12

As long as the triangles are allowed to overlap, I think I can just Zorro through like this: Counting: Triangles with all 3 sides made of grid lines: 0 Triangles with 2 sides made of grid lines: 12 (4 per each drawn line) Triangles with 1 side made of grid lines: 8 (4 for each intersecting red line pair) Triangles with 0 sides made of grid lines: 0 As for ...


8

Unless I'm missing something obvious, the answer to the original question is: And the answer to the bonus question is:


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