A188158 -id:A188158 - cp's OEIS frontend

A230491 Integer areas of the integer-sided triangles such that the length of the inradius is a square.

6, 84, 96, 108, 120, 132, 144, 156, 168, 180, 240, 264, 300, 324, 396, 420, 432, 468, 486, 504, 540, 594, 630, 684, 720, 756, 864, 990, 1026, 1080, 1116, 1134, 1152, 1224, 1332, 1344, 1404, 1440, 1494, 1536, 1584, 1638, 1680, 1710, 1728, 1782, 1824, 1872, 1890

Offset: 1

Michel Lagneau, Oct 20 2013

nonn

The primitive areas are 6, 84, 108, 120, 132, 144, 156, 168, ...
The non-primitive areas 16*a(n) are in the sequence because if r is the inradius corresponding to a(n), then 4*r is the inradius corresponding to 16*a(n).
The following table gives the first values (A, r, a, b, c) where A is the integer area, r the inradius and a, b, c are the integer sides of the triangle.
******************************
*   A *  r  *  a *  b *   c  *
*******************************
*   6 *  1  *  3 *  4 *   5  *
*  84 *  4  * 13 * 14 *  15  *
*  96 *  4  * 12 * 16 *  20  *
* 108 *  4  * 15 * 15 *  24  *
* 120 *  4  * 10 * 24 *  26  *
* 132 *  4  * 11 * 25 *  30  *
* 144 *  4  * 18 * 20 *  34  *
* 156 *  4  * 15 * 26 *  37  *
* 168 *  4  * 10 * 35 *  39  *
* 180 *  4  *  9 * 40 *  41  *
* 240 *  4  * 12 * 50 *  58  *
* 264 *  4  * 33 * 34 *  65  *
* 300 *  4  * 25 * 51 *  74  *
* 324 *  4  *  9 * 75 *  78  *
* 396 *  4  * 11 * 90 *  97  *
* 420 *  4  * 21 * 85 * 104  *
* 432 *  9  * 30 * 30 *  36  *
* 468 *  9  * 25 * 39 *  40  *
.........................

			84 is in the sequence because the area of triangle (13, 14, 15) is given by Heron's formula A = sqrt(21*(21-13)*(21-14)*(21-15))= 84 where the number 21 is the semiperimeter and the inradius is given by r = A/s = 84/21 = 4 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: Inradius

Cf. A000290, A188158, A228383.

nn = 600; 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[Sqrt[area2]/s]], 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) and inradius r = A/s.

A230558 Integer areas of extouch triangles of integer-sided triangles.

Original entry on oeis.org

30, 48, 72, 84, 120, 192, 252, 270, 288, 336, 432, 480, 648, 720, 750, 756, 768, 780, 936, 1008, 1080, 1152, 1200, 1344, 1470, 1728, 1800, 1920, 2100, 2268, 2352, 2400, 2430, 2592, 2784, 2880, 3000, 3024, 3060, 3072, 3120, 3528, 3600, 3630, 3888, 4032, 4116

Offset: 1

Michel Lagneau, Oct 23 2013

nonn

The extouch triangle T1T2T3 is the triangle formed by the points of tangency of a triangle ABC with its excircles J1, J2 and J3. The points T1, T2, and T3 can also be constructed as the points that bisect the perimeter of the triangle ABC starting at A, B, and C.
The side lengths of the extouch triangle are:
a'= sqrt(a^2 - 4*A^2/b*c)
b'= sqrt(b^2 - 4*A^2/a*c)
c'= sqrt(c^2 - 4*A^2/a*b)
where A is the triangle area of the original triangle.
The extouch triangle has area:
A*(a+b-c)*(a-b+c)*(-a+b+c)/4abc = A*2*r^2*s/(a*b*c) where r and s are the inradius and semiperimeter, respectively.
It is interesting to note that the sides of the extouch triangles are irrational numbers (in the general case) but the areas are integers.
The following table gives the first values (A', A, a, b, c,t1,t2,t3) where A' is the area of the extouch triangles, A is the area of the triangles ABC, a, b, c the integer sides of the original triangles ABC and t1, t2, t3 are the integer sides of the extouch triangles.
-------------------------------------------------------------
A'  |   A  | a | b  | c  |     t1    |     t2    |  t3
-------------------------------------------------------------
30  |  150 | 15| 20 | 25 | 3*sqrt(5) | 4*sqrt(10)|5*sqrt(13)
48  |  300 | 25| 25 | 40 | sqrt(265) | sqrt(265) | 32
72  |  300 | 25| 25 | 30 | sqrt(145) | sqrt(145) | 18
84  | 1050 | 35| 75 |100 | 7*sqrt(13)|3*sqrt(385)|8*sqrt(130)
120 |  600 | 30| 40 | 50 | 6*sqrt(5) |8*sqrt(10) |10*sqrt(13)
192 | 1200 | 50| 50 | 80 |2*sqrt(265)|2*sqrt(265)| 64
252 | 2100 | 35|120 |125 |    7      |72*sqrt(2) |5*sqrt(457)
270 | 1350 | 45| 60 | 75 |9*sqrt(5)  |12*sqrt(10)|15*sqrt(13)
288 | 1200 | 50| 50 | 60 |2*sqrt(145)|2*sqrt(145)| 36
336 | 4200 | 70|150 |200 |14*sqrt(13)|6*sqrt(485)|16*sqrt(130)
432 | 2700 | 75| 75 |120 |3*sqrt(265)|3*sqrt(265)| 96
480 | 2400 | 60| 80 |100 |12*sqrt(5) |16*sqrt(10)|20*sqrt(13)
648 | 2700 | 75| 75 | 90 |3*sqrt(145)|3*sqrt(145)| 54
..................................................
Observation: the three altitudes of a majority of initial triangles ABC are integers, except very rare triangles, for example the initial triangle (35, 120, 125) where A = 2100 (see the following table).
This table gives the first values (A',A, h1, h2, h3) where A' is the area of the extouch triangles, A is the area of the initial triangles ABC and h1, h2, h3 are the altitudes of the initial triangles.
-------------------------------
A'  |  A   |  h1 |  h2 |  h3
-------------------------------
30  |  150 |  20 |  15 |  12
48  |  300 |  24 |  24 |  15
72  |  300 |  24 |  24 |  20
84  | 1050 |  60 |  28 |  21
120 |  600 |  40 |  30 |  24
192 | 1200 |  48 |  48 |  30
252 | 2100 | 120 |  35 | 168/5
270 | 1350 |  60 |  45 |  36
288 | 1200 |  48 |  48 |  40
336 | 4200 | 120 |  56 |  42
432 | 2700 |  72 |  72 |  45
480 | 2400 |  80 |  60 |  48
648 | 2700 |  72 |  72 |  60
...............................

