To solve this problem, we need to determine if the origin (0,0) lies inside a given triangle formed by three distinct points in the 2D plane. The solution leverages the cross product to check the orientation of the origin relative to each edge of the triangle.

Approach

The core idea is to use the cross product to determine the orientation of the origin with respect to each edge of the triangle. For the origin to be inside the triangle, it must lie on the same side (either all clockwise or all counter-clockwise) of each edge as the opposite vertex. Here's the step-by-step approach:

1. Cross Product Calculation: For three points (A(x1,y1)), (B(x2,y2)), (C(x3,y3)):

   • Compute the cross product of vectors (OA) and (OB): (cross1 = x1y2 - y1x2)
   • Compute the cross product of vectors (OB) and (OC): (cross2 = x2y3 - y2x3)
   • Compute the cross product of vectors (OC) and (OA): (cross3 = x3y1 - y3x1)

2. Check Orientation: If all cross products are positive or all are negative, the origin is inside the triangle. This indicates the origin is on the same side of each edge as the interior of the triangle.

Solution Code

count = 0

# Read input from the file (adjust the path as needed)
with open('triangles.txt', 'r') as file:
    for line in file:
        # Split the line into coordinates (comma-separated)
        coords = list(map(int, line.strip().split(',')))
        x1, y1, x2, y2, x3, y3 = coords

        # Calculate cross products
        cross1 = x1 * y2 - y1 * x2
        cross2 = x2 * y3 - y2 * x3
        cross3 = x3 * y1 - y3 * x1

        # Check if all cross products have the same sign
        if (cross1 > 0 and cross2 > 0 and cross3 > 0) or (cross1 < 0 and cross2 < 0 and cross3 < 0):
            count += 1

print(count)

Explanation

• Cross Product: The cross product of two vectors gives the signed area of the parallelogram they form. The sign indicates the orientation (clockwise or counter-clockwise).
• Same Orientation: If all cross products are positive, the origin is on the counter-clockwise side of all edges. If all are negative, it's on the clockwise side. Both cases mean the origin is inside the triangle.
• Efficiency: This approach runs in (O(n)) time where (n) is the number of triangles, making it suitable for large inputs (like the 1000 triangles in Project Euler 102).

This method is efficient and straightforward, leveraging basic vector operations to solve the problem. The solution correctly counts the number of triangles containing the origin.

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