Stacking Pythagorean Triangles Background
A Pythagorean triangle is a right triangle where each side length is an integer (that is, the side lengths form a Pythagorean triple):

Using the sides of this triangle, we can attach two more non-congruent Pythagorean triangles as follows:

We can continue with this pattern as we see fit, so long as no two triangles overlap, and connecting sides have equal length:

The question is, how many non-congruent Pythagorean triangles can we fit in a given space?
The Input
You will receive two integers as input, W and H, through function arguments, STDIN, strings, or whatever you like. The integers may be received as decimal, hexadecimal, binary, unary (good luck, Retina), or any other integer base. You may assume that max(W, H) &lt;= 2^15 - 1.
The Output
Your program or function should calculate a list of non-overlapping connected non-congruent Pythagorean triangles and output a list of sets of three coordinates each, where the coordinates in a set form one of the Pythagorean triangles when connected by lines. Coordinates must be real numbers in our space (x must be in the interval [0, W] and y must be in the interval [0, H]), and distance should be accurate to machine precision. The order of the triangles and the exact format of each coordinate is not important.
It must be possible to &quot;walk&quot; from one triangle to any other only stepping over connected boundaries.
Using the above diagram as an example, let our input be W = 60, H = 60.
Our output could then be the following list of coordinates:
(0, 15), (0, 21), (8, 15)
(0, 21), (14.4, 40.2), (8, 15)
(0, 15), (8, 0), (8, 15)
(8, 0), (8, 15), (28, 15)
(8, 15), (28, 15), (28, 36)
(28, 15), (28, 36), (56, 36)

Now, 6 triangles is most certainly not the best we can do given our space. Can you do better?
Rules and Scoring

Your score for this challenge is the number of triangles your program generates given the input of W = 1000, H = 1000. I reserve the right to change this input if I suspect someone hardcoding this case.

You may not use builtins that calculate Pythagorean triples, and you may not hardcode a list of more than 2 Pythagorean triples (if you hardcode your program to always start with a (3, 4, 5) triangle, or some similar starting circumstance, that is okay).

You may write your submission in any language. Readability and commenting are encouraged.

You may find a list of Pythagorean triples here.

Standard Loopholes are disallowed.
Python 3, 109

This was certainly a deceptively hard challenge. There were many times writing the code that I found myself questioning my basic geometry abilities. That being said, I'm pretty happy with the result. I put no effort in to coming up with a complex algorithm for placing the triangles, and instead my code just blunders through always placing the smallest it can find. I hope I can improve this later on, or my answer will spurn others to find a better algorithm! All in all, a very fun problem, and it produces some interesting pictures.

Here's the code:

import time
import math

W = int(input("Enter W: "))
H = int(input("Enter H: "))

middle_x = math.floor(W/2)
middle_y = math.floor(H/2)

sides = [ # each side is in the format [length, [x0, y0], [x1, y1]]
    [3,[middle_x,middle_y],[middle_x+3,middle_y]],
    [4,[middle_x,middle_y],[middle_x,middle_y+4]],
    [5,[middle_x+3,middle_y],[middle_x,middle_y+4]]
    ]

triangles = [[0,1,2]] # each triangle is in the format [a, b, c] where a, b and c are the indexes of sides

used_triangles = [[3,4,5]] # a list of used Pythagorean triples, where lengths are ordered (a &lt; b &lt; c)

max_bounds_length = math.sqrt(W**2 + H**2)

def check_if_pyth_triple(a,b): # accepts two lists of the form [l, [x0,y0], [x1,y1]] defining two line segments
    # returns 0 if there are no triples, 1 if there is a triple with a right angle on a,
    # and 2 if there is a triple with the right angle opposite a
    c = math.sqrt(a[0]**2 + b[0]**2)
    if c.is_integer():
        if not sorted([a[0], b[0], c]) in used_triangles:
            return 1
        return 0
    else:
        if a[0] &gt; b[0]:
            c = math.sqrt(a[0]**2 - b[0]**2)
            if c.is_integer() and not sorted([a[0], b[0], c]) in used_triangles:
                return 2
        return 0

def check_if_out_of_bounds(p):
    out = False
    if p[0] &lt; 0 or p[0] &gt; W:
        out = True
    if p[1] &lt; 0 or p[1] &gt; H:
        out = True
    return out

def in_between(a,b,c):
    maxi = max(a,c)
    mini = min(a,c)
    return mini &lt; b &lt; maxi

def sides_intersect(AB,CD): # accepts two lists of the form [l, [x0,y0], [x1,y1]] defining two line segments
    # doesn't count overlapping lines
    A = AB[1]
    B = AB[2]
    C = CD[1]
    D = CD[2]

