Three different points from the 16 points of this 4 × 4 grid are to be chosen as vertices of a
triangle. How many different triangles can be drawn?



I already found all the right triangles (200) by finding all the rectangles in the grid. But since there's also the obtuse and acute triangles that I forgot when solving this. So is there any easy way to find the other triangles?


I'm only asking about the obtuse and acute triangles, since I already found the right triangles. (I got the question wrong because I didn't find the other triangles.)

 Nov 1, 2022
edited by Chandalier  Nov 1, 2022
edited by Chandalier  Nov 1, 2022

Yes there is an easier way. Assuming that triangles are still distinct even through rotations/reflections, etc, then the total number of points you could choose is 16 choose 3. However that would get some straight lined triangles, so we have to eliminate those. There are 16 + 16 + 4 + 4 = 40 of these cases counting three consecutive, 2 consecutive and 1 two apart, then considering the diagonal cases. Thus there are 16 choose 3 - 40 = 520 total triangles that are non-degenerate. (assuming that rotating the grid is not going to eliminate cases)

 Nov 1, 2022

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