			30 is in the sequence. We use two ways:
First way: the formula A' = A*(a+b-c)*(a-b+c)*(-a+b+c)/(4*a*b*c) gives directly the result: A' = 150*(15+20-25)*(15-20+25)*(-15+20+25)/(4*15*20*25) = 30, with the area A = 150 obtained by Heron's formula A = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(30*(30-15)*(30-20)*(30-25)) = 150, where s is the semiperimeter.
Second way: by calculation of the sides t1, t2, t3 and by using Heron's formula.
The extouch triangle (t1,t2,t3) of the initial triangle (a, b, c) = (15, 20, 25) is the triangle (3*sqrt(5), 4*sqrt(10), 5*sqrt(13)) where:
a' = sqrt(a^2 - 4*A^2/b*c) = sqrt(15^2-4*150^2/(20*25)) = 3*sqrt(5);
b' = sqrt(b^2 - 4*A^2/a*c) = sqrt(20^2-4*150^2/(15*25)) = 4*sqrt(10);
c' = sqrt(c^2 - 4*A^2/a*b) = sqrt(25^2 - 4*150^2/(15*20)) = 5*sqrt(13).
Now, we use Heron's formula with (t1,t2,t3). We find A'=sqrt(s1*(s1-t1)*(s1-t2)*(s1-t3))with:
s1 =(t1+t2+t3)/2 = (3*sqrt(5)+ 4*sqrt(10) + 5*sqrt(13))/2;
We find A'= 30.

T. Dosa, Some Triangle Centers Associated with the Excircles, Forum Geometricorum, Volume 7 (2007) 151-158.
C. Kimberling, Triangle Centers and Central Triangles, Congr. Numer. 129, 1-295, 1998.
Eric Weisstein's World of Mathematics, Extouch Triangles

Cf. A188158, A210643.

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

A230758 Integer areas of the first Neuberg triangles of integer-sided triangles.

Original entry on oeis.org

1275, 4614, 5100, 7275, 11475, 18456, 20400, 29100, 31875, 41526, 45900, 60409, 62475, 65475, 73824, 81600, 103275, 103935, 115350, 116400, 127500, 154275, 166104, 173229, 181875, 183600, 215475, 226086, 241636, 249900, 261900, 265974, 286875, 295296, 434209

