A188158 -id:A188158 - cp's OEIS frontend

A228858 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 exactly 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).

6, 12, 126, 60, 240, 210, 1080, 336, 1260, 924, 2016, 2640, 7980, 6930, 420, 2520, 840, 3696, 10080, 5460, 3360

Offset: 1

Michel Lagneau, Sep 05 2013

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 Maple program examines all triangles having longest side <= 600.

Conjecture: there exists an integer corresponding to the common area of exactly n distinct triangles (x_i, y_i, z_i) for i = 1,2,...,n such that A(x_1, y_1, z_1) = A(x_2, y_2, z_2) = ... = A(x_n, y_n, z_n) for n = 1, 2, ...

			a(6) = 210 because there exists exactly 6 triangles with the same area. We obtain:
(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, A228847.

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 50 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:else fi:od: if ii = n then jj:=1:printf ( "%d %d \n",n,lst[i]):else fi:od:od:

A229098 Smallest area A of Heron triangles with sides (a, b, c) in arithmetic progression of the form b - d(n), b, b + d(n), where d(n) = A091998(n) = 12*n +- 1.

Original entry on oeis.org

6, 156, 126, 546, 3750, 7350, 570, 1176, 14406, 2046, 3216, 4740, 1554, 3354, 43350, 54150, 6180, 3924, 17556, 84966, 3294, 24174, 106134, 7446, 126150, 144150, 28236, 33174, 21294, 10374, 6006, 9264, 16716, 247254, 252150, 277350, 282534, 55944, 75894, 26676

Offset: 1

Michel Lagneau, Sep 13 2013

nonn

a(1) = A011945(1).
According to the reference, d(n) is congruent (mod 12) to 1 or -1.
Let the sides be b - d, b, b + d where 1 <= d <= b. Then the semiperimeter s = 3b/2 and by Heron's formula, the area is A = b*sqrt(3*(b^2 - 4*d^2))/4.
The following table gives the first values (d(n), a, b, c, A):
+------+-----+-----+-----+-------+
| d(n) |  a  |  b  |  c  |   A   |
+------+-----+-----+-----+-------+
|   1  |   3 |   4 |   5 |     6 |
|  11  |  15 |  26 |  37 |   156 |
|  13  |  15 |  28 |  41 |   126 |
|  23  |  29 |  52 |  75 |   546 |
|  25  |  75 | 100 | 125 |  3750 |
|  35  | 105 | 140 | 175 |  7350 |
|  37  |  39 |  76 | 113 |   570 |
|  47  |  51 |  98 | 145 |  1176 |
|  49  | 147 | 196 | 245 | 14406 |

			a(2) = 156 is in the sequence because d(2) = A091998(2) = 11 and (a, b, c) = (15, 26, 37) => the semiperimeter is (15 + 26 + 37)/2 = 39, and A = sqrt(39*(39-15)*(39-26)*(39-37)) = 156.

J. A. MacDougall, Heron Triangles With Sides in Arithmetic Progression, School of Mathematical and Physical Sciences, University of Newcastle, NSW, Australia 2308, February 2, 2005.
Kival Ngaokrajang, Illustration of initial terms

Cf. A011945, A091998, A188158.

with(numtheory):u:=0:nn:=1000:lst:={1}:for k from 1 to 10 do:x:=12*k-1:y:=12*k+1:lst:=lst union {x} union {y}:od:for n from 1 to 20 do:ii:=0:d:=lst[n]:for b from 1 to nn while(ii=0)do:s:= b*sqrt(3*(b^2-4*d^2))/4:if s>0 and s=floor(s) then ii:=1:u:=u+1:printf ( "%d %d %d %d \n",u,d,b,s):else fi:od:od:

A231275 Integer areas of the outer vecten triangles of integer-sided triangles.

Original entry on oeis.org

49, 91, 105, 196, 289, 301, 364, 379, 420, 441, 459, 505, 529, 609, 631, 697, 751, 784, 799, 819, 889, 897, 945, 961, 991, 1009, 1156, 1171, 1204, 1219, 1225, 1351, 1395, 1401, 1456, 1516, 1521, 1611, 1680, 1681, 1689, 1759, 1764, 1836, 1849, 1939, 2020, 2095

