3
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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
  • $\begingroup$ no the points are the vertices of the triangle $\endgroup$ Jul 1 '20 at 11:42
6
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I have arranged for a total of...

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

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7
  • 1
    $\begingroup$ Oooh very nice! This could be our winner. Let's wait and see if anyone can beat it. $\endgroup$ Jul 2 '20 at 3:35
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    $\begingroup$ If you restrict to a triangular grid with 10 points per side, 29 is optimal. $\endgroup$
    – RobPratt
    Jul 2 '20 at 18:29
  • $\begingroup$ @RobPratt how did you prove this? $\endgroup$ Jul 3 '20 at 0:56
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    $\begingroup$ @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
  • $\begingroup$ What is the $\binom{n+1}{2}$? @RobPratt $\endgroup$
    – justhalf
    Jul 3 '20 at 1:37
4
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My attempt:

enter image description here

There is

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

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  • $\begingroup$ I think there are more triangles. Have you counted all the sqrt(3) triangles? See the other answer. $\endgroup$ Jul 1 '20 at 11:51
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    $\begingroup$ @DmitryKamenetsky That's the 6 in the list (3 in each orientation). The final 2 is for the sqrt(7) triangles. $\endgroup$ Jul 1 '20 at 12:42
4
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Here's an attempt for

28

equilateral triangles:

enter image description here

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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3
  • $\begingroup$ This was the solution I had in mind. $\endgroup$ Jul 1 '20 at 11:41
  • $\begingroup$ I believe there are two more sqrt(3) triangles - they have the central point in their centre. $\endgroup$ Jul 1 '20 at 11:51
  • $\begingroup$ thanks ... #icantcount $\endgroup$
    – Glorfindel
    Jul 1 '20 at 11:52

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