Offset: 1

Michel Lagneau, Oct 29 2013

nonn

The triangle LMN formed by joining a set of three Neuberg centers (i.e., centers of the Neuberg circles) is obtained from the edges of a given triangle ABC (see the figure given in the link). Similarly, three reflected Neuberg circles with centers L', M', and N' can be obtained from the main circles by reflection in their respective sides of the triangle, producing a reflected Neuberg triangle L'M'N'.
The Neuberg triangle has area
A' = (a^2*b^2 + a^2*c^2 + b^2*c^2)/(4*sqrt((-a+b+c)*(a-b+c)*(a+b-c)*(a+b+c))).
And the side lengths are
a' = sqrt((b^6 -a^2*b^2*c^2 + b^4*c^2 + b^2*c^4 + c^6)/d)
b' = sqrt((a^6 -a^2*b^2*c^2 + a^4*c^2 + a^2*c^4 + c^6)/d)
c' = sqrt((a^6 -a^2*b^2*c^2 + a^4*b^2 + a^2*b^4 + b^6)/d)
with d = (-a+b+c)*(a-b+c)*(a+b-c)*(a+b+c).
Properties of this sequence:
The areas of the original triangles are integers. The primitive triangles with areas a(n) are 1275, 4614, 7275, 11475, ...
The non-primitive triangles with areas 4*a(n), 9*a(n), ..., p^2*a(n), ... are in the sequence.
It appears that if the original triangle is isosceles, the corresponding Neuberg triangle is also isosceles and the greatest side is an integer divisible by 25.
It is interesting to note that the sides of the Neuberg triangle are irrational numbers (in the general case) and the areas are integers.
The following table gives the first values (A',A,a,b,c,a',b',c') where A' is the area of the Neuberg triangles, A is the area of the reference triangles ABC, a, b, c the integer sides of the original triangles ABC and a', b', c' the integer sides of the Neuberg triangles.
------------------------------------------------------------------------
|   A'|   A |  a |  b |  c |       a'      |       b'      |   c'
------------------------------------------------------------------------
| 1275|  768|  40|  40|  64|  sqrt(42241)/2|  sqrt(42241)/2|   25
| 4614| 3456|  72|  96| 120|  sqrt(32521)  |2*sqrt(5101)   |5*sqrt(193)
| 5100| 3072|  80|  80| 128|  sqrt(42241)  |  sqrt(42241)  |   50
| 7275| 6912| 120| 120| 144|  sqrt(94681)/2|  sqrt(94681)/2|  100
|11475| 6912| 120| 120| 192|3*sqrt(42241)/2|3*sqrt(42241)/2|   75
|18456|13824| 144| 192| 240|2*sqrt(32521)  |4*sqrt(5101)   |10*sqrt(193)
|20400|12288| 160| 160| 256|2*sqrt(42241)  |2*sqrt(42241)  |  100
|29100|27648| 240| 240| 288|  sqrt(94681)  |  sqrt(94681)  |  200
|31875|19200| 200| 200| 320|5*sqrt(42241)/2|5*sqrt(42241)/2|  125
|41526|31104| 216| 288| 360|3*sqrt(32521)  |6*sqrt(5101)   |15*sqrt(193)
|45900|27648| 240| 240| 384|3*sqrt(42241)  |3*sqrt(42241)  |  150
......................................................

			4614 is in the sequence because the area A' = (a^2*b^2 + a^2*c^2 + b^2*c^2)/(4*sqrt((-a+b+c)*(a-b+c)*(a+b-c)*(a+b+c))) of the first Neuberg triangle corresponding to the initial triangle (72,96,120) is A' = (72^2*96^2 + 72^2*120^2 + 96^2*120^2)/(4*sqrt((-72+96+120)*(72-96+120)*(72+96-120)*(72+96+120))) = 4614.

Wolfram MathWorld, First Neuberg Triangles

Cf. A188158.

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

A231328 Integer areas of the reflection triangles of integer-sided triangles.

Original entry on oeis.org

18, 72, 90, 162, 180, 252, 288, 360, 450, 540, 630, 648, 720, 810, 882, 990, 1008, 1152, 1440, 1458, 1512, 1620, 1638, 1800, 1890, 2160, 2178, 2250, 2268, 2520, 2592, 2772, 2880, 2970, 3042, 3240, 3528, 3672, 3960, 4032, 4050, 4158, 4410, 4500, 4608, 4680, 4860

Offset: 1

Michel Lagneau, Nov 07 2013

nonn

The triangle A'B'C' obtained by reflecting the vertices of a reference triangle ABC about the opposite sides is called the reflection triangle (Grinberg 2003).
The area of the reflection triangle is given by
A' = A*t/(a^2*b^2*c^2) where A is the area of the reference triangle of sides (a, b, c) and
t=-(a^6-b^2*a^4-c^2*a^4-b^4*a^2-c^4*a^2-b^2*c^2*a^2+b^6+c^6-b^2*c^4-b^4*c^2)/(a^2*b^2*c^2).
See the link for the side lengths of the reflection triangles.
Properties of this sequence:
The areas corresponding to the primitive reflection triangles are 18, 90, 180, 252, 540,...
The non-primitive triangles of areas 4*a(n),9*a(n),...,p^2*a(n),... are in the sequence.
It appears that one of the side of the reflection triangles equals the greatest side of the initial triangle (see the table below), and the initial triangles are Pythagorean triangles => a(n) = 3*A009112(n).
The following table gives the first values (A, A', a, b, c, a', b', c') where A' is the area of the reflection 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 reflection triangles.
-------------------------------------------------------------------------
|  A'  |   A |  a |  b |   c |       a'         |       b'         |  c'|
-------------------------------------------------------------------------
|  18  |   6 |  3 |  4 |   5 |  9*sqrt(17)/5    |  4*sqrt(97)/5    |  5 |
|  72  |  24 |  6 |  8 |  10 | 18*sqrt(17)/5    |  8*sqrt(97)/5    | 10 |
|  90  |  30 |  5 | 12 |  13 |  5*sqrt(1321)/13 | 36*sqrt(41)/13   | 13 |
| 162  |  54 |  9 | 12 |  15 | 27*sqrt(17)/5    | 12*sqrt(97)/5    | 15 |
| 180  |  60 |  8 | 15 |  17 |  8*sqrt(2089)/17 | 45*sqrt(89)/17   | 17 |
| 252  |  84 |  7 | 24 |  25 |  7*sqrt(5233)/25 | 72*sqrt(113)/25  | 25 |
| 288  |  96 | 12 | 16 |  20 | 36*sqrt(17)/5    | 16*sqrt(97)/5    | 20 |
| 360  | 120 | 10 | 24 |  26 | 10*sqrt(1321)/13 | 72*sqrt(41)/13   | 26 |
| 450  | 150 | 15 | 20 |  25 |  9*sqrt(17)      |  4*sqrt(97)      | 25 |
| 540  | 180 |  9 | 40 |  41 | 27*sqrt(1609)/41 | 40*sqrt(2329)/41 | 41 |
| 630  | 210 | 12 | 35 |  37 | 36*sqrt(1241)/37 | 35*sqrt(2521)/37 | 37 |
| 648  | 216 | 18 | 24 |  30 | 54*sqrt(17)/5    | 24*sqrt(97)/5    | 30 |
.......................................................................