Offset: 1

Michel Lagneau, Nov 06 2013

nonn

Consider the external erection of three squares on the sides of a triangle ABC. These centers form a triangle IJK. The vertices of this triangle satisfy a number of remarkable properties.
The area of the outer Vecten triangle is A' = A + (a^2 + b^2 + c^2)/8, where A is the area of the reference triangle.
Its side lengths are
  a' = sqrt((b^2 + c^2 + 4*A)/2),
  b' = sqrt((a^2 + c^2 + 4*A)/2),
  c' = sqrt((a^2 + b^2 + 4*A)/2).
The circumcircle of the outer Vecten circle is the outer vecten circle.
Properties of this sequence:
The primitive triangles are 49, 91, 105, 289, 301, ...
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 outer vecten triangles is an integer (see the table below).
The following table gives the first values (A, A', a, b, c, a', b', c') where A is the area of the initial triangles, A' is the area of the outer vecten triangles, a, b, c are the integer sides of the initial triangles, and a', b', c' are the sides of the outer vecten triangles.
-----------------------------------------------------------------------
|   A' | A   |  a  |   b |  c   |    a'      |  b'       |  c'        |
-----------------------------------------------------------------------
|   49 |  24  |  6 |   8 | 10  |  sqrt(130) | 2*sqrt(29) | 7*sqrt(2)  |
|   91 |  48  | 10 |  10 | 12  |  sqrt(218) |  sqrt(218) |   14       |
|  105 |  48  | 10 |  10 | 16  |  sqrt(274) |  sqrt(274) |   14       |
|  196 |  96  | 12 |  16 | 20  | 2*sqrt(130)|  4*sqrt(29)| 14*sqrt(2) |
|  289 | 120  | 10 |  24 | 26  |  sqrt(866) | 2*sqrt(157)| 17*sqrt(2) |
|  301 |  96  |  8 |  26 | 30  |14*sqrt(5)  |  sqrt(674) | sqrt(562)  |
|  364 | 192  | 20 |  20 | 24  |2*sqrt(218) | 2*sqrt(218)| 28         |
|  379 | 144  | 18 |  20 | 34  |sqrt(1066)  | 2*sqrt(257)| 5*sqrt(26) |
|  420 |  192 | 20 |  20 | 32  |2*sqrt(274) | 2*sqrt(274)| 28         |
|  441 | 216  | 18 |  24 | 30  |3*sqrt(130) | 6*sqrt(29) | 21*sqrt(2) |
|  459 | 240  | 20 |  26 | 26  |    34      | sqrt(1018) | sqrt(1018) |
|  505 | 168  | 14 |  30 | 40  |sqrt(1586)  | sqrt(1586) | 2*sqrt(221)|
....................................................................

			49 is in the sequence. We use two ways:
First way: with the triangle (6, 8, 10) the formula A' = A + (a^2 + b^2 + c^2)/8 gives directly the result: A' = 24 + (6^2 + 8^2 + 10^2)/8 = 24 + 25 = 49 where the area A = 24 is obtained by Heron's formula A = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(12*(12-6)*(12-8)*(12-10)) = sqrt(576) = 24, where s is the semiperimeter.
Second way: by calculation of the sides a', b', c' and by use of Heron's formula.
  a' = sqrt((b^2 + c^2 + 4*A)/2) = sqrt((8^2 + 10^2 + 4*24)/2) = sqrt(130);
  b' = sqrt((a^2 + c^2 + 4*A)/2) = sqrt((6^2 + 10^2 + 4*24)/2) = 2*sqrt(29);
  c' = sqrt((a^2 + b^2 + 4*A)/2) = sqrt((6^2 + 8^2 + 4*24)/2) = 7*sqrt(2).
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 = (sqrt(130) + 2*sqrt(29) + 7*sqrt(2))/2;
We find A'= 49.

H. S. M. Coxeter and S. L. Greitzer, Points and Lines Connected with a Triangle, Ch. 1 in Geometry Revisited, Washington, DC, Math. Assoc. Amer., pp. 1-26 and 96-97, 1967.

Eric Weisstein's World of Mathematics, Outer Vecten Triangle

Cf. A188158, A231276.

nn=500; lst={}; Do[s =(a + b + c)/2; If[IntegerQ[s], area2 = s (s-a)(s-b) (s-c); If[area2 > 0 && IntegerQ[Sqrt[area2] + (a^2 + b^2 + c^2)/8], AppendTo[lst,  Sqrt[area2] + (a^2 + b^2 + c^2)/8]]], {a,nn}, {b,a}, {c,b}]; Union[lst]

A231276 Integer areas of the inner vecten triangles of integer-sided triangles.

Original entry on oeis.org

5, 20, 21, 23, 29, 39, 41, 45, 59, 63, 80, 83, 84, 92, 116, 125, 131, 156, 164, 173, 180, 189, 203, 207, 227, 236, 237, 245, 252, 257, 261, 269, 320, 329, 332, 336, 351, 368, 369, 371, 405, 464, 479, 497, 500, 524, 525, 531, 567, 575, 605, 623, 624, 656, 663

