A208984 -id:A208984 - cp's OEIS frontend

A210250 Area A of the cyclic quadrilaterals such that A, the sides and the radius of the circumcircle are integers.

48, 192, 240, 432, 480, 672, 768, 936, 960, 1200, 1440, 1680, 1728, 1920, 2160, 2352, 2640, 2688, 2856, 3072, 3744, 3840, 3864, 3888, 4032, 4320, 4368, 4536, 4800, 5016, 5040, 5376, 5712, 5760, 5808, 5880, 6000, 6048, 6072, 6696, 6720, 6912, 7056, 7392, 7560, 7680, 7728, 7752, 7920

Offset: 1

Michel Lagneau, Mar 19 2012

nonn

In Euclidean geometry, a cyclic quadrilateral is a quadrilateral whose vertices all lie on a single circle. This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic.
The area A of a cyclic quadrilateral with sides a, b, c, d is given by Brahmagupta's formula: A = sqrt((s - a)(s - b)(s - c)(s - d)) where s, the semiperimeter is s = (a+b+c+d)/2.
The circumradius R (the radius of the circumcircle) is given by:
R = sqrt[(ab+cd)(ac+bd)(ad+bc)]/4A.
The corresponding R of a(n) are not unique; for example, for a(12) = 1680 => (a,b,c,d) = (24, 24, 70, 70) with R = 37 and (a,b,c,d) = (40, 40, 42,42) with R = 29.
The smallest corresponding R of a(n) is {5, 10, 13, 15, 17, 25, 20, 25, 26, 25, 41, 29, ...}.
Properties of this sequence:
A majority of quadrilaterals [a, b, c, d] have the property that a = b and c = d, and in this case s = a+c, A = a*c and R = sqrt(a^2+c^2)/2. Because a and c are even => a = 2p and c = 2q, then A = 4pq and R = sqrt(p^2+q^2). Consequently, 2*A103251(n) is included in this sequence.
Nevertheless, there also exist quadrilaterals whose four sides are distinct, for example [a, b, c, d] = [14, 30, 40, 48] => A = 936 = a(8) and R = 25. The subset of a(n) with this property is {936, 2856, 3744, 3864, 4536, 5016, 5376, 5712, 5880, 6696, 7056, 7560, ...}.

			48 is in the sequence because, for (a,b,c,d) = (6,6,8,8),
s = (6+6+8+8)/2 = 14;
A = sqrt((14-6)(14-6)(14-8)(14-8)) = 48;
R = sqrt((6*6+8*8)(6*8+6*8)(6*8+6*8))/(4*48) = 960/192 = 5.

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

Mohammad K. Azarian, Solution to Problem S125: Circumradius and Inradius, Math Horizons, Vol. 16, Issue 2, November 2008, p. 32.
E. Gürel, Solution to Problem 1472, Maximal Area of Quadrilaterals, Math. Mag. 69 (1996), 149.
Eric Weisstein's World of Mathematics, Cyclic Quadrilateral

Cf. A103251, A208984.

SMax=8000;
Do[
  Do[
    x=S^2/(u v w);
    If[u+v+w+x//OddQ,Continue[]];
    If[v+w+x<=u,Continue[]];
    r=Sqrt[v w+u x]Sqrt[u w+v x]Sqrt[u v+w x]/(4S);
    If[r//IntegerQ//Not,Continue[]];
    (*{a,b,c,d}=(u+v+w+x)/2-{u,v,w,x};{a,b,c,d,r,S}//Sow*);
    S//Sow;Break[];(*to generate a table, comment out this line and uncomment previous line*)
    ,{u,S^2//Divisors//Select[#,S<=#^2&]&}
    ,{v,S^2/u//Divisors//Select[#,S^2<=u#^3&&#<=u&]&}
    ,{w,S^2/(u v)//Divisors//Select[#,S^2<=u v#^2&&#<=v&]&}
  ]
  ,{S,24,SMax,24}
]//Reap//Last//Last
{x,r,a,b,c,d}=.;
(* Albert Lau, May 25 2016 *)

Incorrect Mathematica program removed by Albert Lau, May 25 2016

Missing term 5880 and more terms from Albert Lau, May 25 2016

A185210 Areas A of the triangles such that A, the sides, the inradius and the radius of the three excircles are integers.

