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Question: Twenty lattice points are arranged along the edges of a $4 \times3$ rectangles as shown. How many triangles have all three of their vertices among these points?

enter image description here

I started off with getting the total number of points possible, which is simply $\binom {20}3$. To get the number of triangles formed, we can first consider the complementary case when the lines are collinear (i.e no triangles formed).

However, I'm having trouble evaluating how many collinear lines can be formed in a $4 \times 3$. I'm not sure how you would consider the tilted lines.

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Crescendo
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  • I would just enumerate all the lines. There can't be that many. There are the seven horizontal/vertical lines. Then there are four lines with slope $\pm1$. Finally, there are two lines of slope $\pm 2$ (or a half, depending on the orientation. – Theo Bendit Apr 15 '18 at 15:07
  • @G.Sassatelli I've added the picture – Crescendo Apr 15 '18 at 15:26

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Yeah, I would do it the way you proposed.

Horiz: $4\binom{5}{3}$

Vert: $5\binom{4}{3}$

slope=$\pm 1: 2\big(2\binom{3}{3}+2\binom{4}{3}\big)$

slope=$\pm2,\pm\cfrac{1}{2}: 4\times 2$

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