Offset: 1

Michel Lagneau, Nov 06 2013

nonn

Consider the internal erection of three squares on the sides of a triangle ABC. These centers form a triangle IJK.
The area of the inner vecten triangle is A' = A - (a^2 + b^2 + c^2)/8, where A is the area of the reference triangle.
Its side lengths are
  a' = sqrt((b^2 + c^2 - 4*A)/2),
  b' = sqrt((a^2 + c^2 - 4*A)/2),
  c' = sqrt((a^2 + b^2 - 4*A)/2).
The circumcircle of the inner vecten circle is the inner vecten circle.
Properties of this sequence:
The primitive triangles are 5, 21, 23, 29, 39, 41, ...
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 inner vecten triangles is an integer (see the table below).
The following table gives the first values (A, A', a, b, c, a', b', c') where A is the area of the initial triangles, A' is the area of the inner vecten triangles, a, b, c are the integer sides of the initial triangles, and a', b', c' are the sides of the inner vecten triangles.
-----------------------------------------------------------------------
|  A' |  A   |  a  |   b  |  c   |     a'      |    b'       |  c'
-----------------------------------------------------------------------
|  5  |   48 |  10 |   10 |  12  |   sqrt(26)  |    sqrt(26) |   2
| 20  |  192 |  20 |   20 |  24  | 2*sqrt(26)  |  2*sqrt(26) |   4
| 21  |  240 |  20 |   20 |  26  |    14       |   sqrt(58)  | sqrt(58)
| 23  | 1680 |  48 |   74 |  74  |    46       |   sqrt(530) | sqrt(530)
| 29  | 1680 |  50 |   68 |  78  |   sqrt(1994)|  2*sqrt(233)| sqrt(202)
| 39  | 1680 |  58 |   58 |  80  |   sqrt(1522)|   sqrt(1522)|   2
| 41  |  336 |  26 |   28 |  30  |   sqrt(170) |  2*sqrt(29) | sqrt(58)
| 45  |  432 |  30 |   30 |  36  | 3*sqrt(26)  |  3*sqrt(26) |   6
| 59  | 1440 |  50 |   58 |  72  |   sqrt(1394)|    sqrt(962)| 2*sqrt(13)
| 63  |  480 |  32 |   34 |  34  |    14       |    sqrt(130)|  sqrt(130)
| 80  |  768 |  40 |   40 |  48  | 4*sqrt(26)  |  4*sqrt(26) |   8
| 83  | 2880 |  74 |   78 | 104  |   sqrt(2690)|   sqrt(2386)| 2*sqrt(5)
.............................................................

			5 is in the sequence. We use two ways:
First way: with the triangle (10, 10, 12) the formula A' = A - (a^2 + b^2 + c^2)/8 gives directly the result: A' = 48 - (10^2 + 10^2 + 12^2)/8 = 5 where the area A = 48 is obtained by Heron's formula A = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(16*(16-10)*(16-10)*(16-12)) = 48, where s is the semiperimeter.
Second way: by calculation of the sides a', b', c' and by use of Heron's formula.
  a’ = sqrt((b^2 + c^2 - 4*A)/2) = sqrt((10^2 + 12^2 - 4*48)/2) = sqrt(26);
  b’ = sqrt((a^2 + c^2 - 4*A)/2) = sqrt((10^2 + 12^2 - 4*48)/2) = sqrt(26);
  c’ = sqrt((a^2 + b^2 - 4*A)/2) = sqrt((10^2 + 10^2 - 4*48)/2) = 2.
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 = (sqrt(26) + sqrt(26) + 2)/2.
We find A' = 5.

H. S. M. Coxeter and S. L. Greitzer, Points and Lines Connected with a Triangle, Ch. 1 in Geometry Revisited, Washington DC, Math. Assoc. Amer., pp. 1-26 and 96-97, 1967.

Eric Weisstein's World of Mathematics, Inner Vecten Triangle

Cf. A188158, A231275.

nn = 500; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); t = (a^2 + b^2 + c^2)/8; If[0 < area2 && Sqrt[area2] - t > 0 && IntegerQ[Sqrt[area2] - t], AppendTo[lst, Sqrt[area2] - t]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

A231739 Integer areas of the Lucas Central triangles of integer-sided triangles.

Original entry on oeis.org

2775, 11100, 24975, 34125, 44400, 69375, 99900, 135975, 136500, 177600, 224775, 277500, 307125, 335775, 399600, 468975, 543900, 546000, 624375, 710400, 801975, 853125, 899100, 1001775, 1110000, 1223775, 1228500, 1343100, 1467975, 1598400, 1672125, 1734375

