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There are $n$ triangles of positive area that have one vertex at $(0,0)$ and their other two vertices with coordinates in $\{0,1,2,3,4\}$. Find the value of $n$.

I know that other than $(0,0)$ the vertices are $$(0,1),(0,2),(0,3),(0,4),(1,0),(1,1),(1,2),(1,3),(1,4),(2,0),(2,1),(2,2),(2,3),(2,4),(3,0),(3,1),(3,2),(3,3),(3,4),(4,0),(4,1),(4,2),(4,3),(4,4)$$ but how do I select the two points which satisfy the given condition.

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closed as off-topic by Saad, Isaac Browne, Strants, Xander Henderson, José Carlos Santos Jul 19 '18 at 22:54

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  • $\begingroup$ Without restricting to say nonoverlapping triangles it seems no limit on $n.$ Also if they're drawn "at random" how can that determine $n$? $\endgroup$ – coffeemath Jul 18 '18 at 9:09
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The necessary and sufficient condition for the triangle to have positive area is to have the other two (non-origin) vertices not lie on the same line through the origin. That is, those vertices cannot be a subset of any of the following sets: $$\{(0,1),(0,2),(0,3),(0,4)\}\qquad(x=0)$$ $$\{(1,0),(2,0),(3,0),(4,0)\}\qquad(y=0)$$ $$\{(1,1),(2,2),(3,3),(4,4)\}\qquad(x=y)$$ $$\{(1,2),(2,4)\}\qquad(2x=y)$$ $$\{(2,1),(4,2)\}\qquad(2y=x)$$ There are $\binom{24}2=276$ possible lattice triangles, and those that have zero area number $3\binom42+2\binom22=20$. Therefore $n=276-20=256$.

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