# Law of cosines - triples

Law of cosines - triples
You are encouraged to solve this task according to the task description, using any language you may know.

The Law of cosines states that for an angle γ, (gamma) of any triangle, if the sides adjacent to the angle are A and B and the side opposite is C; then the lengths of the sides are related by this formula:

A2 + B2 - 2ABcos(γ) = C2
Specific angles

For an angle of of   90º   this becomes the more familiar "Pythagoras equation":

A2 + B2  =  C2

For an angle of   60º   this becomes the less familiar equation:

A2 + B2 - AB  =  C2

And finally for an angle of   120º   this becomes the equation:

A2 + B2 + AB  =  C2

•   Find all integer solutions (in order) to the three specific cases, distinguishing between each angle being considered.
•   Restrain all sides to the integers   1..13   inclusive.
•   Show how many results there are for each of the three angles mentioned above.

Note: Triangles with the same length sides but different order are to be treated as the same.

Optional Extra credit
• How many 60° integer triples are there for sides in the range 1..10_000 where the sides are not all of the same length.

## ALGOL 68

BEGIN
# find all integer sided 90, 60 and 120 degree triangles by finding integer solutions for #
# a^2 + b^2 = c^2, a^2 + b^2 - ab = c^2, a^2 + b^2 + ab = c^2 where a, b, c in 1 .. 13 #
INT max side = 13; # max triangle side to consider #
INT max square = max side * max side; # max triangle side squared to consider #
[ 1 : max square ]INT root; # table of square roots #
FOR s TO UPB root DO root[ s ] := 0 OD;
FOR s TO max side DO root[ s * s ] := s OD;
INT tcount := 0;
[ 1 : max square ]INT ta, tb, tc, tangle;
# prints solutions for the specified angle #
PROC print triangles = ( INT angle )VOID:
BEGIN
INT scount := 0;
FOR t TO tcount DO IF tangle[ t ] = angle THEN scount +:= 1 FI OD;
print( ( whole( scount, -4 ), " ", whole( angle, -3 ), " degree triangles:", newline ) );
FOR t TO tcount DO
IF tangle[ t ] = angle THEN
print( ( " ", whole( ta[ t ], -3 ), whole( tb[ t ], -3 ), whole( tc[ t ], -3 ), newline ) )
FI
OD
END # print triangles # ;
# stores the triangle with sides a, b, root[ c2 ] and the specified angle, #
# if it is a solution #
PROC try triangle = ( INT a, b, c2, angle )VOID:
IF c2 <= max square THEN
# the third side is small enough #
INT c = root[ c2 ];
IF c /= 0 THEN
# the third side is the square of an integer #
tcount +:= 1;
ta[ tcount ] := a; tb[ tcount ] := b; tc[ tcount ] := root[ c2 ];
tangle[ tcount ] := angle
FI
FI # try triangle # ;
# find all triangles #
FOR a TO max side DO
FOR b FROM a TO max side DO
try triangle( a, b, ( a * a ) + ( b * b ) - ( a * b ), 60 );
try triangle( a, b, ( a * a ) + ( b * b ), 90 );
try triangle( a, b, ( a * a ) + ( b * b ) + ( a * b ), 120 )
OD
OD;
# print the solutions #
print triangles( 60 );
print triangles( 90 );
print triangles( 120 )
END
Output:
15  60 degree triangles:
1  1  1
2  2  2
3  3  3
3  8  7
4  4  4
5  5  5
5  8  7
6  6  6
7  7  7
8  8  8
9  9  9
10 10 10
11 11 11
12 12 12
13 13 13
3  90 degree triangles:
3  4  5
5 12 13
6  8 10
2 120 degree triangles:
3  5  7
7  8 13

## C

### A brute force algorithm, O(N^3)

/*
* RossetaCode: Law of cosines - triples
*
* An quick and dirty brute force solutions with O(N^3) cost.
* Anyway it is possible set MAX_SIDE_LENGTH equal to 10000
* and use fast computer to obtain the "extra credit" badge.
*
* Obviously, there are better algorithms.
*/

#include <stdio.h>
#include <math.h>

#define MAX_SIDE_LENGTH 13
//#define DISPLAY_TRIANGLES 1

int main(void)
{
static char description[3][80] = {
"gamma = 90 degrees, a*a + b*b == c*c",
"gamma = 60 degrees, a*a + b*b - a*b == c*c",
"gamma = 120 degrees, a*a + b*b + a*b == c*c"
};
static int coeff[3] = { 0, 1, -1 };

for (int k = 0; k < 3; k++)
{
int counter = 0;
for (int a = 1; a <= MAX_SIDE_LENGTH; a++)
for (int b = 1; b <= a; b++)
for (int c = 1; c <= MAX_SIDE_LENGTH; c++)
if (a * a + b * b - coeff[k] * a * b == c * c)
{
counter++;
#ifdef DISPLAY_TRIANGLES
printf("  %d  %d  %d\n", a, b, c);
#endif
}
printf("%s, number of triangles = %d\n", description[k], counter);
}

return 0;
}

Output:
gamma =  90 degrees,  a*a + b*b       == c*c,  number of triangles = 3
gamma =  60 degrees,  a*a + b*b - a*b == c*c,  number of triangles = 15
gamma = 120 degrees,  a*a + b*b + a*b == c*c,  number of triangles = 2