Offset: 1

Michel Lagneau, Nov 13 2013

nonn

Consider a reference triangle ABC and externally inscribe a square on the side BC. Now join the new vertices S_AB and S_AC of this square with the vertex A, marking the points of intersection Q_ABC and Q_ACB. Next, draw lines perpendicular to the side BC through each of Q_ABC and Q_ACB. These points cross the sides AB and AC at Q_AB and Q_AC, respectively, resulting in an inscribed square Q_ABC Q_ACB Q_AB Q_AC. The circumcircle through A, Q_AB, and Q_AC is then known as the Lucas A-circles (Panakis 1973, p. 458; Yiu and Hatzipolakis 2001), and repeating the process for other sides gives the corresponding B- and C-circles.
The Lucas central triangle is the triangle L1 L2 L3 formed by the centers of the Lucas circles of a given reference triangle ABC.
The Lucas central triangle has side lengths
a' = 2*R*(a*b*c+b^2*R+c^2*R)*a/((a*c+2*b*R)*(a*b+2*c*R));
b' = 2*R*(a*b*c+a^2*R+c^2*R)*b/((b*c+2*a*R)*(a*b+2*c*R));
c' = 2*R*(a*b*c+a^2*R+b^2*R)*c/((b*c+2*a*R)*(a*c+2*b*R)).
Its area is given by
S' = a*b*c*R^2*sqrt(u)/v where:
u=3*a^2*b^2*c^2+4*a*b*c*(a^2+b^2+c^2)*R+4*(a^2*b^2+a^2*c^2+b^2*c^2)*R^2 ;
v=(b*c+2*a*R)*(a*c+2*b*R)*(a*b+2*c*R);
R = a*b*c/sqrt((a+b+c)*(b+c-a)*(c+a-b)*(a+b-c)) = a*b*c/(4*S) where R is the circumradius of the reference triangle ABC and S its area.
Properties of this sequence:
The side lengths of the Lucas central triangles are rational numbers, sometimes integers, for example a(n) = 136500, ...
The primitive Lucas central triangles are 2775, 34125, ...
The non-primitive triangles of areas 4*a(n), 9*a(n), ..., p^2*a(n), ... are in the sequence.
The following table gives the first values (S', S, a, b, c, a', b', c') where S' is the area of the Lucas central triangles, S is the area of the initial triangles ABC, a, b, c are the integer sides of ABC and a', b', c' are the sides of the Lucas central triangles.
+--------+---------+------+------+------+----------+---------+---------+
|    S'  |     S   |   a  |   b  |   c  |    a'    |    b'   |   c'    |
+--------+---------+------+------+------+----------+---------+---------+
|   2775 |    8214 |  111 |  148 |  185 |  975/14  |  580/7  |  185/2  |
|  11100 |   32856 |  222 |  296 |  370 |  975/7   |  1160/7 |  185    |
|  24975 |   73926 |  333 |  444 |  555 |  2925/14 |  1740/7 |  555/2  |
|  34125 |  115248 |  490 |  490 |  588 |  545/2   |   545/2 |  300    |
|  44400 |  131424 |  444 |  592 |  740 |  1950/7  |  2320/7 |  370    |
|  69375 |  205350 |  555 |  740 |  925 |  4875/14 |  2900/7 |  925/2  |
|  99900 |  295704 |  666 |  888 | 1110 |  2925/7  |  3480/7 |  555    |
| 135975 |  402486 |  777 | 1036 | 1295 |   975/2  |   580   | 1295/2  |
| 136500 |  460992 |  980 |  980 | 1176 |   545    |   545   |  600    |
| 177600 |  525696 |  888 | 1184 | 1480 |  3900/7  |  4640/7 |  740    |
| 224775 |  665334 |  999 | 1332 | 1665 |  8775/14 |  5220/7 | 1665/2  |
| 277500 |  821400 | 1110 | 1480 | 1850 | 48755/7  |  5800/7 |  925    |
| 307125 | 1037232 | 1470 | 1470 | 1764 |  1635/2  |  1635/2 |  900    |
+--------+---------+------+------+------+----------+---------+---------+

			2775 is in the sequence. We use two ways:
First way: from the initial triangle (111, 148, 185) the formula in the comments gives directly the area of the Lucas central triangle: S' = a*b*c*R^2*sqrt(u)/v where:
R = a*b*c/4S = a*b*c/(4*sqrt(s(s-a)(s-b)(s-c))) = 111*148*185/(4*sqrt(222(222-111)(222-148)(222-185))) = 111*148*185/(4*8214) = 185/2 with S=8214.
sqrt(u) = sqrt(3*a^2*b^2*c^2+4*a*b*c*(a^2+b^2+c^2)*R+4*(a^2*b^2+a^2*c^2+b^2*c^2)*R^2) = sqrt(154007727700225) = 12409985.
v = (b*c+2*a*R)*(a*c+2*b*R)*(a*b+2*c*R) = 116291549487925
And S' = 111*148*185*(185/2)^2*12409985/116291549487925 = 2775.
Second way: by calculation of the sides a', b', c' and by using Heron's formula. With the formulas given in the link, we find
a' = 975/14;
b' = 580/7;
c' = 185/2.
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 = (975/14 + 580/7 + 185/2)/2 = 245/2;
We find S'= 2775.

Peter J. C. Moses, Circles and Triangle Centers Associated with the Lucas Circles, Forum Geom. 5, 97-106, 2005.
Wolfram MathWorld, Lucas Central Triangle
Wolfram MathWorld, Lucas Circles
P. Yiu and A. P. Hatzipolakis, The Lucas Circles of a Triangle, Amer. Math. Monthly 108, 444-446, 2001.

