Most number of equilateral triangles formed by 13 points

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.

• "pass through 3 points": so they don't have to be the triangle's vertices? Jul 1, 2020 at 11:27
• no the points are the vertices of the triangle Jul 1, 2020 at 11:42

I have arranged for a total of...

$$14+7+5+3=29$$ triangles
(edge lengths $$1,\sqrt{3},2,\sqrt{7}$$)

• Oooh very nice! This could be our winner. Let's wait and see if anyone can beat it. Jul 2, 2020 at 3:35
• If you restrict to a triangular grid with 10 points per side, 29 is optimal. Jul 2, 2020 at 18:29
• @RobPratt how did you prove this? Jul 3, 2020 at 0:56
• @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 $t$. Jul 3, 2020 at 1:26
• What is the $\binom{n+1}{2}$? @RobPratt Jul 3, 2020 at 1:37

My attempt:

There is

$$28$$ Equilateral Triangles. (Can you spot all of them?)

• I think there are more triangles. Have you counted all the sqrt(3) triangles? See the other answer. Jul 1, 2020 at 11:51
• @DmitryKamenetsky That's the 6 in the list (3 in each orientation). The final 2 is for the sqrt(7) triangles. Jul 1, 2020 at 12:42

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

• This was the solution I had in mind. Jul 1, 2020 at 11:41
• I believe there are two more sqrt(3) triangles - they have the central point in their centre. Jul 1, 2020 at 11:51