What is the most number of equilateral triangles you can form by drawing 13 points on a piece of paper? Each triangle must have 3 equal sides and pass through 3 points. Only equilateral triangles can be counted, while other triangles must be ignored. Triangles can be of different size.
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$\begingroup$ "pass through 3 points": so they don't have to be the triangle's vertices? $\endgroup$ – msh210 Jul 1 '20 at 11:27
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$\begingroup$ no the points are the vertices of the triangle $\endgroup$ – Dmitry Kamenetsky Jul 1 '20 at 11:42
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1$\begingroup$ Oooh very nice! This could be our winner. Let's wait and see if anyone can beat it. $\endgroup$ – Dmitry Kamenetsky Jul 2 '20 at 3:35
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1$\begingroup$ If you restrict to a triangular grid with 10 points per side, 29 is optimal. $\endgroup$ – RobPratt Jul 2 '20 at 18:29
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2$\begingroup$ @DmitryKamenetsky, I used integer linear programming with a binary decision variable $x_i$ for each of the $\binom{n+1}{2}$ nodes and a binary decision variable $y_t$ for each equilateral triangle. The problem is to maximize $\sum_t y_t$ subject to $\sum_i x_i = 13$ and $y_t \le x_i$ if node $i$ is a vertex of triangle $t$. $\endgroup$ – RobPratt Jul 3 '20 at 1:26
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$\begingroup$ I think there are more triangles. Have you counted all the sqrt(3) triangles? See the other answer. $\endgroup$ – Dmitry Kamenetsky Jul 1 '20 at 11:51
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1$\begingroup$ @DmitryKamenetsky That's the 6 in the list (3 in each orientation). The final 2 is for the sqrt(7) triangles. $\endgroup$ – Jaap Scherphuis Jul 1 '20 at 12:42
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Here's an attempt for
28
equilateral triangles:
12 of size 1
8 of size $\sqrt3$, 6 using the central point and two of the outermost points, and 2 with the central point in the middle (thanks @hexomino and @DmitryKamenetsky)
6 of size 2
2 of size 3
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$\begingroup$ This was the solution I had in mind. $\endgroup$ – Dmitry Kamenetsky Jul 1 '20 at 11:41
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$\begingroup$ I believe there are two more sqrt(3) triangles - they have the central point in their centre. $\endgroup$ – Dmitry Kamenetsky Jul 1 '20 at 11:51
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