			18 is in the sequence. We use two ways:
First way: with the triangle (3, 4, 5) the formula A' = A*t/(a^2*b^2*c^2) gives directly the result: A'= 18 where the area A = 6 is obtained by Heron's formula A =sqrt(s*(s-a)*(s-b)*(s-c))= sqrt(6*(6-3)*(6-4)*(6-5)) = 6, where s is the semiperimeter.
Second way: by calculation of the sides a', b', c' and by using Heron's formula. We obtain from the formulas given in the link:
a' = 9*sqrt(17)/5;
b' = 4*sqrt(97/5);
c' = 5.
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(17)/5+ 4*sqrt(97/5)+ 5)/2. We find A'= 18.

D. Grinberg, On the Kosnita Point and the Reflection Triangle, Forum Geom. 3, 105-111, 2003.

Eric W. Weisstein, MathWorld: Reflection Triangle

Cf. A188158, A009111, A009112.

nn = 300 ; 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]

A231740 Integer areas of the intangents triangle of integer-sided triangles.

Original entry on oeis.org

672, 2688, 6048, 10752, 15435, 16800, 19250, 24192, 32928, 37730, 43008, 54432, 56133, 61740, 67200, 77000, 81312, 96768, 113568, 131712, 138915, 150920, 151200, 172032, 173250, 194208, 217728, 221130, 224532, 242592, 246960, 268800, 296352, 308000, 325248, 339570

Offset: 1

Michel Lagneau, Nov 13 2013

nonn

Given triangle ABC, there are four lines simultaneously tangent to the incircle (with center I) and the excircle (with center J). Of these, three correspond to the sidelines of the triangle, and the fourth is known as the intangent (Kimberling 1998, p. 161), illustrated in the first link. The intangents intersect one another pairwise, and their points of intersection form the so-called intangents triangle A'B'C' illustrated in the second link.
The area of the intangents triangle is given by
S' = S*(a+b-c)^2*(a-b+c)^2*(-a+b+c)^2 /((-a^2+b^2+c^2)*(a^2+b^2-c^2)*(a^2-b^2+c^2)) where (a, b, c) are the sides of the initial triangle of ABC and S is the area of ABC.
Properties of this sequence:
The primitive triangles are 672, 15435, 19250, ...
The nonprimitive triangles of areas 4*a(n), 9*a(n), ..., p^2*a(n), ... are in the sequence.
The areas of the non-isosceles triangles are 15435, 19250, 37730, 56133, ...
The following table gives the first values (S', S, a, b, c, a', b', c') where S' is the area of the intangents triangle, S is the area of the triangles ABC, a, b, c the integer sides of the triangles ABC and a', b', c' are the integer sides of the intangents triangle.
---------------------------------------
|   S'  |     S  |   a  |   b   |  c  |
---------------------------------------
|   672 |   588  |   35 |   35  |  42 |
|  2688 |  2352  |   70 |   70  |  84 |
|  6048 |  5292  |  105 |  105  | 126 |
| 10752 |  9408  |  140 |  140  | 168 |
| 15435 | 12600  |  130 |  200  | 210 |
| 16800 | 14700  |  175 |  175  | 210 |
| 19250 |  9240  |  102 |  182  | 200 |
| 24192 | 21168  |  210 |  210  | 252 |
| 32928 | 28812  |  245 |  245  | 294 |
| 37730 | 36960  |  272 |  300  | 308 |
| 43008 | 37632  |  280 |  280  | 336 |
| 54432 | 47628  |  315 |  315  | 378 |
| 56133 | 44352  |  220 |  416  | 420 |

			19250 is in the sequence from the initial triangle (102, 182, 200); we use the formula S' = S*(a+b-c)^2*(a-b+c)^2*(-a+b+c)^2 /((-a^2+b^2+c^2)*(a^2+b^2-c^2)*(a^2-b^2+c^2)) = 9240*(102+182-200)^2*(102-182+200)^2*(-102+182+200)^2 /((-102^2+182^2+200^2)*(102^2+182^2-200^2)*(102^2-182^2+200^2)) = 19250 where the area S = 9240 is obtained by Heron's formula S = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(242*(242-102)*(242-182)*(242-200)) = 9240 with the semiperimeter s = (a+b+c)/2 = (102+182+200)/2 = 242.

