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

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

1,1

Comments

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 = 2A/a, Hb = 2A/b, Hc = 2A/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(2q+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(2a-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 |

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

Links

Table of n, a(n) for n=1..41.

Eric Weisstein, Altitude

Eric Weisstein, Isosceles Triangle

Eric Weisstein, Right Triangle

Formula

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

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

Maple

# 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):

Mathematica

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]

Crossrefs

Cf. A188158, A210643, A210645, A226453.

Sequence in context: A011987 A036690 A229746 * A357197 A322374 A351694

Adjacent sequences: A256576 A256577 A256578 * A256580 A256581 A256582

Keyword

nonn

Author

Michel Lagneau, Apr 02 2015

Status

approved