### An algorithm with O(N^2) cost

/*
* RossetaCode: Law of cosines - triples
*
* A solutions with O(N^2) cost.
*/

#include <stdio.h>
#include <math.h>

#define MAX_SIDE_LENGTH 10000
//#define DISPLAY_TRIANGLES

int main(void)
{
static char description[3][80] = {
"gamma = 90 degrees, a*a + b*b == c*c",
"gamma = 60 degrees, a*a + b*b - a*b == c*c",
"gamma = 120 degrees, a*a + b*b + a*b == c*c"
};
static int coeff[3] = { 0, 1, -1 };

printf("MAX SIDE LENGTH = %d\n\n", MAX_SIDE_LENGTH);

for (int k = 0; k < 3; k++)
{
int counter = 0;
for (int a = 1; a <= MAX_SIDE_LENGTH; a++)
for (int b = 1; b <= a; b++)
{
int cc = a * a + b * b - coeff[k] * a * b;
int c = (int)(sqrt(cc) + 0.5);
if (c <= MAX_SIDE_LENGTH && c * c == cc)
{
#ifdef DISPLAY_TRIANGLES
printf("%d %d %d\n", a, b, c);
#endif
counter++;
}
}
printf("%s, number of triangles = %d\n", description[k], counter);
}

return 0;
}

Output:
MAX SIDE LENGTH = 10000

gamma =  90 degrees,  a*a + b*b       == c*c,  number of triangles = 12471
gamma =  60 degrees,  a*a + b*b - a*b == c*c,  number of triangles = 28394
gamma = 120 degrees,  a*a + b*b + a*b == c*c,  number of triangles = 10374

## Factor

USING: backtrack formatting kernel locals math math.ranges
sequences sets sorting ;
IN: rosetta-code.law-of-cosines

:: triples ( quot -- seq )
[
V{ } clone :> seen
13 [1,b] dup dup [ amb-lazy ] [email protected] :> ( a b c )
a sq b sq + a b quot call( x x x -- x ) c sq =
{ b a c } seen member? not and
must-be-true { a b c } dup seen push
] bag-of ;

: show-solutions ( quot angle -- )
[ triples { } like dup length ] dip rot
"%d solutions for %d degrees:\n%u\n\n" printf ;

[ * + ] 120
[ 2drop 0 - ] 90
[ * - ] 60 [ show-solutions ] [email protected]
Output:
2 solutions for 120 degrees:
{ { 3 5 7 } { 7 8 13 } }

3 solutions for 90 degrees:
{ { 3 4 5 } { 5 12 13 } { 6 8 10 } }

15 solutions for 60 degrees:
{
{ 1 1 1 }
{ 2 2 2 }
{ 3 3 3 }
{ 3 8 7 }
{ 4 4 4 }
{ 5 5 5 }
{ 5 8 7 }
{ 6 6 6 }
{ 7 7 7 }
{ 8 8 8 }
{ 9 9 9 }
{ 10 10 10 }
{ 11 11 11 }
{ 12 12 12 }
{ 13 13 13 }
}

## FreeBASIC

' version 03-03-2019
' compile with: fbc -s console

#Define max 13

#Define Format(_x) Right(" " + Str(_x), 4)

Dim As UInteger a, b, c, a2, b2, c2, c60 , c90, c120
Dim As String s60, s90, s120

For a = 1 To max
a2 = a * a
For b = a To max
b2 = b * b
' 60 degrees
c2 = a2 + b * b - a * b
c = Sqr(c2)
If c * c = c2 AndAlso c <= max Then
s60 += Format(a) + Format(b) + Format(c) + Chr(10, 13)
c60 += 1
End If
' 90 degrees
c2 = a2 + b * b
c = Sqr(c2)
If c * c = c2 AndAlso c <= max Then
s90 += Format(a) + Format(b) + Format(c) + Chr(10, 13)
c90 += 1
End If
' 120 degrees
c2 = a2 + b * b + a * b
c = Sqr(c2)
If c * c = c2 AndAlso c <= max Then
s120 += Format(a) + Format(b) + Format(c) + Chr(10, 13)
c120 += 1
End If
Next
Next

Print Using "###: 60 degree triangles"; c60
Print s60
Print

Print Using "###: 90 degree triangles"; c90
Print s90
Print

Print Using "###: 120 degree triangles"; c120
Print s120
Print

#Undef max
#Define max 10000

c60 = 0
For a = 1 To max
a2 = a * a
For b = a +1 To max
c2 = a2 + b * (b - a)
c = Sqr(c2)
If c * c = c2 AndAlso c <= max Then
c60 += 1
End If
Next
Next

Print "For 60 degree triangles in the range [1, 10000]"
Print "There are "; c60; " triangles that have different length for a, b and c"

' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
Output:
15: 60 degree triangles
1   1   1
2   2   2
3   3   3
3   8   7
4   4   4
5   5   5
5   8   7
6   6   6
7   7   7
8   8   8
9   9   9
10  10  10
11  11  11
12  12  12
13  13  13

3: 90 degree triangles
3   4   5
5  12  13
6   8  10

2: 120 degree triangles
3   5   7
7   8  13

For 60 degree triangles in the range [1, 10000]
There are 18394 triangles that have different length for a, b and c

