Heronian triangles

Heronian triangles is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Hero's formula for the area of a triangle given the length of its three sides a, b, and c is given by:

$A={\sqrt {s(s-a)(s-b)(s-c)}},$

where s is half the perimeter of the triangle; that is,

$s={\frac {a+b+c}{2}}.$

Heronian triangles are triangles whose sides and area are all integers.

An example is the triangle with sides 3, 4, 5 whose area is 6 (and whose perimeter is 12).

Note that any triangle whose sides are all an integer multiple of 3,4,5; such as 6,8,10, will also be a heronian triangle.

Define a Primitive Heronian triangle as a heronian triangle where the greatest common divisor of all three sides is 1. this will exclude, for example triangle 6,8,10

The task is to:

1. Create a named function/method/proceedure/... that implements Hero's formula.
2. Use the function to generate all the primitive heronian triangles with sides <= 200.
3. Show the count of how many triangles are found.
4. Order the triangles by first increasing area, then by increasing perimeter, then by increasing maximum side lengths
5. Show the first ten ordered triangles in a table of sides, perimeter, and area.
6. Show a similar ordered table for those triangles with area = 210

Show all output here.

Note: when generating triangles it may help to restrict $a<=b<=c$

Perl 6

<lang perl6>sub hero($a, $b, $c) {

   my $s = ($a + $b + $c) / 2;
   my $a2 = $s * ($s - $a) * ($s - $b) * ($s - $c);
   $a2.sqrt;

}

sub heronian-area($a, $b, $c) {

   given hero $a, $b, $c { .floor == .ceiling ?? .floor !! 0 }

}

sub primitive-heronian-area($a, $b, $c) {

   heronian-area $a, $b, $c
       if 1 == [gcd] $a, $b, $c;

}

sub show {

   say "   Area Perimeter   Sides";
   for @_ -> [$area, $perim, $c, $b, $a] {

printf "%6d %6d %12s\n", $area, $perim, "$a×$b×$c";

   }

}

sub MAIN ($maxside = 200, $first = 10, $witharea = 210) {

   my \h = sort *.fmt('%05d'), gather
       for 1 .. $maxside -> $c {
           for 1 .. $c -> $b {
               for $c - $b + 1 .. $b -> $a {
                   if primitive-heronian-area($a,$b,$c) -> $area {
                       take [$area, $a+$b+$c, $c, $b, $a];
                   }
               }
           }
       }

   say "Primitive Heronian triangles with sides up to $maxside: ", +h;

   say "\nFirst $first:";
   show h[^$first];

   say "\nArea $witharea:";
   show h.grep: *[0] == $witharea;

}</lang>

Output:

Primitive Heronian triangles with sides up to 200: 517

First 10:
   Area Perimeter   Sides
     6     12        3×4×5
    12     16        5×5×6
    12     18        5×5×8
    24     32      4×13×15
    30     30      5×12×13
    36     36      9×10×17
    36     54      3×25×26
    42     42      7×15×20
    60     36     10×13×13
    60     40      8×15×17

Area 210:
   Area Perimeter   Sides
   210     70     17×25×28
   210     70     20×21×29
   210     84     12×35×37
   210     84     17×28×39
   210    140      7×65×68
   210    300    3×148×149

Python

<lang python>from math import sqrt from fractions import gcd from itertools import product

def hero(a, b, c):

   s = (a + b + c) / 2
   a2 = s*(s-a)*(s-b)*(s-c)
   return sqrt(a2) if a2 > 0 else 0
   
   

def is_heronian(a, b, c):

   a = hero(a, b, c)
   return a > 0 and a.is_integer()
   

def gcd3(x, y, z):

   return gcd(gcd(x, y), z)

if __name__ == '__main__':

   maxside = 200
   h = [(a, b, c) for a,b,c in product(range(1, maxside + 1), repeat=3) 
        if a <= b <= c and a + b > c and gcd3(a, b, c) == 1 and is_heronian(a, b, c)]
   h.sort(key = lambda x: (hero(*x), sum(x), x[::-1]))   # By increasing area, perimeter, then sides
   print('Primitive Heronian triangles with sides up to %i:' % maxside, len(h))
   print('\nFirst ten when ordered by increasing area, then perimeter,then maximum sides:')
   print('\n'.join('  %14r perim: %3i area: %i' 
                   % (sides, sum(sides), hero(*sides)) for sides in h[:10]))
   print('\nAll with area 210 subject to the previous ordering:')
   print('\n'.join('  %14r perim: %3i area: %i' 
                   % (sides, sum(sides), hero(*sides)) for sides in h
                   if hero(*sides) == 210))</lang>

Output:

Primitive Heronian triangles with sides up to 200: 517

First ten when ordered by increasing area, then perimeter,then maximum sides:
       (3, 4, 5) perim:  12 area: 6
       (5, 5, 6) perim:  16 area: 12
       (5, 5, 8) perim:  18 area: 12
     (4, 13, 15) perim:  32 area: 24
     (5, 12, 13) perim:  30 area: 30
     (9, 10, 17) perim:  36 area: 36
     (3, 25, 26) perim:  54 area: 36
     (7, 15, 20) perim:  42 area: 42
    (10, 13, 13) perim:  36 area: 60
     (8, 15, 17) perim:  40 area: 60

All with area 210 subject to the previous ordering:
    (17, 25, 28) perim:  70 area: 210
    (20, 21, 29) perim:  70 area: 210
    (12, 35, 37) perim:  84 area: 210
    (17, 28, 39) perim:  84 area: 210
     (7, 65, 68) perim: 140 area: 210
   (3, 148, 149) perim: 300 area: 210