Original entry on oeis.org

6, 24, 30, 42, 48, 54, 60, 84, 96, 108, 120, 144, 150, 156, 168, 180, 192, 210, 216, 240, 270, 294, 330, 336, 378, 384, 390, 420, 432, 462, 480, 486, 504, 510, 528, 540, 546, 576, 594, 600, 624, 630, 672, 714, 720, 726, 750, 756, 768, 810, 840, 864, 924, 930

Offset: 1

Michel Lagneau, Mar 21 2012

nonn

Theorem 1: Consider a triangle whose area A, sides (a,b,c), inradius r and the radius of whose three excircles r1, r2, r3 are integers. Then the sum a^2 + b^2 + c^2 + r^2 + r1^2 + r2^2 + r3^2 is a perfect square equal to 16R^2, where R is the circumradius.
Proof: (r1 + r2 + r3 - r)^2 = r1^2 + r2^2 + r3^2 + r^2 + a^2 + b^2 + c^2 because: r1*r2 + r2*r3 + r3*r1 - r*r1 - r*r2 - r*r3 = (a^2 + b^2 + c^2)/2 (formula from Feuerbach - see the link). But r1 + r2 + r3 - r = 4*R (see the reference: Johnson 1929, pp. 190-191), hence the result. Remark: R is not necessarily an integer; for example, at a(1) = 6 with (a,b,c) = (3, 4, 5) we obtain r = 1, r1 = 2, r2 = 3, r3 = 6 and R = 5/2. Then 3^2 + 4^2 + 5^2 + 1^2 + 2^2 + 3^2 + 6^2 = 16*(5/2)^2 = 10^2. Nevertheless, if R is an integer, then r, r1, r2 and r3 are necessarily integers (see the following theorem). The subset of a(n) with R integer is A208984 = {24, 96, 120, 168, 216, 240, 336, 384, 432, 480, 600, ...}
Theorem 2: Consider a triangle whose area A, sides (a,b,c) and circumradius R are integers. Then the inradius r and the radius of the three excircles r1, r2, r3 are also integers.
Proof: Let s be the semiperimeter, let s*A = r1*r2*r3 be an integer, and let r*r1*r2*r3 = A^2 also be an integer => r is an integer. r1 = A/(s-a), r2 = A/(s-b), r3 = A/(s-c) => r1*r2 = s*(s-c), r1*r3=s*(s-b), r2*r3 = s*(s-a) are integers. Because r1*r2*r3 is an integer => r1, r2, r3 are integers.

			24 is in the sequence because for (a, b, c) = (6, 8, 10) => s =(6+8+10)/2 = 12; A = sqrt(12(12-6)(12-8)(12-10)) = sqrt(576) = 24; r = A/s = 2; r1 = 2*24(-6+8+10) = 4; r2 = 2*24(6-8+10) = 6; r3 = 2*24(6+8-10) = 12.

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.
Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.

Michel Lagneau, Table of n, a(n) for n = 1..54
Eric W. Weisstein's World of Mathematics, Excircles
Eric W. Weisstein's World of Mathematics, Exradius
Eric W. Weisstein's World of Mathematics, Inradius

Cf. A120572, 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 <= nn^2 && IntegerQ[Sqrt[area2]] && IntegerQ[Sqrt[area2]/s] && IntegerQ[2*Sqrt[area2]/(-a+b+c)] &&  IntegerQ[2*Sqrt[area2]/(a-b+c)] && IntegerQ[2*Sqrt[area2]/(a+b-c)], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

A = sqrt(s*(p-a)*(s-b)*(s-c)) with s = (a+b+c)/2 (Heron's formula);
the inradius is r=A/s;
the exradii of the excircles are r1 = 2*A/(-a+b+c), x2 = 2*A*b/(a-b+c), and x3 = 2*A*c/(a+b-c).

A210207 Area A of the non-right triangles such that A, the sides, and the circumradius are integers.

Original entry on oeis.org

168, 432, 480, 624, 672, 768, 1320, 1512, 1536, 1560, 1680, 1728, 1848, 1920, 2040, 2304, 2376, 2496, 2520, 2688, 2856, 3024, 3072, 3240, 3696, 3720, 3840, 3864, 3888, 4104, 4200, 4320, 4536, 5280, 5376, 5616, 5712, 6000, 6048, 6144, 6240, 6552, 6720, 6912

Offset: 1

Michel Lagneau, Mar 18 2012

nonn

A103251 gives the areas of right triangles with the same property (the area, the sides, and the circumradius are integers). Thus the intersection of this sequence with A103251 will give the areas of 2 families of triangles with the same property: one family of right triangles and one family of non-right triangles.
For example a(3) = A103251(8) = 480 generates two triangles whose sides are
(a,b,c) = (32, 50, 78) = > A = 480, R = 65, and 32^2 + 50^2 is no square;
(a,b,c) = (20, 48, 52) = > A = 480, R = 26, and 20^2 + 48^2  = 52^2 is square.
{a(n) intersection A103251} = {480, 1320, 1536, 1920, 2520, 3024, 3696, 3840, ...}

			168 is in the sequence because, for (a,b,c) = (14,30,40), A = sqrt(42*(42-14)*(42-30)*(42-40)) = 168, and 14^2 + 30^2 is no square.

