A120572 -id:A120572 - cp's OEIS frontend

A120062 Number of triangles with integer sides a <= b <= c having integer inradius n.

1, 5, 13, 18, 15, 45, 24, 45, 51, 52, 26, 139, 31, 80, 110, 89, 33, 184, 34, 145, 185, 103, 42, 312, 65, 96, 140, 225, 36, 379, 46, 169, 211, 116, 173, 498, 38, 123, 210, 328, 44, 560, 60, 280, 382, 134, 64, 592, 116, 228, 230, 271, 47, 452, 229, 510, 276, 134, 54

Offset: 1

Hugo Pfoertner, Jun 11 2006

It is conjectured that the longest possible side c of a triangle with integer sides and inradius n is given by A057721(n) = n^4 + 3*n^2 + 1.
For n >= 1, a(n) >= 1 because triangle (a, b, c) = (n^2 + 2, n^4 + 2*n^2 + 1, n^4 + 3*n^2 + 1) has inradius n. - David W. Wilson, Jun 17 2006
Previous name was "Number of triangles with integer sides a<=bA362669); so, now effectively, a(10) = 52. - Bernard Schott, Apr 24 2023

			a(1)=1: {3,4,5} is the only triangle with integer sides and inradius 1.
a(2)=5: {5,12,13}, {6,8,10}, {6,25,29}, {7,15,20}, {9,10,17} are the only triangles with integer sides and inradius 2.
a(4)=A120252(1)+A120252(2)+A120252(4)=1+4+13 because 1, 2 and 4 are the factors of 4. The 1 primitive triangle with inradius n=1 is (3,4,5). The 4 primitive triangles with n=2 are (5,12,13), (9,10,17), (7,15,20), (6,25,29). The 13 primitive triangles with n=4 are (13,14,15), (15,15,24), (11,25,30), (15,26,37), (10,35,39), (9,40,41), (33,34,65), (25,51,74), (9,75,78), (11,90,97), (21,85,104), (19,153,170), (18,289,305). (Primitive means GCD(a, b, c, n)=1.)

David W. Wilson, Table of n, a(n) for n = 1..10000
Alan F. Beardon and Paul Stephenson, The Heron parameters of a triangle, Mathematical Gazette May 8, 2014.
Frank M Jackson, Mathematica program
Thomas Mautsch, Additional terms

Cf. A078644 [Pythagorean triangles with inradius n], A057721 [n^4+3*n^2+1].
Let S(n) be the set of triangles with integer sides a<=b<=c and inradius n. Then:
A120062(n) gives number of triangles in S(n).
A120261(n) gives number of triangles in S(n) with gcd(a, b, c) = 1.
A120252(n) gives number of triangles in S(n) with gcd(a, b, c, n) = 1.
A005408(n) = 2n+1 gives shortest short side a of triangles in S(n).
A120064(n) gives shortest middle side b of triangles in S(n).
A120063(n) gives shortest long side c of triangles in S(n).
A120570(n) gives shortest perimeter of triangles in S(n).
A120572(n) gives smallest area of triangles in S(n).
A058331(n) = 2n^2+1 gives longest short side a of triangles in S(n).
A082044(n) = n^4+2n^2+1 gives longest middle side b of triangles in S(n).
A057721(n) = n^4+3n^2+1 gives longest long side c of triangles in S(n).
A120571(n) = 2n^4+6n^2+4 gives longest perimeter of triangles in S(n).
A120573(n) = gives largest area of triangles in S(n).
Cf. A120252 [primitive triangles with integer inradius], A120063 [minimum of longest sides], A057721 [maximum of longest sides], A120064 [minimum of middle sides], A082044 [maximum of middle sides], A005408 [minimum of shortest sides], A058331 [maximum of shortest sides], A007237 [number of triangles with integer sides and area = n times perimeter].
Cf. A362669, A362670.

Mathematica
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(* See link above. *)
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The even-numbered terms are given by a(2*n)=A007237(n).

a(n) = Sum_{k|n} A120252(k).

More terms from Graeme McRae and Hugo Pfoertner, Jun 12 2006

Name corrected by Bernard Schott, Apr 24 2023

A070201 Number of integer triangles with perimeter n having integral inradius.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 3, 0, 0, 0, 2, 0, 1, 0, 1, 0, 2, 0, 2, 0, 0, 0, 1, 0, 1, 0, 2, 0, 0, 0, 8, 0, 0, 0, 1, 0, 3

Offset: 1

Reinhard Zumkeller, May 05 2002

nonn

a(n) = #{k | A070083(k) = n and A070200(k) = exact inradius};
a(n) = A070203(n) + A070204(n);
a(n) = A070205(n) + A070206(n) + A024155(n);
a(odd) = 0.

			a(36)=2, as there are two integer triangles with integer inradius having perimeter=32:
First: [A070080(368), A070081(368), A070082(368)] = [9,10,17], for s = A070083(368)/2 = (9+10+17)/2 = 18: inradius = sqrt((s-9)*(s-10)*(s-17)/s) = sqrt(9*8*1/18) = sqrt(4) = 2; therefore A070200(368) = 2.
2nd: [A070080(370), A070081(370), A070082(370)] = [9,12,15], for s = A070083(370)/2 = (9+12+15)/2 = 18: inradius = sqrt((s-9)*(s-12)*(s-15)/s) = sqrt(9*6*3/18) = sqrt(9) = 3; therefore A070200(370) = 3.

Seiichi Manyama, Table of n, a(n) for n = 1..5000
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, Incircle.
Eric Weisstein's World of Mathematics, Heron's Formula.
Reinhard Zumkeller, Integer-sided triangles

Cf. A070209, A070202, A070208, A005044, A070140.