Cf. A188158.

nn=2000;lst={};Do[s =(a + b + c)/2;area2=s (s-a)(s-b)(s-c);If[area2>0,R = a*b*c /(4*Sqrt[area2]);u = a*b*c*R^2 * Sqrt[3*a^2*b^2*c^2 + 4*a*b*c*(a^2 + b^2 + c^2)*R + 4*(a^2*b^2 + a^2*c^2 + b^2*c^2)*R^2];v = (b*c + 2*a*R)*(a*c + 2*b*R)*(a*b + 2*c*R);If[IntegerQ[u/v], AppendTo[lst, u/v]]],{a,nn},{b,a},{c,b}];Union[lst]

A232274 Integer areas A of the integer-sided triangles such that the length of the inradius and the circumradius are both a perfect square.

Original entry on oeis.org

168, 2688, 13608, 43008, 105000, 108000, 217728, 403368, 688128, 1102248, 1680000, 1728000

Offset: 1

Michel Lagneau, Nov 22 2013

Subset of A208984.
The areas of the primitive triangles of sides (a, b, c) and inradius, circumradius equals respectively to r and R are 672, 108000, ...  The sides of the nonprimitive triangles are of the form (a*k^2, b*k^2, c*k^2) with r' = r*k^2 and R' = R*k^2 where r', R' are respectively the inradius and the circumradius of the nonprimitive triangles. The areas A' of the nonprimitive triangles are A' = A*k^4.
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 inradius r is given by r = A/s and the circumradius is given by R = abc/4A.
The following table gives the first values (A, a, b, c, r, R).
+---------+------+------+------+-----+------+
|    A    |   a  |   b  |   c  |  r  |   R  |
+---------+------+------+------+-----+------+
|     168 |   14 |   30 |   40 |   4 |   25 |
|    2688 |   56 |  120 |  160 |  16 |  100 |
|   13608 |  126 |  270 |  360 |  36 |  225 |
|   43008 |  224 |  480 |  640 |  64 |  400 |
|  105000 |  350 |  750 | 1000 | 100 |  625 |
|  108000 |  480 |  510 |  510 | 144 |  289 |
|  217728 |  504 | 1080 | 1440 | 144 |  900 |
|  403368 |  686 | 1470 | 1960 | 196 | 1225 |
|  688128 |  896 | 1920 | 2560 | 256 | 1600 |
| 1102248 | 1134 | 2430 | 3240 | 324 | 2025 |
| 1680000 | 1400 | 3000 | 4000 | 400 | 2500 |
| 1728000 | 1920 | 2040 | 2040 | 576 | 1156 |
+---------+------+------+------+-----+------+

			a(1) = 168 because, for (a,b,c) = (14, 30, 40) => s= (14 + 30 + 40)/2 = 42, and
A = sqrt(42*(42-14)*(42-30)*(42-40)) = sqrt(28224) = 168;
R = abc/4A = 14*30*40/(4*168) = 25;
r = A/s = 168/42 = 4.

Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), p. 32.
Eric W. Weisstein, MathWorld: Circumradius
Eric W. Weisstein, MathWorld: Inradius

Cf. A188158, A208984.

nn=2000;Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s (s-a)(s-b)(s-c);If[0

A232329 Integer areas A of the integer-sided triangles such that the product of the inradius and the circumradius is a square.

Original entry on oeis.org

42, 168, 378, 672, 1050, 1512, 2058, 2088, 2688, 3000, 3402, 4200, 5082, 6048, 6960, 7098, 8232, 8352, 9450, 10752, 12000, 12138, 13608, 15162, 16800, 18522, 18792, 20328, 22218, 24192, 26250, 27000, 27840, 28392, 30618, 31416, 32928, 33408, 35322, 36000, 37800, 40362

Offset: 1

Michel Lagneau, Nov 22 2013

The areas of the primitive triangles of sides (a, b, c) and inradius, circumradius equals respectively to r and R are 42, 3000,...  The sides of the nonprimitive triangles are of the form (a*k, b*k, c*k) with r’ = r*k and R’=R*k where r’, R’ are respectively the inradius and the circumradius of the nonprimitive triangles. The areas A’ of the nonprimitive triangles are A’ = A*k^2. The set {A016850} (numbers (5n)^2) is included in the set of the products r*R (see the table below).
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 inradius r is given by r = A/s and the circumradius is given by R = abc/4A.
The product r*R is given by r*R = abc/2(a+b+c).
The following table gives the first values (A, a, b, c, r, R, r*R).
  -----------------------------------------------------
  |    A  |   a  |   b |   c   |   r  |   R    | r*R  |
  -----------------------------------------------------
  |    42 |   7  |  15  |  20  |    2 |  25/2  |  5^2 |
  |   168 |  14  |  30  |  40  |    4 |  25    | 10^2 |
  |   378 |  21  |  45  |  60  |    6 |  75/2  | 15^2 |
  |   672 |  28  |  60  |  80  |    8 |  50    | 20^2 |
  |  1050 |  35  |  75  | 100  |   10 | 125/2  | 25^2 |
  |  1512 |  42  |  90  | 120  |   12 |  75    | 30^2 |
  |  2058 |  49  | 105  | 140  |   14 | 175/2  | 35^2 |
  |  2688 |  56  | 120  | 160  |   16 | 100    | 40^2 |
  |  3000 |  80  |  85  |  85  |   24 | 289/6  | 34^2 |
  |  3402 |  63  | 135  | 180  |   18 | 225/2  | 45^2 |
  |  4200 |  70  | 150  | 200  |   20 | 125    | 50^2 |
  |  5082 |  77  | 165  | 220  |   22 | 275/2  | 55^2 |
  |  6048 |  84  | 180  | 240  |   24 | 150    | 60^2 |
  |  6960 |  58  | 300  | 338  |   20 | 845/4  | 65^2 |
  |  7098 |  91  | 195  | 260  |   26 | 325/2  | 65^2 |
  ....................................................