Clark Kimberling, Triangle centers and central triangles, Congressus Numerantium, 129 (1998)1-285.

Clark Kimberling, Central points and central lines in the plane of a triangle, Math. Magazine, 67(1994) 163-187.
Eric Weisstein's World of Mathematics, Intangents Triangle.
Eric Weisstein's World of Mathematics, Intangents Circle.
Eric Weisstein's World of Mathematics, Intangent.

Cf. A188158.

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

A237576 Smallest integer areas of integer-sided triangles such that the perimeter equals n times the smallest side.

Original entry on oeis.org

0, 0, 0, 6, 60, 30, 210, 24, 84, 60, 198, 330, 1716, 546, 2730, 252, 4080, 36, 5814, 210, 7980, 2310, 10626, 924, 1380, 1248, 90, 4914, 4176, 6090, 26970, 480, 32736, 1224, 39270, 1938, 46620, 2394, 54834, 4560, 63960, 4620, 74046, 19866, 85140, 22770, 97290

Offset: 1

Michel Lagneau, Feb 09 2014

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 sequence a(n) is the union of four subsequences A, B, C and D where:
A is the subsequence with areas 60, 210, 1716, 2730, 4080, 5814, 7980, 10626, ... where n is odd, and the corresponding sides are of the form (4k, 4k^2-1, 4k^2+1) with areas 2k(4k^2-1) for k = 2, 3, 6, 7, 8, 9, 11, ... These areas are in the sequence A069072 (areas of primitive Pythagorean triangles whose odd sides differ by 2).
B is the subsequence with areas 6, 30, 84, 330, 546, 2310, 4914, 6090, ... where n is even, and the corresponding sides are of the form (2k+1, 2k(k+1), 2k(k+1)+1) with areas k(k+1)(2k+1) for k = 1, 2, 3, 5, 6, 7, 10, 13, 14, ... These areas are in the sequence A055112 (Areas of Pythagorean triangles (a, b, c) with c = b+1).
C is the subsequence with areas 84, 198, 1380, 4176, ... where n is odd but the areas are not Pythagorean triangles.
D is the subsequence with areas 24, 60, 210, 924, 1248, 480, 1224, 1938, ... where n is even but the areas are not Pythagorean triangles.
The triangles with the same areas are not unique; for example:
(8, 15, 17) and (6, 25, 29) => A = 60; the first is a Pythagorean triangle, the second is not.
(12, 35, 37) and (7, 65, 68) => A = 210; the first is a Pythagorean triangle, the second is not.
The following table gives the first values (n, A, p, a, b, c) where A is the area of the triangles, p is the perimeter and a, b, c are the sides.
+----+------+-------------+----+-----+-----+
|  n |   A  |      p      |  a |  b  |  c  |
+----+------+-------------+----+-----+-----+
|  4 |    6 |  12 = 4*3   |  3 |   4 |   5 |
|  5 |   60 |  40 = 5*8   |  8 |  15 |  17 |
|  6 |   30 |  30 = 6*5   |  5 |  12 |  13 |
|  7 |  210 |  84 = 7*12  | 12 |  35 |  37 |
|  8 |   24 |  32 = 8*4   |  4 |  13 |  15 |
|  9 |   84 |  72 = 9*8   |  8 |  29 |  35 |
| 10 |   60 |  60 = 10*6  |  6 |  25 |  29 |
| 11 |  198 | 132 = 11*12 | 12 |  55 |  65 |
| 12 |  330 | 132 = 12*11 | 11 |  60 |  61 |
| 13 | 1716 | 312 = 13*24 | 24 | 143 | 145 |
| 14 |  546 | 182 = 14*13 | 13 |  84 |  85 |
| 15 | 2730 | 420 = 15*28 | 28 | 195 | 197 |
+----+------+-------------+----+-----+-----+

Cf. A188158.

with(numtheory):nn:=600:for n from 4 to 50 do: ii:=0:for a from 1
  to nn while(ii=0) do: 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) and 2*p=n*a then ii:=1:printf ( "%d %d %d %d %d \n",n,x0,a,b,c):else fi:od:od:od:od:

nn=600;lst={};Do[k=0;Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s (s-a) (s-b) (s-c);If[0

A253686 a(n) is the smallest integer area of the triangle having the sides in the commutative ring Z[sqrt(q)] where q = A005117(n) is a squarefree number.

