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

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

Offset

1,1

Comments

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 |

+---------+------+------+------+-----+------+

Links

Table of n, a(n) for n=1..12.

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

Example

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.

Mathematica

nn=2000; 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]]&&IntegerQ[Sqrt[a*b*c/(4*Sqrt[area2])]], Print[Sqrt[area2], " ", a " ", b, " ", c, " ", Sqrt[area2]/s, " ", a*b*c/(4*Sqrt[area2])]]], {a, nn}, {b, a}, {c, b}]

Crossrefs

Cf. A188158, A208984.

Sequence in context: A279725 A234730 A234815 * A110075 A360623 A331908

Adjacent sequences: A232271 A232272 A232273 * A232275 A232276 A232277

Keyword

nonn,hard

Author

Michel Lagneau, Nov 22 2013

Status

approved