Contest Math Possible Triangles In the xy-plane, how many triangles have each of their vertices at points (a,b) where a,b are integers satisfying 1 ≤ a ≤ 5 and 1≤b≤5?
I got twenty-five, but something tells me this isn't right. I set up an inequality, square root of a squared plus b squared is less than a plus b but more than the absolute value of b minus a. Squaring, we can get rid of a squared plus b squared, leaving -2ab<0<2ab, which is always true. So, each pairing works, and we have five times five for twenty-five. 
I feel that the answer is way too simple. If it is indeed wrong, where did I go wrong?

Focus on counting the ways to choose three collinear points.  If the line is horizontal or vertical, then there are $5\times{{5}\choose{3}}=50$ possibilities, for a total of $100$ ways.  If the slope of the line is $\pm 1$, then there are $9$ ways to place a segment containing three points, $4$ ways to place a segment containing four points (each with $2$ ways to choose the intermediate point), and one way to place a segment containing five points (with $3$ ways to choose the intermediate point), for a total of $40$ ways to choose the points with a slope of $\pm1$.  If the slope is $\pm 2$ or $\pm 1/2$, then there are just $3$ ways to place a segment containing three points, for a total of $12$ with these slopes.  In total there are $152$ ways to choose three collinear points.
Subtracting this from the total number of ways of choosing three points gives
$$
{{25}\choose{3}}-152=2300 - 152=2148
$$
distinct triangles.
