Solution
Triangles on a Four-by-Four Grid
Show the problem again
Consider the 16 points $(x, y)$ with $x, y \in \{0, 1, 2, 3\}$. How many triangles with positive area have all three vertices among these points? Give an integer.
Worked solution
The answer is 516. There are $\binom{16}{3} = 560$ triples; subtract the collinear ones. The 4 rows and 4 columns each contain 4 points: $8 \times \binom{4}{3} = 32$. The two main diagonals contain 4 points each: $2 \times 4 = 8$. The four diagonals of length 3 (parallel to the main ones, just off them) contribute $4 \times 1 = 4$. Lines with other slopes (such as 1/2 or 2) meet the grid in at most 2 points, so they contribute nothing. Collinear triples: $32 + 8 + 4 = 44$, leaving $560 - 44 = 516$ triangles.
Source: Classic exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.