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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
 #1
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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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