			a(1) = 42 because, for (a,b,c) = (7, 15, 20):
  the semiperimeter s = (7+15+20)/2 =21, and
  A = sqrt(21*(21-7)*(21-15)*(21-20)) = 42
  R = abc/4A = 7*15*20/(4*42) = 25/2
  r = A/s = 42/21 = 2, hence r*R = 25 is a square.

Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32.

Zak Seidov, Table of n, a(n) for n = 1..100
Zak Seidov, Table of 814 values of area A, sides a>=b>=c, semiprime s, and sqrt(Rr), in the order of increasing "a" from 20 up to 10000.
Mohammad K. Azarian, Solution to Problem S125: Circumradius and Inradius, Math Horizons, Vol. 16, Issue 2, November 2008, p. 32.
Eric Weisstein's World of Mathematics, Circumradius
Eric Weisstein's World of Mathematics, Inradius

Cf. A016850, A188158, A208984, A232274.

nn=800;lst={};Do[s=(a+b+c)/2;rr=a*b*c/(2*(a+b+c))
;If[IntegerQ[s],area2=s(s-a)(s-b)(s-c);If[0

lista(nn)=lst=[]; for (a = 1, nn, for (b=1, a, for (c=1, b, s=(a+b+c)/2; rr=a*b*c/(2*(a+b+c)); if ((type(s) == "t_INT") && (type(rr) == "t_INT"), area2=s*(s-a)*(s-b)*(s-c); if ((0Michel Marcus, Jun 09 2015

{for(a=20,10000,forstep(b=a,2,-1,forstep(c=min(b,a+b-1),a-b+1,-1,if((a+b+c)%2<1,s=(a+b+c)/2;if(issquare(s*(s-a)*(s-b)*(s-c),&A),
if((a*b*c)%(2*(a+b+c))<1&&if(issquare(a*b*c/(2*(a+b+c)),&d),
print([A,a,b,c,s,d]))))))))} \\ Faster version used for afile. Zak Seidov, Jun 06 2015

Missing term 33408 added by Zak Seidov, Jun 08 2015

A232461 Integer areas of integer-sided triangles where two sides are of square length.

Original entry on oeis.org

120, 168, 300, 360, 1920, 2016, 2688, 4680, 4800, 5760, 9720, 10140, 13608, 14280, 18720, 19080, 23256, 24300, 29160, 30720, 32760, 34440, 34680, 38640, 42120, 43008, 57720, 74880, 75000, 76800, 92160, 94080, 105000, 128700, 162240, 177072, 187500, 217728

Offset: 1

Michel Lagneau, Nov 24 2013

nonn

Subset of A188158.
The areas of the triangles (a,b,c) are given by Heron's formula, A = sqrt(s(s-a)(s-b)(s-c)), where its side lengths are a, b, c and semiperimeter s = (a+b+c)/2.
The areas A of the primitive triangles of sides (a,b,c) are 120, 168, 300, 360, 4680, ...
The areas of the nonprimitive triangle of sides (a*p^2, b*p^2, c*p^2) are in the sequence with the value A*p^4.
It is possible to find integer-sided triangles having two square sides, for example:
a(2) = 168 with sides (25,25,48) and (14,25,25);
a(3) = 300 with sides (25,25,30) and (25,25,40);
a(14) = 14280 with sides (169,169,238), (169,169,240), (100,289,291).
The following table gives the first values (A, a, b, c):
+-------+-----+-----+-----+
|   A   |  a  |  b  |  c  |
+-------+-----+-----+-----+
|   120 |  16 |  25 |  39 |
|   168 |  14 |  25 |  25 |
|   300 |  25 |  25 |  40 |
|   360 |  25 |  29 |  36 |
|  1920 |  64 | 100 | 156 |
|  2016 |  64 | 225 | 287 |
|  2688 | 100 | 100 | 192 |
|  4680 |  74 | 169 | 225 |
|  4800 | 100 | 100 | 160 |
|  5760 | 100 | 116 | 144 |
|  9720 | 144 | 225 | 351 |
| 10140 | 169 | 169 | 312 |
| 13608 | 225 | 225 | 432 |
+-------+-----+-----+-----+

			120 is in the sequence because the triangle (4^2, 5^2, 39) has semiperimeter s = (16+25+39)/2 = 40, and A = sqrt(40*(40-16)*(40-25)*(40-39)) = 120.

