 4*R is the circumradius corresponding to 16*a(n).
Each circumradius belongs to the sequence {25, 100, 169, 225, 289, 400, 625, 676, ...}, and it seems that this last sequence is A198385 (second of a triple of squares in arithmetic progression).
The following table gives the first values (A, R, a, b, c) where A is the integer area, R the radius of the circumcircle, and a, b, c are the integer sides of the triangle.
**************************************
*     A  *    R  *   a  *   b  *  c  *
**************************************
*   168  *   25  *  14  *  30 *  40  *
*   336  *   25  *  14  *  48 *  50  *
*   432  *   25  *  30  *  30 *  48  *
*   600  *   25  *  30  *  40 *  50  *
*   768  *   25  *  40  *  40 *  48  *
*  2688  *  100  *  56  * 120 * 160  *
*  5376  *  100  *  56  * 192 * 200  *
*  6912  *  100  * 120  * 120 * 192  *
*  9600  *  100  * 120  * 160 * 200  *
* 12288  *  100  * 160  * 160 * 192  *
* 13608  *  225  * 126  * 270 * 360  *
* 14280  *  169  * 130  * 238 * 312  *
* 20280  *  169  * 130  * 312 * 338  *
* 27216  *  225  * 126  * 432 * 450  *
.............................

			168 is in the sequence because the area of the triangle (14, 30, 40) is given by Heron's formula A = sqrt(42*(42-14)*(42-30)*(42-40))= 168 where the number 42 is the semiperimeter, and the circumcircle is given by R = a*b*c/(4*A) = 14*30*40/(4*168) = 25, which is a square.

Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32. Solution published in Vol. 16, Issue 2, November 2008, p. 32.

Eric W. Weisstein, MathWorld: Circumradius

Cf. A188158, A208984, A210207.

nn = 1000; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 && IntegerQ[Sqrt[area2]] && IntegerQ[Sqrt[a*b*c/(4*Sqrt[area2])]], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

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

Circumradius R = a*b*c/4A.

cp's OEIS Frontend

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