Original entry on oeis.org

6, 1, 3, 1, 6, 3, 3, 4, 3, 5, 3, 2, 6, 6, 6, 6, 3, 5, 6, 6, 6, 6, 6, 3, 6, 6, 4, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 4, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 3, 6, 6, 4, 6, 6, 6, 6, 6, 5, 6, 6, 6, 6, 6, 6, 6, 6, 4, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 1

Michel Lagneau, Jan 09 2015

nonn

Generalized integer areas triangles in the ring Z[sqrt(q)] = {a + b sqrt(q)| a,b in Z}.
Introduction:
The study of triangles having their sides with values in a ring Z[sqrt(q)] and having integer area gives remarkable properties probably still unexplored today.
Property:
a(1) = 6 because the ring Z[sqrt(1)] = Z => the smallest area of integer sides is A188158(1) = 6 => a(n) <=6.
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 instance, in the ring Z[sqrt(19)] the area of each following triangle:
(3, 4, 5),
(8, -2+sqrt(19), 2+sqrt(19)),
(6-sqrt(q), 3+sqrt(19), -1+2*sqrt(19)),
(-3+sqrt(19), 6+sqrt(19), 1+2*sqrt(19)) is A=6.
Conjecture: the set of squarefree numbers q such that the integer area A of the triangles with sides in the commutative ring Z[sqrt(q)] is finite if A < 6.
It follows that a(n)= 6 for n > 384 where A005117(384) = 629 and a(384)=5.
The corresponding values q such that a(n)<6 are 2, 3, 5, 7, 10, 11, 13, 14, 15, 17, 26, 29, 37, 41, 65, 85, 89, 101, 113, 221 and 629 with the corresponding index in the sequence a(n): 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 17, 18, 24, 27, 40, 53, 56, 62, 71, 137 and 384.
We observe a subset of numbers q such that the sides of the triangles are of the particular form (a, b, sqrt(q)) with a, b integers. This subset is {5, 13, 17, 29, 37, 41, 65, 85, 89, 101, 113, 221, 629}. See the table below for the examples.
We find also five isosceles triangles with area less than 6 and it is observed that they are of the form (2,sqrt(q),sqrt(q)) with q = 2, 5, 10, 17 and 26. The corresponding areas A are 1, 2, 3, 4 and 5 respectively with the formula A = sqrt(q-1).
The following table gives the first values (A, sqrt(q), a, b, c) where A is the smallest area of the triangle (a, b, c), Z[sqrt(q)] is the commutative ring and a, b, c are the sides in Z[sqrt(q)]= Z[sqrt(A005117(n))].
+---+----------+-------------+--------------+---------------+
| A |  sqrt(q) |      a      |       b      |       c       |
+---+----------+-------------+--------------+---------------+
| 6 |  sqrt(1) |      2      |       3      |       4       |
| 1 |  sqrt(2) |      2      |    sqrt(2)   |    sqrt(2)    |
| 3 |  sqrt(3) | 3 - sqrt(3) | 2 + 2*sqrt(3)| 1 + 3*sqrt(3) |
| 1 |  sqrt(5) |      1      |       2      |    sqrt(5)    |
| 6 |  sqrt(6) |  2*sqrt(6)  |-2 + 2*sqrt(6)| 2 + 2*sqrt(6) |
| 3 |  sqrt(7) |      4      | -1 + sqrt(7) |  1 + sqrt(7)  |
| 3 | sqrt(10) |      2      |   sqrt(10)   |    sqrt(10)   |
| 4 | sqrt(11) |      6      |-1 + sqrt(11) |  1 + sqrt(11) |
| 3 | sqrt(13) |      2      |       3      |    sqrt(13)   |
| 5 | sqrt(14) |      6      |-2 + sqrt(14) |  2 + sqrt(14) |
| 3 | sqrt(15) |      8      | 5 - sqrt(15) |  5 + sqrt(15) |
| 2 | sqrt(17) |      1      |       4      |    sqrt(17)   |
| 6 | sqrt(19) |      8      |-2 + sqrt(19) |  2 + sqrt(19) |
| 6 | sqrt(21) |      3      |       4      |       5       |
| 6 | sqrt(22) |      3      |       4      |       5       |
| 6 | sqrt(23) |      3      |       4      |       5       |
| 3 | sqrt(26) |     10      |-4 + sqrt(26) | 4 + sqrt(26)  |
| 5 | sqrt(29) |      2      |       5      |    sqrt(29)   |
| 6 | sqrt(30) |      3      |       4      |       5       |
.............................................................

			a(384)=5 because q = A005117(384) = 629 and the area A of the triangle (1, 26, sqrt(629)) is given by Heron's formula: A = sqrt(s*(s-1)*(s-26)*(s-sqrt(629))) where s = (1+26+sqrt(629))/2. We find A = 5.

Eric Weisstein's World of Mathematics, Triangle.
Eric Weisstein's World of Mathematics, Ring.

Cf. A005117, A188158, A238369.

(* take q=sqrt(2), sqrt(3), ..., A005117(k), ... successively *)
err=1/10^10;nn=10;q=Sqrt[2];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]]]]

A254075 Integer area A of triangles with side lengths in the commutative ring Z[sqrt(5)].