J. Peng, Y. Zhang, Heron triangles with figurate number sides, Acta Mathematica Hungarica (2019) 1-11.

Cf. A000290, A188158.

nn=1000;lst={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s (s-a) (s-b) (s-c);If[0 0 && IntegerQ[ar]]; nn = 80; t = {}; ps = sqr[Range[2, nn]]; mx = 3*ps[[-1]]; Do[If[p <= q && goodQ[p, q, e], aa = area[p, q, e]; If[aa <= mx, AppendTo[t, aa]]], {p, ps}, {q, ps}, {e, q - p + 2, p + q - 2, 2}]; t = Union[t] (* program from T. D. Noe adapted for this sequence - see A229746 *)

A238368 Integer area A of triangles having their side lengths in the commutative ring Z[phi] where phi is the golden ratio.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 16, 18, 20, 22, 24, 25, 27, 28, 29, 30, 32, 33, 34, 35, 36, 38, 40, 42, 44, 45, 48, 49, 50, 54, 55, 56, 58, 60, 62, 63, 64, 66, 68, 70, 72, 75, 76, 77, 78, 80, 81, 84, 88, 90, 95, 96, 98, 99, 100, 108, 110, 112, 114, 116

Offset: 1

Michel Lagneau, Feb 25 2014

nonn

Generalized integer areas triangles in the ring Z[phi] = {a + b*phi| a,b in Z}. Z[phi] is a ring because if x = a + b*phi and y = c + d*phi are in the ring, the sum x+y = a+c + (b+d)*phi is in the ring, and the product x*y = (a*c + b*d) + (a*d + b*c + b*d)*phi is also in the ring.
This sequence is tested with a and b in the range [-40, ..., +40]. For the values of areas > 150 it is necessary to expand the range of variation, but nevertheless the calculations become very long.
The sequence A188158 is included in this sequence. The numbers 5*a(n) are in the sequence because if the integer area of the integer-sided triangle (a, b, c) is A,  the area of the triangle of sides (a*sqrt(5), b*sqrt(5), c*sqrt(5)) is 5*A, where sqrt(5)= -1 + 2*phi.
The primitive areas are p = 1, 2, 3, 6, 7, 11, 22, ...  and the areas p^2*a(n) are also in the sequence.
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.
For the same area, the number of triangles is not unique, for example the area of the triangles (1,5,-2+4*phi), (2,2*phi-1, 2*phi-1),(3,3-phi,2+phi) and (4,2*phi-1, 2*phi-1) is A = 2.
It is possible to obtain rational values and also values in the ring Z[phi] for the circumradius (see the table below).
The following table gives the first values (A, a, b, c, R) where A is the integer area, a,b,c are the sides in Z[phi] and R = a*b*c/(4*A) are the values of circumradius.
----------------------------------------------------------------
|  A |  a |       b    |       c     |        R                |
----------------------------------------------------------------
|  1 |  1 |       2    | -1 +  2*phi |   sqrt(5)/2             |
|  2 |  1 |       5    | -2 +  4*phi | 5*sqrt(5)/4             |
|  3 |  3 | -1 + 2*phi | -2 +  4*phi |    5/2                  |
|  4 |  2 |       4    | -2 +  4*phi |    sqrt(5) = -1 + 2*phi |
|  5 |  2 |      13    | -5 + 10*phi | 13*sqrt(5)/2            |
|  6 |  3 |       4    |       5     |    5/2                  |
|  7 |  7 | -2 + 4*phi | -5 + 10*phi |   25/2                  |
|  8 |  5 |      13    | -8 + 16*phi | 65*sqrt(5)/4            |
| 10 |  5 |       5    | -2 +  4*phi |  5*sqrt(5)/4            |
| 11 |  2 |      11    | -5 + 10*phi |  5*sqrt(5)/2            |
| 12 |  4 |      10    | -6 + 12*phi |  5*sqrt(5) = -5 + 10*phi|
| 15 |  5 |      10    | -3 +  6*phi |  5*sqrt(5)/2            |
| 16 |  4 |       8    | -4 +  8*phi |  2*sqrt(5) = -2 + 4*phi |
| 18 |  3 |      15    | -6 + 12*phi | 15*sqrt(5)/4            |
| 20 | 10 | -2 + 4*phi | -4 +  8*phi | 15*sqrt(5)/4            |
----------------------------------------------------------------

Eric Weisstein's World of Mathematics, Ring

Cf. A188158.

err=1/10^10;nn=40;q=(Sqrt[5]+1)/2;lst={};lst1={};Do[If[u+q*v>0,lst=Union[lst,{u+q*v}]],{u,nn,-nn,-1},{v,nn,-nn,-1}];n1=Length[lst];Do[a=Part[lst,i];b=Part[lst,j];c=Part[lst,k];s=(a+b+c)/2;area2=s*(s-a)*(s-b)*(s-c);If[a*b*c!=0&&N[area2]>0&&Abs[N[Sqrt[area2]]-Round[N[Sqrt[area2]]]]

A253277 Integer area A of triangles with side lengths in the commutative ring Z[sqrt(3)].