## Go

package main

import "fmt"

type triple struct{ a, b, c int }

var squares13 = make(map[int]int, 13)
var squares10000 = make(map[int]int, 10000)

func init() {
for i := 1; i <= 13; i++ {
squares13[i*i] = i
}
for i := 1; i <= 10000; i++ {
squares10000[i*i] = i
}
}

func solve(angle, maxLen int, allowSame bool) []triple {
var solutions []triple
for a := 1; a <= maxLen; a++ {
for b := a; b <= maxLen; b++ {
lhs := a*a + b*b
if angle != 90 {
switch angle {
case 60:
lhs -= a * b
case 120:
lhs += a * b
default:
panic("Angle must be 60, 90 or 120 degrees")
}
}
switch maxLen {
case 13:
if c, ok := squares13[lhs]; ok {
if !allowSame && a == b && b == c {
continue
}
solutions = append(solutions, triple{a, b, c})
}
case 10000:
if c, ok := squares10000[lhs]; ok {
if !allowSame && a == b && b == c {
continue
}
solutions = append(solutions, triple{a, b, c})
}
default:
panic("Maximum length must be either 13 or 10000")
}
}
}
return solutions
}

func main() {
fmt.Print("For sides in the range [1, 13] ")
fmt.Println("where they can all be of the same length:-\n")
angles := []int{90, 60, 120}
var solutions []triple
for _, angle := range angles {
solutions = solve(angle, 13, true)
fmt.Printf(" For an angle of %d degrees", angle)
fmt.Println(" there are", len(solutions), "solutions, namely:")
fmt.Printf("  %v\n", solutions)
fmt.Println()
}
fmt.Print("For sides in the range [1, 10000] ")
fmt.Println("where they cannot ALL be of the same length:-\n")
solutions = solve(60, 10000, false)
fmt.Print(" For an angle of 60 degrees")
fmt.Println(" there are", len(solutions), "solutions.")
}
Output:
For sides in the range [1, 13] where they can all be of the same length:-

For an angle of 90 degrees there are 3 solutions, namely:
[{3 4 5} {5 12 13} {6 8 10}]

For an angle of 60 degrees there are 15 solutions, namely:
[{1 1 1} {2 2 2} {3 3 3} {3 8 7} {4 4 4} {5 5 5} {5 8 7} {6 6 6} {7 7 7} {8 8 8} {9 9 9} {10 10 10} {11 11 11} {12 12 12} {13 13 13}]

For an angle of 120 degrees there are 2 solutions, namely:
[{3 5 7} {7 8 13}]

For sides in the range [1, 10000] where they cannot ALL be of the same length:-

For an angle of 60 degrees there are 18394 solutions.

import qualified Data.Map.Strict as Map
import qualified Data.Set as Set
import Data.Monoid ((<>))

triangles
:: (Map.Map Int Int -> Int -> Int -> Int -> Int -> Maybe Int)
-> Int
-> [(Int, Int, Int)]
triangles f n =
let mapRoots = Map.fromList \$ ((,) =<< (^ 2)) <\$> [1 .. n]
in Set.elems \$
foldr
(\(suma2b2, a, b) triSet ->
(case f mapRoots suma2b2 (a * b) a b of
Just c -> Set.insert (a, b, c) triSet
_ -> triSet))
(Set.fromList [])
([1 .. n] >>=
(\a -> (flip (,,) a =<< (a * a +) . (>>= id) (*)) <\$> [1 .. a]))

-- TESTS ------------------------------------------------------------------------

f90, f60, f60ne, f120 :: Map.Map Int Int -> Int -> Int -> Int -> Int -> Maybe Int
f90 dct x2 ab a b = Map.lookup x2 dct

f60 dct x2 ab a b = Map.lookup (x2 - ab) dct

f120 dct x2 ab a b = Map.lookup (x2 + ab) dct

f60ne dct x2 ab a b
| a == b = Nothing
| otherwise = Map.lookup (x2 - ab) dct

main :: IO ()
main = do
putStrLn
(unlines \$
"Triangles of maximum side 13\n" :
zipWith
(\f n ->
let solns = triangles f 13
in show (length solns) <> " solutions for " <> show n <>
" degrees:\n" <>
unlines (show <\$> solns))
[f120, f90, f60]
[120, 90, 60])
putStrLn "60 degrees - uneven triangles of maximum side 10000. Total:"
print \$ length \$ triangles f60ne 10000
Output:
Triangles of maximum side 13

2 solutions for 120 degrees:
(5,3,7)
(8,7,13)

3 solutions for 90 degrees:
(4,3,5)
(8,6,10)
(12,5,13)

15 solutions for 60 degrees:
(1,1,1)
(2,2,2)
(3,3,3)
(4,4,4)
(5,5,5)
(6,6,6)
(7,7,7)
(8,3,7)
(8,5,7)
(8,8,8)
(9,9,9)
(10,10,10)
(11,11,11)
(12,12,12)
(13,13,13)

60 degrees - uneven triangles of maximum side 10000. Total:
18394

## J

Solution:

RHS=: *: NB. right-hand-side of Cosine Law
LHS=: +/@:*:@] - [email protected]@[ * 2 * */@] NB. Left-hand-side of Cosine Law

solve=: 4 :0
oppside=. >: i. y
idx=. (RHS oppside) i. x LHS"1 adjsides
adjsides ((#~ idx ~: #) ,. ({~ idx -. #)@]) oppside
)

Example:

60 90 120 solve&.> 13
+--------+-------+------+
| 1 1 1|3 4 5|3 5 7|
| 2 2 2|5 12 13|7 8 13|
| 3 3 3|6 8 10| |
| 3 8 7| | |
| 4 4 4| | |
| 5 5 5| | |
| 5 8 7| | |
| 6 6 6| | |
| 7 7 7| | |
| 8 8 8| | |
| 9 9 9| | |
|10 10 10| | |
|11 11 11| | |
|12 12 12| | |
|13 13 13| | |
+--------+-------+------+
60 #@(solve -. _3 ]\ 3 # >:@[email protected]]) 10000 NB. optional extra credit
18394