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 19, 20, 22, 24, 25, 27, 28, 29, 30, 32, 33, 35, 36, 38, 40, 42, 44, 45, 48, 49, 50, 52, 54, 55, 57, 58, 60, 63, 64, 65, 66, 72, 75, 76, 77, 80, 81, 84, 88, 90, 95, 96, 98, 99, 100, 108, 110, 112, 114, 116

Offset: 1

Michel Lagneau, May 03 2015

nonn

Extension of A188158 with triangles of sides in the ring Z[sqrt(5)] = {x + y sqrt(5)| x,y in Z}.
The numbers 5*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(5), b*sqrt(5), c*sqrt(5)) is 5*A. The numbers a(n)*5^p and a(n)*q^2 are in the sequence. Because a(1)=1, the squares are in the sequence. The primitive areas of the sequence are {1, 2, 3, 6, 7, 11, 13, 19, ...}.
The values shown were obtained with a and b in the range [-40, ..., +40]. For the areas > 120 it would be necessary to expand the range of variation, but then the calculations would become very slow.
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(5)]:
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(5)] 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(5)and irrat. = irrational numbers of the form u+v*q.
+----+---------+----------+----------+-------+---------+
|  A |    a    |      b   |     c    |   r   |  R      |
+----+---------+----------+----------+-------+---------+
|  1 |       1 |       2  |       q  | irrat.|  irrat. |
|  2 |       1 |       5  |      2q  | irrat.|  irrat. |
|  2 |       2 |       q  |       q  | irrat.|   5/4   |
|  2 |       4 |       q  |       q  | irrat.|   5/2   |
|  3 |       2 |       5  |      3q  | irrat.|  irrat. |
|  3 |       3 |       q  |      2q  | irrat.|   5/2.  |
|  4 |       1 |      17  |      8q  | irrat.|  irrat. |
|  4 |       2 |       4  |      2q  | irrat.|  irrat. |
|  5 |       2 |      13  |      5q  | irrat.| irrat.  |
|  5 |       5 |       q  |      2q  | irrat.|   5/2   |
|  6 |       3 |       4  |       5  |    1  |   5/2   |
|  6 |       1 |      13  |      6q  | irrat.| irrat.  |
|  7 |       7 |      2q  |      5q  | irrat.|  25/2   |
|  8 |       2 |      10  |      4q  | irrat.|  irrat. |
|  8 |       4 |      2q  |      2q  | irrat.|   5/2   |
|  8 |       5 |      13  |      8q  | irrat.|  irrat. |
|  8 |       6 |     5-q  |     5+q  |    1  |  15/4   |
|  8 |       8 |      2q  |      2q  | irrat.|    5    |
+----+---------+----------+----------+-------+---------+

Eric Weisstein's World of Mathematics, Ring

Cf. A188158, A238368, A238369, A253277.

err=1/10^10;nn=40;q=Sqrt[5];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]]]]

A256579 Integer areas of integer-sided triangles where at least one of the three altitudes is of prime length.

Original entry on oeis.org

6, 12, 30, 60, 84, 168, 330, 546, 660, 1092, 1224, 1710, 2448, 3036, 3420, 6072, 6090, 7440, 12180, 12654, 14880, 17220, 19866, 25308, 25944, 34440, 37206, 39732, 51330, 51888, 56730, 74412, 75174, 89460, 97236, 102660, 113460, 123240, 142926, 150348, 176220

Offset: 1

Michel Lagneau, Apr 02 2015

nonn

Subset of A226453.
The corresponding primes are: 3, 3, 5, 5, 7, 7, 11, 13, 11, 13, 17, 19, 17, 23, 19, 23, 29, 31, 29, 37, 31, 41, 43, 37, 47, 41, 53, 43, 59, 47, 61, 53, 67, 71, 73, 59, 61, 79, 83, 67, 89, ...
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 altitudes of a triangle with sides length a, b, c and area A have length given by Ha = 2*A/a, Hb = 2*A/b, Hc = 2*A/c.
Properties of this sequence:
- The sequence is infinite (see the formula below);
- The prime altitude of a triangle is the greatest prime divisor of a(n) (the proof is easy if we observe the formula);
- There exists two subsets of numbers included into a(n):
Case (i): A subset with right triangles (a,b,c) where a^2 + b^2 = c^2 with area a1(n) = {6, 30, 84, 330, 546, 1224, ...}. The lengths of the prime altitudes are Ha or Hb = a = p. The sides are of the form (p, q, q+1) with p = sqrt(2*q+1) => the sides are equal to (p, (p^2-1)/2, (p^2+1)/2) and a(n) = (p^3-p)/4.
Case (ii): A subset with isosceles triangles formed by two right triangles of the sequence. So, the areas are a2(n) = {12, 60, 168, 660, 1092, 2448, ...} = 2*a1(n). The sides are of the form (a, a, 2*(a-1)) = ((p^2+1)/2, (p^2+1)/2, p^2-1) and Ha = sqrt(2*a-1) = p, a2(n) = 2*a1(n) = (p^3-p)/2.
We did not find a class of non-isosceles and non-right triangles (a, b, c) whose three altitudes include one of prime length.
The following table gives the first values (A, a, b, c, Ha, Hb, Hc) where A is the integer area, a, b, c are the sides and Ha <= Hb <= Hc are the altitudes.
+------+-----+-----+-----+----------+----------+---------+
|   A  |  a  |  b  |  c  |    Ha    |   Hb     |   Hc    |
+------+-----+-----+-----+----------+----------+---------+
|    6 |   3 |   4 |   5 |   12/5   |    3     |    4    |
|   12 |   5 |   5 |   8 |    3     |   24/5   |   24/5  |
|   30 |   5 |  12 |  13 |    5     |   60/13  |   12    |
|   60 |  13 |  13 |  24 |    5     |  120/13  |  120/13 |
|   84 |   7 |  24 |  25 |  168/25  |    7     |   24    |
|  168 |  25 |  25 |  48 |    7     |  336/25  |  336/25 |
|  330 |  11 |  60 |  61 |  660/61  |   11     |   60    |
|  546 |  13 |  84 |  85 | 1092/85  |   13     |   84    |
|  660 |  61 |  61 | 120 |   11     | 1320/61  | 1320/61 |
| 1092 |  85 |  85 | 168 |   13     | 2184/85  | 2184/85 |
| 1224 |  17 | 144 | 145 | 2448/145 |   17     |  144    |
| 1710 |  19 | 180 | 181 | 3420/181 |   19     |  180    |
| 2448 | 145 | 145 | 288 | 4896/145 | 4896/145 |   17    |
+------+-----+-----+-----+----------+----------+---------+

