2017 AMC 10A Problem 23

We can use complementary counting to find the total number of triangles and subtract out the ones that don’t work. There are a total of $5^2 = 25$ points, so there are $\binom{25}{3} = 2300$ possible triangles. Note that the only way a triangle doesn’t work is if all the $3$ points are in a straight line. There are $5$ rows, $5$ columns, and $2$ long diagonals. Each of these $12$ lines have $5$ points, which means they contribute $$12 \cdot \binom{5}{3} = 12 \cdot 10 = 120$$ degenerate triangles. There are also the diagonal lines with $4$ points, such as the line from $(1,2)$ to $(4,5).$ There are $4$ of these lines, so they have $$4 \cdot \binom{4}{3} = 4 \cdot 4 = 16$$ degenerate triangles. Similarly, there are $4$ diagonal lines with $3$ points. These give us $4 \cdot 1 = 4$ extra triangles that don’t work. Now, we have to look at the lines with slopes of $\dfrac{1}{2}, 2, -\dfrac{1}{2},$ and $-2.$ There are $3$ such lines for each slope, and they all have $3$ points on them. Therefore, they contribute $$4 \cdot 3 \cdot 1 = 12$$ more triangles to discount. The total number of working triangles is then $$2300 - 120 - 16 - 4 - 12$$$$= 2148.$$ Thus, B is the correct answer.