## JavaScript

(() => {
'use strict';

// main :: IO ()
const main = () => {

const
f90 = dct => x2 => dct[x2],
f60 = dct => (x2, ab) => dct[x2 - ab],
f120 = dct => (x2, ab) => dct[x2 + ab],
f60unequal = dct => (x2, ab, a, b) =>
(a !== b) ? (
dct[x2 - ab]
) : undefined;

// triangles :: Dict -> (Int -> Int -> Int -> Int -> Maybe Int)
// -> [String]
const triangles = (f, n) => {
const
xs = enumFromTo(1, n),
fr = f(xs.reduce((a, x) => (a[x * x] = x, a), {})),
gc = xs.reduce((a, _) => a, {}),
setSoln = new Set();
return (
xs.forEach(
a => {
const a2 = a * a;
enumFromTo(1, 1 + a).forEach(
b => {
const
suma2b2 = a2 + b * b,
c = fr(suma2b2, a * b, a, b);
if (undefined !== c) {
};
}
);
}
),
Array.from(setSoln.keys())
);
};

const
result = 'Triangles of maximum side 13:\n\n' +
unlines(
zipWith(
(s, f) => {
const ks = triangles(f, 13);
return ks.length.toString() + ' solutions for ' + s +
' degrees:\n' + unlines(ks) + '\n';
},
['120', '90', '60'],
[f120, f90, f60]
)
) + '\nUneven triangles of maximum side 10000. Total:\n' +
triangles(f60unequal, 10000).length

return (
//console.log(result),
result
);
};

// GENERIC FUNCTIONS ----------------------------

// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = (f, xs) =>
xs.reduce((a, x) => a.concat(f(x)), []);

// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
m <= n ? iterateUntil(
x => n <= x,
x => 1 + x,
m
) : [];

// iterateUntil :: (a -> Bool) -> (a -> a) -> a -> [a]
const iterateUntil = (p, f, x) => {
const vs = [x];
let h = x;
while (!p(h))(h = f(h), vs.push(h));
return vs;
};

// Returns Infinity over objects without finite length
// this enables zip and zipWith to choose the shorter
// argument when one non-finite like cycle, repeat etc

// length :: [a] -> Int
const length = xs => xs.length || Infinity;

// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = (n, xs) =>
xs.constructor.constructor.name !== 'GeneratorFunction' ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));

// unlines :: [String] -> String
const unlines = xs => xs.join('\n');

// Use of `take` and `length` here allows zipping with non-finite lists
// i.e. generators like cycle, repeat, iterate.

// Use of `take` and `length` here allows zipping with non-finite lists
// i.e. generators like cycle, repeat, iterate.

// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = (f, xs, ys) => {
const
lng = Math.min(length(xs), length(ys)),
as = take(lng, xs),
bs = take(lng, ys);
return Array.from({
length: lng
}, (_, i) => f(as[i], bs[i], i));
};

// MAIN ---
return main();
})();
Output:
Triangles of maximum side 13:

2 solutions for 120 degrees:
3,5,7
13,7,8

3 solutions for 90 degrees:
3,4,5
10,6,8
12,13,5

15 solutions for 60 degrees:
1,1,1
2,2,2
3,3,3
4,4,4
5,5,5
6,6,6
7,7,7
3,7,8
5,7,8
8,8,8
9,9,9
10,10,10
11,11,11
12,12,12
13,13,13

Uneven triangles of maximum side 10000. Total:
18394
[Finished in 3.444s]

## Julia

Translation of: zkl
sqdict(n) = Dict([(x*x, x) for x in 1:n])
numnotsame(arrarr) = sum(map(x -> !all(y -> y == x[1], x), arrarr))

function filtertriangles(N)
sqd = sqdict(N)
t60 = Vector{Vector{Int}}()
t90 = Vector{Vector{Int}}()
t120 = Vector{Vector{Int}}()
for x in 1:N, y in 1:x
xsq, ysq, xy = (x*x, y*y, x*y)
if haskey(sqd, xsq + ysq - xy)
push!(t60, sort([x, y, sqd[xsq + ysq - xy]]))
push!(t90, sort([x, y, sqd[xsq + ysq]]))
elseif haskey(sqd, xsq + ysq + xy)
push!(t120, sort([x, y, sqd[xsq + ysq + xy]]))
end
end
t60, t90, t120
end

tri60, tri90, tri120 = filtertriangles(13)
println("Integer triples for 1 <= side length <= 13:\n")
println("Angle 60:"); for t in tri60 println(t) end
println("Angle 90:"); for t in tri90 println(t) end
println("Angle 120:"); for t in tri120 println(t) end
println("\nFor sizes N through 10000, there are \$(numnotsame(filtertriangles(10000)[1])) 60 degree triples with nonequal sides.")

Output:

Integer triples for 1 <= side length <= 13:

Angle 60: [1, 1, 1] [2, 2, 2] [3, 3, 3] [4, 4, 4] [5, 5, 5] [6, 6, 6] [7, 7, 7] [3, 7, 8] [5, 7, 8] [8, 8, 8] [9, 9, 9] [10, 10, 10] [11, 11, 11] [12, 12, 12] [13, 13, 13] Angle 90: [3, 4, 5] [6, 8, 10] [5, 12, 13] Angle 120: [3, 5, 7] [7, 8, 13]

For sizes N through 10000, there are 18394 60 degree triples with nonequal sides.