Eric Weisstein's World of Mathematics, Altitude
Eric Weisstein's World of Mathematics, Isosceles Triangle
Eric Weisstein's World of Mathematics, Right Triangle

Cf. A188158, A210643, A210645, A226453.

# program using the formula
lst:={}:for n from 2 to 50 do:p:=ithprime(n):p1:=(p^3-p)/4:p2:=(p^3-p)/2:lst:=lst union {p1} union {p2}:od:print(lst):

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

a(n) = (prime(n)^3 - prime(n))/4 for the right triangles;

a(n) = (prime(n)^3 - prime(n))/2 for the isosceles triangles.

A256629 Integer areas A of integer-sided triangles such that the length of the circumradius is a prime number.

Original entry on oeis.org

24, 120, 240, 720, 840, 1320, 2520, 3360, 3960, 5280, 6240, 6840, 9360, 10920, 14280, 15600, 16320, 17160, 18480, 22440, 24360, 26520, 31920, 35880, 38760, 42840, 43680, 46200, 50160, 55200, 57960, 59280, 70200, 73920, 91080, 93840, 100800, 107640, 117600, 118320, 122400

Offset: 1

Michel Lagneau, Apr 05 2015

nonn

Subsequence of A208984.
For the same area, the number of triangles such that the length of the circumradius is a prime number is not unique; for example, from a(5)= 840 we find two triangles of sides (a,b,c)=(40,42,58) and (24,70,74) where R = 29 and 37, respectively.
The circumradius R values corresponding to the terms of the sequence are 5, 13, 17, 41, (29 or 37), 61, 53, 113, 101, 73, 89, 181, 97, (109 or 197), 149, ...
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 circumradius R is given by R = abc/4A.
Observation:
- all sides of the triangles are even;
- the inradius values are also even;
- the first triangle, of sides (6,8,10), is the unique triangle in which the lengths of the inradius and the circumradius are both prime numbers (r = A/p = 24/12 = 2 and R = abc/4A = 480/4*24 = 5).
For the same area, it is possible to find a prime inradius (see A230195), but the corresponding circumradius is generally rational. For example, for a(2) = 120, we find two triangles:
(10,24,26) => r = 4 and R = 13;
(16,25,39) => r = 3 prime and R = 65/2.
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 and r = A/p, R = a*b*c/4*A are respectively the values of the inradius and the circumradius.
+--------+------+-------+-------+------+-------+
|    A   |   a  |   b   |   c   |   r  |   R   |
+--------+------+-------+-------+------+-------+
|    24  |   6  |    8  |   10  |   2  |    5  |
|   120  |  10  |   24  |   26  |   4  |   13  |
|   240  |  16  |   30  |   34  |   6  |   17  |
|   720  |  18  |   80  |   82  |   8  |   41  |
|   840  |  40  |   42  |   58  |  12  |   29  |
|   840  |  24  |   70  |   74  |  10  |   37  |
|  1320  |  22  |  120  |  122  |  10  |   61  |
|  2520  |  56  |   90  |  106  |  20  |   53  |
|  3360  |  30  |  224  |  226  |  14  |  113  |
|  3960  |  40  |  198  |  202  |  18  |  101  |
|  5280  |  96  |  110  |  146  |  30  |   73  |
|  6240  |  78  |  160  |  178  |  30  |   89  |
+--------+------+-------+-------+------+-------+

			a(1) = 24 because, for (a,b,c) = (6, 8, 10) => s= (6+8+10)/2 =12, and
A = sqrt(12(12-6)(12-8)(12-10)) = sqrt(576) = 24;
R = abc/4A = 480/4*24 = 5 is prime.

Cf. A188158, A208984, A230195.

Missing terms 91080 and 117600 added by Zachary Sizer, Jan 02 2025

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A253686 a(n) is the smallest integer area of the triangle having the sides in the commutative ring Z[sqrt(q)] where q = A005117(n) is a squarefree number.

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A254075 Integer area A of triangles with side lengths in the commutative ring Z[sqrt(5)].

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