    if A[0] == B[0]: # AB is vertical
        if C[0] == D[0]: # CD is vertical
            return False
        else:
            m1 = (C[1] - D[1])/(C[0] - D[0]) # slope of CD
            y = m1*(A[0] - C[0]) + C[1] # the y value of CD at AB's x value
            return in_between(A[1], y, B[1]) and in_between(C[0], A[0], D[0])
    else:
        m0 = (A[1] - B[1])/(A[0] - B[0]) # slope of AB
        if C[0] == D[0]: # CD is vertical
            y = m0*(C[0] - A[0]) + A[1] # the y value of CD at AB's x value
            return in_between(C[1], y, D[1]) and in_between(A[0],C[0],B[0])
        else:
            m1 = (C[1] - D[1])/(C[0] - D[0]) # slope of CD
            if m0 == m1:
                return False
            else:
                x = (m0*A[0] - m1*C[0] - A[1] + C[1])/(m0 - m1)
                return in_between(A[0], x, B[0]) and in_between(C[0], x, D[0])

def check_all_sides(b,triangle):
    no_intersections = True
    for side in sides:
        if sides_intersect(side, b):
            no_intersections = False
            break

    return no_intersections

def check_point_still_has_room(A): # This function is needed for the weird case when all 2pi degrees
    # around a point are filled by triangles, but you could fit in a small triangle into another one
    # already built around the point. Doing this won't cause sides_intersect() to detect it because
    # the sides will all be parallel. Crazy stuff.
    connecting_sides = []
    for side in sides:
        if A in side:
            connecting_sides.append(side)

    match_count = 0
    slopes = []
    for side in connecting_sides:
        B = side[1]
        if A == B:
            B = side[2]
        if not A[0] == B[0]:
            slope = round((A[1]-B[1])/(A[0]-B[0]),4)
        else:
            if A[1] &lt; B[1]:
                slope = "infinity"
            else:
                slope = "neg_infinity"
        if slope in slopes:
            match_count -= 1
        else:
            slopes.append(slope)
            match_count += 1

    return match_count != 0

def construct_b(a,b,pyth_triple_info,straight_b_direction,bent_b_direction):
    # this function finds the correct third point of the triangle given a and the length of b
    # pyth_triple_info determines if a is a leg or the hypotenuse
    # the b_directions determine on which side of a the triangle should be formed
    a_p = 2 # this is the index of the point in a that is not the shared point with b
    if a[1] != b[1]:
        a_p = 1

    vx = a[a_p][0] - b[1][0] # v is our vector, and these are the coordinates, adjusted so that
    vy = a[a_p][1] - b[1][1] # the shared point is the origin

    if pyth_triple_info == 1:
        # because the dot product of orthogonal vectors is zero, we can use that and the Pythagorean formula
        # to get this simple formula for generating the coordinates of b's second point
        if vy == 0:
            x = 0
            y = b[0]
        else:
            x = b[0]/math.sqrt(1+((-vx/vy)**2)) # b[0] is the desired length
            y = -vx*x/vy

        x = x*straight_b_direction # since the vector is orthagonal, if we want to reverse the direction,
        y = y*straight_b_direction # it just means finding the mirror point

    elif pyth_triple_info == 2: # this finds the intersection of the two circles of radii b[0] and c 
        # around a's endpoints, which is the third point of the triangle if a is the hypotenuse
        c = math.sqrt(a[0]**2 - b[0]**2)
        D = a[0]
        A = (b[0]**2 - c**2 + D**2 ) / (2*D)
        h = math.sqrt(b[0]**2 - A**2)
        x2 = vx*(A/D)
        y2 = vy*(A/D)        
        x = x2 + h*vy/D
        y = y2 - h*vx/D

        if bent_b_direction == -1: # this constitutes reflection of the vector (-x,-y) around the normal vector n,
            # which accounts for finding the triangle on the opposite side of a
            dx = -x
            dy = -y
            v_length = math.sqrt(vx**2 + vy**2)
            nx = vx/v_length
            ny = vy/v_length

            d_dot_n = dx*nx + dy*ny

            x = dx - 2*d_dot_n*nx
            y = dy - 2*d_dot_n*ny

    x = x + b[1][0] # adjust back to the original frame
    y = y + b[1][1]

    return [x,y]

def construct_triangle(side_index):
    a = sides[side_index] # a is the base of the triangle
    a_p = 1
    b = [1, a[a_p], []] # side b, c is hypotenuse

    for index, triangle in enumerate(triangles):
        if side_index in triangle:
            triangle_index = index
            break

    triangle = list(triangles[triangle_index])
    triangle.remove(side_index)

    add_tri = False

    straight_b = construct_b(a,b,1,1,1)

    bent_b = construct_b(a,b,2,1,1)