## Kotlin

Translation of: Go
// Version 1.2.70

val squares13 = mutableMapOf<Int, Int>()
val squares10000 = mutableMapOf<Int, Int>()

class Trio(val a: Int, val b: Int, val c: Int) {
override fun toString() = "(\$a \$b \$c)"
}

fun init() {
for (i in 1..13) squares13.put(i * i, i)
for (i in 1..10000) squares10000.put(i * i, i)
}

fun solve(angle :Int, maxLen: Int, allowSame: Boolean): List<Trio> {
val solutions = mutableListOf<Trio>()
for (a in 1..maxLen) {
inner@ for (b in a..maxLen) {
var lhs = a * a + b * b
if (angle != 90) {
when (angle) {
60 -> lhs -= a * b
120 -> lhs += a * b
else -> throw RuntimeException("Angle must be 60, 90 or 120 degrees")
}
}
when (maxLen) {
13 -> {
val c = squares13[lhs]
if (c != null) {
if (!allowSame && a == b && b == c) continue@inner
}
}

10000 -> {
val c = squares10000[lhs]
if (c != null) {
if (!allowSame && a == b && b == c) continue@inner
}
}

else -> throw RuntimeException("Maximum length must be either 13 or 10000")
}
}
}
return solutions
}

fun main(args: Array<String>) {
init()
print("For sides in the range [1, 13] ")
println("where they can all be of the same length:-\n")
val angles = intArrayOf(90, 60, 120)
lateinit var solutions: List<Trio>
for (angle in angles) {
solutions = solve(angle, 13, true)
print(" For an angle of \${angle} degrees")
println(" there are \${solutions.size} solutions, namely:")
println(" \${solutions.joinToString(" ", "[", "]")}\n")
}
print("For sides in the range [1, 10000] ")
println("where they cannot ALL be of the same length:-\n")
solutions = solve(60, 10000, false)
print(" For an angle of 60 degrees")
println(" there are \${solutions.size} solutions.")
}
Output:
For sides in the range [1, 13] where they can all be of the same length:-

For an angle of 90 degrees there are 3 solutions, namely:
[(3 4 5) (5 12 13) (6 8 10)]

For an angle of 60 degrees there are 15 solutions, namely:
[(1 1 1) (2 2 2) (3 3 3) (3 8 7) (4 4 4) (5 5 5) (5 8 7) (6 6 6) (7 7 7) (8 8 8) (9 9 9) (10 10 10) (11 11 11) (12 12 12) (13 13 13)]

For an angle of 120 degrees there are 2 solutions, namely:
[(3 5 7) (7 8 13)]

For sides in the range [1, 10000] where they cannot ALL be of the same length:-

For an angle of 60 degrees there are 18394 solutions.

## Perl

Translation of: Perl 6
use utf8;
binmode STDOUT, "utf8:";
use Sort::Naturally;

sub triples {
my(\$n,\$angle) = @_;
my(@triples,%sq);
\$sq{\$_**2}=\$_ for 1..\$n;
for \$a (1..\$n-1) {
for \$b (\$a+1..\$n) {
my \$ab = \$a*\$a + \$b*\$b;
my \$cos = \$angle == 60 ? \$ab - \$a * \$b :
\$angle == 120 ? \$ab + \$a * \$b :
\$ab;
if (\$angle == 60) {
push @triples, "\$a \$sq{\$cos} \$b" if exists \$sq{\$cos};
} else {
push @triples, "\$a \$b \$sq{\$cos}" if exists \$sq{\$cos};
}
}
}
@triples;
}

\$n = 13;
print "Integer triangular triples for sides 1..\$n:\n";
for my \$angle (120, 90, 60) {
my @itt = triples(\$n,\$angle);
if (\$angle == 60) { push @itt, "\$_ \$_ \$_" for 1..\$n }
printf "Angle %3d° has %2d solutions: %s\n", \$angle, scalar @itt,
join ', ', nsort @itt;
}

printf "Non-equilateral n=10000/60°: %d\n", scalar triples(10000,60);
Output:
Integer triangular triples for sides 1..13:
Angle 120° has  2 solutions: 3 5 7, 7 8 13
Angle  90° has  3 solutions: 3 4 5, 6 8 10, 5 12 13
Angle  60° has 15 solutions: 1 1 1, 2 2 2, 3 3 3, 3 7 8, 4 4 4, 5 5 5, 5 7 8, 6 6 6, 7 7 7, 8 8 8, 9 9 9, 10 10 10, 11 11 11, 12 12 12, 13 13 13
Non-equilateral n=10000/60°: 18394

## Perl 6

In each routine, race is used to allow concurrent operations, requiring the use of the atomic increment operator, ⚛++, to safely update @triples, which must be declared fixed-sized, as an auto-resizing array is not thread-safe. At exit, default values in @triples are filtered out with the test !eqv Any.

