Euler Problem 39

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If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.

$\{20,48,52\}, \{24,45,51\}, \{30,40,50\}$

For which value of p ≤ 1000, is the number of solutions maximised?

We are looking for right angle triangles so Pythagoras' theorem $a^2 + b^2 = c^2$ can be used. Also it must be true that $a <= b <= c$ is true for every solution. Let's start with a function that tests the given examples.

In [13]:
def is_right_angle_triangle(a, b, c, p):
    if a + b + c != p:
        return False
    if a**2 + b**2 != c**2:
        return False
    return True

given = [(20, 48, 52, 120), (24, 45, 51, 120), (30, 40, 50, 120)]
for g in given:
    assert(is_right_angle_triangle(*g))
assert(is_right_angle_triangle(29, 41, 50, 120) == False)

This seems bruteforceable. Let's try to find the solutions for p = 120.

In [16]:
def brute_force(p):
    solutions = [(a,b, p - a - b)
        for b in range(1, p // 2 + 1)
        for a in range(1, b)
        if a*a + b*b == (p - a - b) * (p - a - b)
    ]
    return solutions

print(brute_force(120))

[(30, 40, 50), (24, 45, 51), (20, 48, 52)]

Looks good. Let's try for all $p <= 1000$.

In [18]:
solutions = [(len(brute_force(p)), p) for p in range(1, 1001)]

Well, not too fast, but also not too slow. Let's get the max and we have our solution.

In [24]:
s, p = max(solutions)
print(p)
assert(p == 840)

840