    A = sides[triangle[0]][1]
    if A in a:
        A = sides[triangle[0]][2]

    Ax = A[0] - b[1][0] # adjusting A so that it's a vector
    Ay = A[1] - b[1][1]

    # these are for determining if construct_b() is going to the correct side
    triangle_on_side = (a[2][0]-a[1][0])*(A[1]-a[1][1]) - (a[2][1]-a[1][1])*(A[0]-a[1][0])
    straight_b_on_side = (a[2][0]-a[1][0])*(straight_b[1]-a[1][1]) - (a[2][1]-a[1][1])*(straight_b[0]-a[1][0])
    bent_b_on_side = (a[2][0]-a[1][0])*(bent_b[1]-a[1][1]) - (a[2][1]-a[1][1])*(bent_b[0]-a[1][0])

    straight_b_direction = 1
    if (triangle_on_side &gt; 0 and straight_b_on_side &gt; 0) or (triangle_on_side &lt; 0 and straight_b_on_side &lt; 0):
        straight_b_direction = -1

    bent_b_direction = 1
    if (triangle_on_side &gt; 0 and bent_b_on_side &gt; 0) or (triangle_on_side &lt; 0 and bent_b_on_side &lt; 0):
        bent_b_direction = -1


    a_ps = []
    for x in [1,2]:
        if check_point_still_has_room(a[x]): # here we check for that weird exception
            a_ps.append(x)

    while True:
        out_of_bounds = False
        if b[0] &gt; max_bounds_length:
            break

        pyth_triple_info = check_if_pyth_triple(a,b)

        for a_p in a_ps:
            if a_p == 1: # this accounts for the change in direction when switching a's points
                new_bent_b_direction = bent_b_direction
            else:
                new_bent_b_direction = -bent_b_direction

            b[1] = a[a_p]
            if pyth_triple_info &gt; 0:
                b[2] = construct_b(a,b,pyth_triple_info,straight_b_direction,new_bent_b_direction)

                if check_if_out_of_bounds(b[2]): # here is the check to make sure we don't go out of bounds
                    out_of_bounds = True
                    break

                if check_all_sides(b,triangle):
                    if pyth_triple_info == 1:
                        c = [math.sqrt(a[0]**2 + b[0]**2), a[3-a_p], b[2]]
                    else:
                        c = [math.sqrt(a[0]**2 - b[0]**2), a[3-a_p], b[2]]

                    if check_all_sides(c,triangle):
                        add_tri = True
                        break

        if out_of_bounds or add_tri:
            break

        b[0] += 1 # increment the length of b every time the loop goes through

    if add_tri: # this adds a new triangle
        sides.append(b)
        sides.append(c)
        sides_len = len(sides)
        triangles.append([side_index, sides_len - 2, sides_len - 1])
        used_triangles.append(sorted([a[0], b[0], c[0]])) # so we don't use the same triangle again

def build_all_triangles(): # this iterates through every side to see if a new triangle can be constructed
    # this is probably where real optimization would take place so more optimal triangles are placed first
    t0 = time.clock()

    index = 0
    while index &lt; len(sides):
        construct_triangle(index)
        index += 1

    t1 = time.clock()

    triangles_points = [] # this is all for printing points
    for triangle in triangles:
        point_list = []
        for x in [1,2]:
            for side_index in triangle:
                point = sides[side_index][x]
                if not point in point_list:
                    point_list.append(point)
        triangles_points.append(point_list)

    for triangle in triangles_points:
        print(triangle)

    print(len(triangles), "triangles placed in", round(t1-t0,3), "seconds.")

def matplotlib_graph(): # this displays the triangles with matplotlib
    import pylab as pl
    import matplotlib.pyplot as plt
    from matplotlib import collections as mc

    lines = []
    for side in sides:
        lines.append([side[1],side[2]])

    lc = mc.LineCollection(lines)
    fig, ax = pl.subplots()
    ax.add_collection(lc)
    ax.autoscale()
    ax.margins(0.1)
    plt.show()

build_all_triangles()


Here's a graph of the output for W = 1000 and H = 1000 with 109 triangles:


Here is W = 10000 and H = 10000 with 724 triangles:


Call the matplotlib_graph() function after build_all_triangles() to graph the triangles.

I think the code runs reasonably fast: at W = 1000 and H = 1000 it takes 0.66 seconds, and at W = 10000 and H = 10000 it takes 45 seconds using my crappy laptop.