Cf. A120062, A120572, A331040.

def A(n)
  cnt = 0
  (1..n / 3).each{|a|
    (a..(n - a) / 2).each{|b|
      c = n - a - b
      if a + b > c
        s = n / 2r
        t = (s - a) * (s - b) * (s - c) / s
        if t.denominator == 1
          t = t.to_i
          cnt += 1 if Math.sqrt(t).to_i ** 2 == t
        end
      end
    }
  }
  cnt
end
def A070201(n)
  (1..n).map{|i| A(i)}
end
p A070201(100) # Seiichi Manyama, Oct 06 2017

A331040 Numerator of squared radius of inscribed circle of a triangle with integer sides i <= j <= k, such that the number of triangles with this radius sets a new record. Denominators are A331041.

Original entry on oeis.org

1, 35, 3, 7, 3, 15, 8, 35, 55, 63, 95, 119, 135, 56, 231, 255, 80, 351, 455, 495, 855, 216, 224, 1071, 360

Offset: 1

Hugo Pfoertner, Jan 11 2020

The radius rho of the inscribed circle of a triangle (a,b,c) is rho = sqrt((s-a)*(s-b)*(s-c)/s), with s=(a+b+c)/2. For given integer values of a <= b and a rational target value r2 of the squared incircle radius, c is given by the two positive real roots of the polynomial P(a,b,x,r2) = x^3 - x^2 * (a+b) + x * (4*r2-(b-a)^2) + (a+b)^3 + 4*(a+b)*(r2-a*b). P(a,b,x,r2) = 0 may have 0, 1 or 2 positive integer solutions.

The potential ranges of the side lengths of the triangles can be determined in analogy to the ranges for the case of integer radii of the incircles, see A120062 for the relevant formulas and sequences.

			b(1) = a(1)/A331041(1) = 1/12: Triangle (1,1,1) has the least possible radius of incircle = sqrt(1/12).
b(2) = a(2)/A331041(2) = 35/52: Triangles (2,18,19) and (3,4,6) are the first occurrence of more than one triangle with the same radius of their incircles. rho = sqrt(35/52) in both cases.
b(3) = a(3)/A331041(3) = 3/4: Triangles are (2,7,7), (3,3,3), and (3,5,7).
b(4) = a(4)/A331041(4) = 7/4: (3,12,12), (3,22,23), (4,5,6), (5,18,22), (6,11,16) are the A331043(4) = 5 triangles with rho^2 = b(4).
b(15) = 231/4 includes the rare case, where two distinct integer solutions for the same pair of sides a and b exist: (20,37,38) and (20,37,39), both with rho^2=231/4 and thus contributing 2 of the A331043(15)=84 triangles with this squared radius of the incircle.

Hugo Pfoertner, Illustration of side lengths for terms a(13)-a(25).

Cf. A057721, A082044, A120062, A120572, A331012.

Cf. A331041 (corresponding denominators), A331042 (4*a(n)/A331041(n)), A331043 (records of numbers of triangles).

\\ Only suitable for demonstration of initial terms
rh2(a,b,c)={my(s=(a+b+c)/2);(s-a)*(s-b)*(s-c)/s};
lim_a(x)=ceil(4*(x^2+2));
lim_b(x)=ceil(4*(x^4+2*x^2+1));
target=35/4; v=vector(333938); n=0;
for(a=1,lim_a(sqrt(target)), for(b=a,lim_b(sqrt(target)), for(c=b,a+b-1, f=rh2(a,b,c);v[n++]=f)));
v=vecsort(v); print("A331040 A331041 A331043"); print(numerator(v[1])," ",denominator(v[1])," ",1); m=0; mm=0; for(k=2,#v, if(v[k]>target,break); if(v[k]==v[k-1], m++; if(m>mm&&v[k+1]>v[k], print(numerator(v[k])," ",denominator(v[k])," ",m);mm=m),m=1));

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).

A289155 Smallest area of triangle with integer sides and area = n times perimeter.

Original entry on oeis.org

24, 84, 192, 336, 540, 756, 1134, 1344, 1710, 2100, 2640, 3000, 4056, 4116, 4680, 5376, 6936, 6804, 8664, 8400, 9240, 10164, 12696, 12000, 13500, 14196, 15390, 16296, 20184, 18720, 23064, 21504, 23232, 24276, 26040, 27000, 32856, 30324

Offset: 1

Zhining Yang, Jun 26 2017

nonn

			For n = 4, a(4)=336 means for the smallest triangle (a,b,c) = (26,28,30), the area is 336, which is 4 times the perimeter 84.

Ray Chandler, Table of n, a(n) for n = 1..5000 (first 300 terms from Zhining Yang)

Cf. A007237, A120572, A188158, A228383.

a(n) is the leading entry in row n of the triangle in A290451.

for(k=1, 50, n=0;A=10^9; d=4*k^2; e=3*d; for(b=1, sqrt(e), for (c=2*k, e/b, if(b*c>d&&c>=b, f = (b + c)*d / (b * c - d); if(f>=c, a=floor(f); if(a==f, n++; s=2*(a+b+c)*k;if(s

a(n) = A120572(2n). - Ray Chandler, Jul 27 2017

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]

cp's OEIS Frontend

A120062 Number of triangles with integer sides a <= b <= c having integer inradius n.

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A070201 Number of integer triangles with perimeter n having integral inradius.

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A331040 Numerator of squared radius of inscribed circle of a triangle with integer sides i <= j <= k, such that the number of triangles with this radius sets a new record. Denominators are A331041.

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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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A289155 Smallest area of triangle with integer sides and area = n times perimeter.

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

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