multi triples (60, \$n) {
my %sq = (1..\$n).map: { .² => \$_ };
my atomicint \$i = 0;
my @triples[2*\$n];
(1..^\$n).race(:8degree).map: -> \$a {
for \$a^..\$n -> \$b {
my \$cos = \$a * \$a + \$b * \$b - \$a * \$b;
@triples[\$i++] = \$a, %sq{\$cos}, \$b if %sq{\$cos}:exists;
}
}
@triples.grep: so *;
}

multi triples (90, \$n) {
my %sq = (1..\$n).map: { .² => \$_ };
my atomicint \$i = 0;
my @triples[2*\$n];
(1..^\$n).race(:8degree).map: -> \$a {
for \$a^..\$n -> \$b {
my \$cos = \$a * \$a + \$b * \$b;
@triples[\$i++] = \$a, \$b, %sq{\$cos} and last if %sq{\$cos}:exists;
}
}
@triples.grep: so *;
}

multi triples (120, \$n) {
my %sq = (1..\$n).map: { .² => \$_ };
my atomicint \$i = 0;
my @triples[2*\$n];
(1..^\$n).race(:8degree).map: -> \$a {
for \$a^..\$n -> \$b {
my \$cos = \$a * \$a + \$b * \$b + \$a * \$b;
@triples[\$i++] = \$a, \$b, %sq{\$cos} and last if %sq{\$cos}:exists;
}
}
@triples.grep: so *;
}

use Sort::Naturally;

my \$n = 13;
say "Integer triangular triples for sides 1..\$n:";
for 120, 90, 60 -> \$angle {
my @itt = triples(\$angle, \$n);
if \$angle == 60 { push @itt, "\$_ \$_ \$_" for 1..\$n }
printf "Angle %3d° has %2d solutions: %s\n", \$angle, +@itt, @itt.sort(*.&naturally).join(', ');
}

my (\$angle, \$count) = 60, 10_000;
say "\nExtra credit:";
say "\$angle° integer triples in the range 1..\$count where the sides are not all the same length: ", +triples(\$angle, \$count);
Output:
Integer triangular triples for sides 1..13:
Angle 120° has  2 solutions: 3 5 7, 7 8 13
Angle  90° has  3 solutions: 3 4 5, 5 12 13, 6 8 10
Angle  60° has 15 solutions: 1 1 1, 2 2 2, 3 3 3, 3 7 8, 4 4 4, 5 5 5, 5 7 8, 6 6 6, 7 7 7, 8 8 8, 9 9 9, 10 10 10, 11 11 11, 12 12 12, 13 13 13

Extra credit:
60° integer triples in the range 1..10000 where the sides are not all the same length: 18394

## Phix

Using a simple flat sequence of 100 million elements (well within the language limits) proved significantly faster than a dictionary (5x or so).

sequence squares = repeat(0,10000*10000)
for c=1 to 10000 do
squares[c*c] = c
end for

function solve(integer angle, maxlen, bool samelen=true)
sequence res = {}
for a=1 to maxlen do
integer a2 = a*a
for b=a to maxlen do
integer c2 = a2+b*b
if angle!=90 then
if angle=60 then c2 -= a*b
elsif angle=120 then c2 += a*b
else crash("angle must be 60/90/120")
end if
end if
integer c = iff(c2>length(squares)?0:squares[c2])
if c!=0 and c<=maxlen then
if samelen or a!=b or b!=c then
res = append(res,{a,b,c})
end if
end if
end for
end for
return res
end function

procedure show(string fmt,sequence res, bool full=true)
printf(1,fmt,{length(res),iff(full?sprint(res):"")})
end procedure

puts(1,"Integer triangular triples for sides 1..13:\n")
show("Angle 60 has %2d solutions: %s\n",solve( 60,13))
show("Angle 90 has %2d solutions: %s\n",solve( 90,13))
show("Angle 120 has %2d solutions: %s\n",solve(120,13))
show("Non-equilateral angle 60 triangles for sides 1..10000: %d%s\n",solve(60,10000,false),false)
Output:
Integer triangular triples for sides 1..13:
Angle  60 has 15 solutions: {{1,1,1},{2,2,2},{3,3,3},{3,8,7},{4,4,4},{5,5,5},{5,8,7},{6,6,6},{7,7,7},{8,8,8},{9,9,9},{10,10,10},{11,11,11},{12,12,12},{13,13,13}}
Angle  90 has  3 solutions: {{3,4,5},{5,12,13},{6,8,10}}
Angle 120 has  2 solutions: {{3,5,7},{7,8,13}}
Non-equilateral angle 60 triangles for sides 1..10000: 18394

## Python

### Sets

N = 13

def method1(N=N):
squares = [x**2 for x in range(0, N+1)]
sqrset = set(squares)
tri90, tri60, tri120 = (set() for _ in range(3))
for a in range(1, N+1):
a2 = squares[a]
for b in range(1, a + 1):
b2 = squares[b]
c2 = a2 + b2
if c2 in sqrset:
ab = a * b
c2 -= ab
if c2 in sqrset:
c2 += 2 * ab
if c2 in sqrset:
return sorted(tri90), sorted(tri60), sorted(tri120)
#%%
if __name__ == '__main__':
print(f'Integer triangular triples for sides 1..{N}:')
for angle, triples in zip([90, 60, 120], method1(N)):
print(f' {angle:3}° has {len(triples)} solutions:\n {triples}')
_, t60, _ = method1(10_000)
notsame = sum(1 for a, b, c in t60 if a != b or b != c)
print('Extra credit:', notsame)
Output:
Integer triangular triples for sides 1..13:
90° has 3 solutions:
[(3, 4, 5), (5, 12, 13), (6, 8, 10)]
60° has 15 solutions:
[(1, 1, 1), (2, 2, 2), (3, 3, 3), (3, 7, 8), (4, 4, 4), (5, 5, 5), (5, 7, 8), (6, 6, 6), (7, 7, 7), (8, 8, 8), (9, 9, 9), (10, 10, 10), (11, 11, 11), (12, 12, 12), (13, 13, 13)]
120° has 2 solutions:
[(3, 5, 7), (7, 8, 13)]
Extra credit: 18394

