|
%I #14 Apr 09 2017 02:57:06
%S 6,12,30,60,84,168,330,546,660,1092,1224,1710,2448,3036,3420,6072,
%T 6090,7440,12180,12654,14880,17220,19866,25308,25944,34440,37206,
%U 39732,51330,51888,56730,74412,75174,89460,97236,102660,113460,123240,142926,150348,176220
%N Integer areas of integer-sided triangles where at least one of the three altitudes is of prime length.
%C Subset of A226453.
%C The corresponding primes are:
%C 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, ...
%C 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.
%C 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.
%C Properties of this sequence:
%C - The sequence is infinite (see the formula below);
%C - The prime altitude of a triangle is the greatest prime divisor of a(n) (the proof is easy if we observe the formula);
%C - There exists two subsets of numbers included into a(n):
%C 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.
%C 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.
%C We did not find a class of non-isosceles and non-right triangles (a, b, c) whose three altitudes include one of prime length.
%C 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.
%C +------+-----+-----+-----+----------+----------+---------+
%C | A | a | b | c | Ha | Hb | Hc |
%C +------+-----+-----+-----+----------+----------+---------+
%C | 6 | 3 | 4 | 5 | 12/5 | 3 | 4 |
%C | 12 | 5 | 5 | 8 | 3 | 24/5 | 24/5 |
%C | 30 | 5 | 12 | 13 | 5 | 60/13 | 12 |
%C | 60 | 13 | 13 | 24 | 5 | 120/13 | 120/13 |
%C | 84 | 7 | 24 | 25 | 168/25 | 7 | 24 |
%C | 168 | 25 | 25 | 48 | 7 | 336/25 | 336/25 |
%C | 330 | 11 | 60 | 61 | 660/61 | 11 | 60 |
%C | 546 | 13 | 84 | 85 | 1092/85 | 13 | 84 |
%C | 660 | 61 | 61 | 120 | 11 | 1320/61 | 1320/61 |
%C | 1092 | 85 | 85 | 168 | 13 | 2184/85 | 2184/85 |
%C | 1224 | 17 | 144 | 145 | 2448/145 | 17 | 144 |
%C | 1710 | 19 | 180 | 181 | 3420/181 | 19 | 180 |
%C | 2448 | 145 | 145 | 288 | 4896/145 | 4896/145 | 17 |
%C +------+-----+-----+-----+----------+----------+---------+
%H Eric Weisstein, <a href="http://mathworld.wolfram.com/Altitude.html">Altitude</a>
%H Eric Weisstein, <a href="http://mathworld.wolfram.com/IsoscelesTriangle.html">Isosceles Triangle</a>
%H Eric Weisstein, <a href="http://mathworld.wolfram.com/RightTriangle.html">Right Triangle</a>
%F a(n) = (prime(n)^3 - prime(n))/4 for the right triangles;
%F a(n) = (prime(n)^3 - prime(n))/2 for the isosceles triangles.
%p # program using the formula
%p 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):
%t 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]
%Y Cf. A188158, A210643, A210645, A226453.
%K nonn
%O 1,1
%A _Michel Lagneau_, Apr 02 2015
|
