Add solution for Project Euler problem 91. (TheAlgorithms#3144)

* Added solution for Project Euler problem 91. Reference: TheAlgorithms#2695

* Added doctest for solution() in project_euler/problem_91/sol1.py

* Update docstring and 0-padding in directory name. Reference: TheAlgorithms#3256

* Update sol1.py

Co-authored-by: Dhruv <dhruvmanila@gmail.com>

project_euler/problem_091/sol1.py

@@ -0,0 +1,59 @@
+"""
+Project Euler Problem 91: https://projecteuler.net/problem=91
+
+The points P (x1, y1) and Q (x2, y2) are plotted at integer coordinates and
+are joined to the origin, O(0,0), to form ΔOPQ.
+
+There are exactly fourteen triangles containing a right angle that can be formed
+when each coordinate lies between 0 and 2 inclusive; that is,
+0 ≤ x1, y1, x2, y2 ≤ 2.
+
+Given that 0 ≤ x1, y1, x2, y2 ≤ 50, how many right triangles can be formed?
+"""
+
+
+from itertools import combinations, product
+
+
+def is_right(x1: int, y1: int, x2: int, y2: int) -> bool:
+    """
+    Check if the triangle described by P(x1,y1), Q(x2,y2) and O(0,0) is right-angled.
+    Note: this doesn't check if P and Q are equal, but that's handled by the use of
+    itertools.combinations in the solution function.
+
+    >>> is_right(0, 1, 2, 0)
+    True
+    >>> is_right(1, 0, 2, 2)
+    False
+    """
+    if x1 == y1 == 0 or x2 == y2 == 0:
+        return False
+    a_square = x1 * x1 + y1 * y1
+    b_square = x2 * x2 + y2 * y2
+    c_square = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)
+    return (
+        a_square + b_square == c_square
+        or a_square + c_square == b_square
+        or b_square + c_square == a_square
+    )
+
+
+def solution(limit: int = 50) -> int:
+    """
+    Return the number of right triangles OPQ that can be formed by two points P, Q
+    which have both x- and y- coordinates between 0 and limit inclusive.
+
+    >>> solution(2)
+    14
+    >>> solution(10)
+    448
+    """
+    return sum(
+        1
+        for pt1, pt2 in combinations(product(range(limit + 1), repeat=2), 2)
+        if is_right(*pt1, *pt2)
+    )
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")