### Dictionaries

A variant Python draft based on dictionaries. (Test functions are passed as parameters to the main function.)

from itertools import (starmap)

def f90(dct):
return lambda x2, ab, a, b: dct.get(x2, None)

def f60(dct):
return lambda x2, ab, a, b: dct.get(x2 - ab, None)

def f120(dct):
return lambda x2, ab, a, b: dct.get(x2 + ab, None)

def f60unequal(dct):
return lambda x2, ab, a, b: (
dct.get(x2 - ab, None) if a != b else None
)

# triangles :: Dict -> (Int -> Int -> Int -> Int -> Maybe Int)
# -> [String]
def triangles(f, n):
upto = enumFromTo(1)
xs = upto(n)
dctSquares = dict(zip(xs, [x**2 for x in xs]))
dctRoots = {v: k for k, v in dctSquares.items()}
fr = f(dctRoots)
dct = {}
for a in xs:
a2 = dctSquares[a]
for b in upto(a):
suma2b2 = a2 + dctSquares[b]
c = fr(suma2b2, a * b, a, b)
if (c is not None):
dct[str(sorted([a, b, c]))] = 1
return list(dct.keys())

def main():
print(
'Triangles of maximum side 13\n\n' +
unlines(
zipWith(
lambda f, n: (
lambda ks=triangles(f, 13): (
str(len(ks)) + ' solutions for ' +
str(n) + ' degrees:\n' +
unlines(ks) + '\n'
)
)()
)([f120, f90, f60])
([120, 90, 60])
) + '\n\n' +
'60 degrees - uneven triangles of maximum side 10000. Total:\n' +
str(len(triangles(f60unequal, 10000)))
)

# GENERIC --------------------------------------------------------------

# enumFromTo :: Int -> Int -> [Int]
def enumFromTo(m):
return lambda n: list(range(m, 1 + n))

# unlines :: [String] -> String
def unlines(xs):
return '\n'.join(xs)

# zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
def zipWith(f):
return lambda xs: lambda ys: (
list(starmap(f, zip(xs, ys)))
)

if __name__ == '__main__':
main()
Output:
Triangles of maximum side 13

2 solutions for 120 degrees:
[3, 5, 7]
[7, 8, 13]

3 solutions for 90 degrees:
[3, 4, 5]
[6, 8, 10]
[5, 12, 13]

15 solutions for 60 degrees:
[1, 1, 1]
[2, 2, 2]
[3, 3, 3]
[4, 4, 4]
[5, 5, 5]
[6, 6, 6]
[7, 7, 7]
[3, 7, 8]
[5, 7, 8]
[8, 8, 8]
[9, 9, 9]
[10, 10, 10]
[11, 11, 11]
[12, 12, 12]
[13, 13, 13]

60 degrees - uneven triangles of maximum side 10000. Total:
18394

## REXX

### using some optimization

Instead of coding a general purpose subroutine (or function) to solve all of the task's requirements,   it was decided to
write three very similar   do   loops (triple nested) to provide the answers for the three requirements.

Three arguments   (from the command line)   can be specified which indicates the maximum length of the triangle sides
(the default is   13,   as per the task's requirement)   for each of the three types of angles   (60º, 90º, and 120º)   for
the triangles.   If the maximum length of the triangle's number of sides is positive,   it indicates that the triangle sides are
displayed,   as well as a total number of triangles found.

If the maximum length of the triangle sides is negative,   only the   number   of triangles are displayed   (using the
absolute value of the negative number).

/*REXX pgm finds integer sided triangles that satisfy Law of cosines for 60º, 90º, 120º.*/
parse arg s1 s2 s3 . /*obtain optional arguments from the CL*/
if s1=='' | s1=="," then s1= 13 /*Not specified? Then use the default.*/
if s2=='' | s2=="," then s2= 13 /* " " " " " " */
if s3=='' | s3=="," then s3= 13 /* " " " " " " */
w= max( length(s1), length(s2), length(s3) ) /*W is used to align the side lengths.*/

if s1>0 then do; call head 120 /*────120º: a² + b² + ab ≡ c² */
do a=1 for s1; aa = a*a
do b=a+1 to s1; x= aa + b*b + a*b
do c=b+1 to s1 until c*c>x
if x==c*c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s1
end

if s2>0 then do; call head 90 /*────90º: a² + b² ≡ c² */
do a=1 for s2; aa = a*a
do b=a+1 to s2; x= aa + b*b
do c=b+1 to s2 until c*c>x
if x==c*c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s2
end

if s3>0 then do; call head 60 /*────60º: a² + b² ─ ab ≡ c² */
do a=1 for s3; aa = a*a
do b=a to s3; x= aa + b*b - a*b
do c=a to s3 until c*c>x
if x==c*c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s3
end
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
foot: say right(# ' solutions found for' angle "(sides up to" arg(1)')', 65); say; return
head: #= 0; parse arg deg; angle= ' 'deg"º "; say center(angle, 65, '═'); return
show: #= # + 1; say ' ('right(a, w)"," right(b, w)"," right(c, w)')'; return
output   when using the default number of sides for the input:     13
═════════════════════════════ 120º ══════════════════════════════
( 3,  5,  7)
( 7,  8, 13)
2  solutions found for  120º  (sides up to 13)

