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The problem is the following: enter image description here Please include your steps. Thanks!

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  • $\begingroup$ Is this a HW/Exam problem? $\endgroup$ – user99680 Jun 18 '14 at 19:29
  • $\begingroup$ No. (10 characters needed) $\endgroup$ – Pikapikachu Jun 18 '14 at 19:49
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Pick any two points in the triangular lattice. Then, there are two equilateral triangles in the plane with those two points as vertices. This gives a total of $2 \cdot \dbinom{15}{2} = 210$ triangles.

But this overcounts equilateral triangles with all three vertices in the triangular lattice. There are $1+3+6+10+15+8 = 43$ equilateral triangles with all three vertices in the triangular lattice (do you see why?). Each triangle got counted $3$ times, but should have only been counted once.

Therefore, the answer is $210 - 2\cdot 43 = 124$.

EDIT: This solution was incorrect, but I think I've fixed it. Originally, I was only counting those with sides parallel to the main directions of the lattice. I believe there are $8$ more that I missed ($6$ with side length $\sqrt{3}$, and $2$ with side length $\sqrt{7}$). Someone please check this.

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  • $\begingroup$ YAY, that's the answer I got. Thanks so much. $\endgroup$ – Pikapikachu Jun 18 '14 at 19:50

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