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

Cf. A103251, A208984, A188158.

T:=array(1..4000):nn:=400: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): u:=a^2+b^2:if  x>0 then x1:=sqrt(x) : y:=a*b*c/(4*x1):
else fi:if x1=floor(x1) and y = floor(y) and u <> c^2 then k:=k+1:T[k]:=x1:else fi:od:od:od: L := [seq(T[i],i=1..k)]:L1:=convert(T,set):A:=sort(L1, `<`): print(A):

nn=400; lst={}; Do[s=(a+b+c)/2; If[IntegerQ[s], area2=s (s-a) (s-b) (s-c); If[0 < area2 && a^2 != b^2+c^2 && IntegerQ[Sqrt[area2]] && IntegerQ[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.

A228383 Area A of the triangle such that A, the sides, and the inradius are integers.

Original entry on oeis.org

6, 24, 30, 36, 42, 48, 54, 60, 66, 84, 96, 108, 114, 120, 126, 132, 144, 150, 156, 168, 180, 192, 198, 210, 216, 240, 252, 264, 270, 294, 300, 324, 330, 336, 360, 378, 384, 390, 396, 408, 420, 432, 456, 462, 468, 480, 486, 504, 510, 522, 528, 540, 546, 570

Offset: 1

Michel Lagneau, Aug 21 2013

nonn

The sequences A208984 and A185210 are subsequences of this sequence. The corresponding inradius r are 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 3, 4, 3, ...
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.
a(n) is divisible by 6 and the squares of the form 36k^2 are in the sequence.

			24 is in the sequence because for (a, b, c) = (6, 8, 10) => s =(6 + 8 + 10)/2 = 12; A = sqrt(12*(12-6)*(12-8)*(12-10)) = sqrt(576)= 24; r = A/s = 2.

Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), pp. 32-34.
Eric W. Weisstein, MathWorld: Inradius

Cf. A188158, A120572, A210250, A208984, A185210.

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

A350378 Integer areas of integer-sided triangles such that the distance d between the incenter and the circumcenter is a prime number.

Original entry on oeis.org

48, 768, 3840, 108000, 1134000, 200202240, 4382077920

Offset: 1

Michel Lagneau, Dec 28 2021

Subsequence of A231174. The sequence is probably infinite.
The corresponding sequence of the primes d is {5, 5, 13, 17, 53, 193, 241,...} (see A350379).
In geometry, Euler's theorem states that the distance between the incenter and circumcenter can be expressed as d = sqrt(R(R-2r)), where R is the circumradius and r is the inradius.
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, d) where A is the area of the triangles, a, b, c are the integer sides of the triangles, r is the inradius, R is the circumradius and d is the distance between the incenter and circumcenter with d = sqrt(R(R-2r)).
+------------+--------+--------+---------+---------+---------+-----+
|         A  |     a  |     b  |     c   |    r    |    R    |  d  |
+------------+--------+--------+---------+---------+---------+-----+
|         48 |     10 |     10 |     16  |    8/3  |    25/3 |   5 |
|        768 |     40 |     40 |     48  |   12    |    25   |   5 |
|       3840 |     80 |    104 |    104  |   80/3  |   169/3 |  13 |
|     108000 |    480 |    510 |    510  |  144    |   289   |  17 |
|    1134000 |   1590 |   1590 |   1680  |  1400/3 |  2809/3 |  53 |
|  200202240 |  21280 |  21616 |  21616  | 18620/3 | 37249/3 | 193 |
| 4382077920 | 100320 | 100738 | 100738  | 29040   | 58081   | 241 |
....................................................................
From the previous table, we observe that the triangles are isosceles, the distance between the incenter and the circumcenter is d = sqrt(R) if R is a perfect square, or d = sqrt(3R) if R is of the form k^2/3, k integer. We also observe that d divides the two equal sides of the isosceles triangles: 10/5 = 2, 40/5 = 8, 104/13 = 8, 510/17 = 30, 1590/853 = 30, 21616/193 = 112, 100738/241 = 418, ....

Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32.
R. A. Johnson, Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.

Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), p. 32.
Eric Weisstein's World of Mathematics, Exradius
Eric Weisstein's World of Mathematics, Inradius

Cf. A185210, A208984, A210207, A231174, A350379.

nn=520; lst={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s (s-a)(s-b)(s-c); If[area2>0&&IntegerQ[Sqrt[area2]]&&PrimeQ[Sqrt[a*b*c/(4*Sqrt[area2])*(a*b*c/(4*Sqrt[area2])-2*Sqrt[area2]/s)]],Print[Sqrt[area2]," ",c," ",b," ",a," ",Sqrt[area2]/s," ",a*b*c/(4*Sqrt[area2])," ",Sqrt[a*b*c/(4*Sqrt[area2])*(a*b*c/(4*Sqrt[area2])-2*Sqrt[area2]/s)]]]],{a,nn},{b,a},{c,b}]

A350379 Consider integer-sided triangles having integer area and let d be the distance between the center of the inscribed circle and the center of the circumscribed circle. The sequence lists the set of distances d which are prime numbers.

Original entry on oeis.org

5, 13, 17, 53, 193, 241

Offset: 1

Michel Lagneau, Dec 28 2021

The corresponding integer areas of integer-sided triangles such that the distance between the incenter and the circumcenter is a prime number is given by the sequence A350378.
In geometry, Euler's theorem states that the distance between the incenter and circumcenter can be expressed as d = sqrt(R(R-2r)), where R is the circumradius and r is the inradius.
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, d) where A is the area of the triangles, a, b, c are the integer sides of the triangles, r is the inradius, R is the circumradius and d is the distance between the incenter and circumcenter with d = sqrt(R(R-2r)).
+------------+--------+--------+---------+---------+---------+-----+
|         A  |     a  |     b  |     c   |    r    |    R    |  d  |
+------------+--------+--------+---------+---------+---------+-----+
|         48 |     10 |     10 |     16  |    8/3  |    25/3 |   5 |
|        768 |     40 |     40 |     48  |   12    |    25   |   5 |
|       3840 |     80 |    104 |    104  |   80/3  |   169/3 |  13 |
|     108000 |    480 |    510 |    510  |  144    |   289   |  17 |
|    1134000 |   1590 |   1590 |   1680  |  1400/3 |  2809/3 |  53 |
|  200202240 |  21280 |  21616 |  21616  | 18620/3 | 37249/3 | 193 |
| 4382077920 | 100320 | 100738 | 100738  | 29040   | 58081   | 241 |
....................................................................
From the previous table, we observe that the triangles are isosceles, the distance between the incenter and the circumcenter is d = sqrt(R) if R is a perfect square, or d = sqrt(3R) if R is of the form k^2/3, k integer. We also observe that d divides the two equal sides of the isosceles triangle: 10/5 = 2, 40/5 = 8, 104/13 = 8, 510/17 = 30, 1590/853 = 30, 21616/193 = 112, 100738/241 = 418, ....

Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32.
R. A. Johnson, Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.

Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), p. 32.
Eric Weisstein's World of Mathematics, Exradius
Eric Weisstein's World of Mathematics, Inradius

Cf. A185210, A208984, A210207, A231174, A350378.

nn=520;lst={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s (s-a)(s-b)(s-c);If[area2>0&&IntegerQ[Sqrt[area2]]&&PrimeQ[Sqrt[a*b*c/(4*Sqrt[area2])*(a*b*c/(4*Sqrt[area2])-2*Sqrt[area2]/s)]],Print[Sqrt[area2]," ",c," ",b," ",a," ",Sqrt[area2]/s," ",a*b*c/(4*Sqrt[area2])," ",Sqrt[a*b*c/(4*Sqrt[area2])*(a*b*c/(4*Sqrt[area2])-2*Sqrt[area2]/s)]]]],{a,nn},{b,a},{c,b}]

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

Original entry on oeis.org

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

Offset: 1

Michel Lagneau, Mar 09 2012

nonn

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

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

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

Cf. A120062, A188158, A208984.

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

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

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

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

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

nonn

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 W. Weisstein, MathWorld: Circumradius
Eric W. Weisstein, MathWorld: 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 uesd for afile. Zak Seidov, Jun 06 2015

Missing term 33408 added by Zak Seidov, Jun 08 2015

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

cp's OEIS Frontend

A210250 Area A of the cyclic quadrilaterals such that A, the sides and the radius of the circumcircle are integers.

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A185210 Areas A of the triangles such that A, the sides, the inradius and the radius of the three excircles are integers.

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A210207 Area A of the non-right triangles such that A, the sides, and the circumradius are integers.

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A228383 Area A of the triangle such that A, the sides, and the inradius are integers.

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A350378 Integer areas of integer-sided triangles such that the distance d between the incenter and the circumcenter is a prime number.

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A350379 Consider integer-sided triangles having integer area and let d be the distance between the center of the inscribed circle and the center of the circumscribed circle. The sequence lists the set of distances d which are prime numbers.

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

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