══════════════════════════════ 90º ══════════════════════════════
( 3,  4,  5)
( 5, 12, 13)
( 6,  8, 10)
3  solutions found for  90º  (sides up to 13)

══════════════════════════════ 60º ══════════════════════════════
( 1,  1,  1)
( 2,  2,  2)
( 3,  3,  3)
( 3,  8,  7)
( 4,  4,  4)
( 5,  5,  5)
( 5,  8,  7)
( 6,  6,  6)
( 7,  7,  7)
( 8,  8,  8)
( 9,  9,  9)
(10, 10, 10)
(11, 11, 11)
(12, 12, 12)
(13, 13, 13)
15  solutions found for  60º  (sides up to 13)

### using memoization

/*REXX pgm finds integer sided triangles that satisfy Law of cosines for 60º, 90º, 120º.*/
parse arg s1 s2 s3 s4 . /*obtain optional arguments from the CL*/
if s1=='' | s1=="," then s1= 13 /*Not specified? Then use the default.*/
if s2=='' | s2=="," then s2= 13 /* " " " " " " */
if s3=='' | s3=="," then s3= 13 /* " " " " " " */
if s4=='' | s4=="," then s4= -10000 /* " " " " " " */
parse value s1 s2 s3 s4 with os1 os2 os3 os4 . /*obtain the original values for sides.*/
s1=abs(s1); s2=abs(s2); s3=abs(s3); s4=abs(s4) /*use absolute values for the # sides. */
@.=
do j=1 for max(s1, s2, s3, s4); @.j = j*j
end /*j*/ /*build memoization array for squaring.*/

if s1>0 then do; call head 120,,os1 /*────120º: a² + b² + ab ≡ c² */
do a=1 for s1
do b=a+1 to s1; x= @.a + @.b + a*b
if x>z then iterate a
do c=b+1 to s1 until @.c>x
if [email protected].c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s1
end

if s2>0 then do; call head 90,, os2 /*────90º: a² + b² ≡ c² */
do a=1 for s2
do b=a+1 to s2; x= @.a + @.b
if x>z then iterate a
do c=b+1 to s2 until @.c>x
if [email protected].c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s2
end

if s3>0 then do; call head 60,, os3 /*────60º: a² + b² ─ ab ≡ c² */
do a=1 for s3
do b=a to s3; x= @.a + @.b - a*b
if x>z then iterate a
do c=a to s3 until @.c>x
if [email protected].c then do; call show; iterate b; end
end /*c*/
end /*b*/
end /*a*/
call foot s3
end

if s4>0 then do; call head 60, 'unique', os4 /*────60º: a² + b² ─ ab ≡ c² */
do a=1 for s4
do b=a to s4; x= @.a + @.b - a*b
if x>z then iterate a
do c=a to s4 until @.c>x
if [email protected].c then do; if a==b&a==c then iterate b
call show; iterate b
end
end /*c*/
end /*b*/
end /*a*/
call foot s4
end
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
foot: say right(# ' solutions found for' ang "(sides up to" arg(1)')', 65); say; return
head: #=0; arg d,,s;z=s*s;w=length(s); ang=' 'd"º " arg(2); say center(ang,65,'═'); return
show: #= # + 1; if s>0 then say ' ('right(a,w)"," right(b,w)"," right(c,w)')'; return
output   when using the inputs of:     0   0   0   -10000

Note that the first three computations are bypassed because of the three zero (0) numbers,   the negative ten thousand indicates to find all the triangles with sides up to 10,000,   but not list the triangles, it just reports the   number   of triangles found.

══════════════════════════ 60º  unique═══════════════════════════
18394  solutions found for  60º  unique (sides up to 10000)

## zkl

fcn tritri(N=13){
tri90, tri60, tri120 := List(),List(),List();
foreach a,b in ([1..N],[1..a]){
aa,bb := a*a,b*b;
ab,c  := a*b, aa + bb - ab; // 60*
if(sqrset.holds(c)){ tri60.append(abc(a,b,c)); continue; }

c=aa + bb; // 90*
if(sqrset.holds(c)){ tri90.append(abc(a,b,c)); continue; }

c=aa + bb + ab; // 120*
if(sqrset.holds(c)) tri120.append(abc(a,b,c));
}
List(tri60,tri90,tri120)
}
fcn abc(a,b,c){ List(a,b).sort().append(c.toFloat().sqrt().toInt()) }
fcn triToStr(tri){ // ((c,d,e),(a,b,c))-->"(a,b,c),(c,d,e)"
tri.sort(fcn(t1,t2){ t1[0]<t2[0] })
.apply("concat",",").apply("(%s)".fmt).concat(",")
}
N:=13;
println("Integer triangular triples for sides 1..%d:".fmt(N));
foreach angle, triples in (T(60,90,120).zip(tritri(N))){
println(" %3d\U00B0; has %d solutions:\n  %s"
.fmt(angle,triples.len(),triToStr(triples)));
}
Output:
Integer triangular triples for sides 1..13:
60° has 15 solutions:
(1,1,1),(2,2,2),(3,8,7),(3,3,3),(4,4,4),(5,8,7),(5,5,5),(6,6,6),(7,7,7),(8,8,8),(9,9,9),(10,10,10),(11,11,11),(12,12,12),(13,13,13)
90° has 3 solutions:
(3,4,5),(5,12,13),(6,8,10)
120° has 2 solutions:
(3,5,7),(7,8,13)

Extra credit:

fcn tri60(N){	// special case 60*