Original entry on oeis.org

3, 6, 9, 12, 18, 21, 24, 27, 30, 33, 36, 39, 42, 48, 49, 54, 60, 63, 66, 72, 75, 78, 81, 84, 90, 96, 99, 108, 114, 117, 120, 126, 132, 138, 144, 147, 150, 156, 162, 168, 180, 189, 192, 196, 198, 204, 210, 216, 222, 225, 227, 228, 234, 240, 243, 252, 264, 270

Offset: 1

Michel Lagneau, May 02 2015

nonn

Extension of A188158 with triangles of sides in the ring Z[sqrt(3)] = {a + b sqrt(3)| a,b in Z}.
The numbers 3*A188158(n) are in the sequence because if the integer area of the integer-sided triangle (a, b, c) is A,  the area of the triangle of sides (a*sqrt(3), b*sqrt(3), c*sqrt(3)) is 3*A. The primitive areas of the sequence are in the subsequence b(n)={3, 6, 21, 30, 33, 39, 42, 49, ...} => the numbers b(n)*3^p and b(n)*q^2 are in the sequence.
The squares of the sequence are 9, 36, 49, 81, 144, 196, 225, ...
This sequence is tested with a and b in the range [-40, ..., +40]. For the values of areas > 400 it is necessary to expand the range of variation, but nevertheless the calculations become very long.
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. For the same area, the number of triangles is not unique (see the table below).
Geometric property of the triangles in the ring Z[sqrt(3)]:
It is possible to obtain integers values (or rational values) for the inradius (and/or) the circumradius of the triangles (see the table below).
The following table gives the first values (A, a, b, c, r, R) where A is the integer area, a,b,c are the sides in Z[sqrt(3)] and r = A/p, R = a*b*c/(4*A) are the values of the inradius and the circumradius respectively.
Notation in the table:
q=sqrt(3)and irrat. = irrational numbers of the form u+v*q.
+----+---------+----------+----------+-------+---------+
|  A |    a    |      b   |     c    |   r   |  R      |
+----+---------+----------+----------+-------+---------+
|  3 |   3 - q |  2 + 2q  |  1 + 3q  | irrat.|  irrat. |
|  3 |   3 + q | -2 + 2q  | -1 + 3q  | irrat.|  irrat. |
|  6 |   3     |  4       |  5       |   1   |   5/2   |
|  6 |   8     |  5 - 2q  |  5 + 2q  | 2/3   |  13/3   |
|  6 |  4q     |  4 - q   |  4 + q   | irrat.|  irrat. |
|  6 |  8q     |  7 - 2q  |  7 + 2q  | irrat.|  irrat. |
|  9 |  3 + 3q |  6 - 2q  |  9 - q   |   1   |  irrat. |
| 12 |   5     |  5       |  6       |  3/2  |  25/8   |
| 12 |   5     |  5       |  8       |  4/3  |  25/6   |
| 12 |  2q     | -1 + 5q  |  1 + 5q  | irrat.|  irrat. |
| 12 |   6     | -1 + 3q  |  1 + 3q  | irrat.|  13/4   |
| 18 |  12     | -3 + 4q  |  3 + 4q  | irrat.|  13/2   |
| 21 |  9 + q  | -2 + 6q  | -7 + 7q  | irrat.|  irrat. |
+----+---------+----------+----------+-------+---------+

Eric Weisstein's World of Mathematics, Ring.

Cf. A188158, A238369.

err=1/10^10;nn=10;q=Sqrt[3];lst={};lst1={};Do[If[u+q*v>0,lst=Union[lst,{u+q*v}]],{u,-nn,nn},{v,-nn,nn}];n1=Length[lst];Do[a=Part[lst,i];b=Part[lst,j];c=Part[lst,k];s=(a+b+c)/2;area2=s*(s-a)*(s-b)*(s-c);If[a*b*c !=0&&N[area2]>0&&Abs[N[Sqrt[area2]]-Round[N[Sqrt[area2]]]]

cp's OEIS Frontend

A228858 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 exactly 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).

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Maple

A229098 Smallest area A of Heron triangles with sides (a, b, c) in arithmetic progression of the form b - d(n), b, b + d(n), where d(n) = A091998(n) = 12*n +- 1.

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Maple

A231275 Integer areas of the outer vecten triangles of integer-sided triangles.

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Mathematica

A231276 Integer areas of the inner vecten triangles of integer-sided triangles.

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Mathematica

A231739 Integer areas of the Lucas Central triangles of integer-sided triangles.

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Mathematica

A232274 Integer areas A of the integer-sided triangles such that the length of the inradius and the circumradius are both a perfect square.

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Mathematica

A232329 Integer areas A of the integer-sided triangles such that the product of the inradius and the circumradius is a square.

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PARI

PARI

Extensions

A232461 Integer areas of integer-sided triangles where two sides are of square length.

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