3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ "pass through 3 points": so they don't have to be the triangle's vertices? $\endgroup$ – msh210 Jul 1 at 11:27
  • $\begingroup$ no the points are the vertices of the triangle $\endgroup$ – Dmitry Kamenetsky Jul 1 at 11:42
6
$\begingroup$

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

| improve this answer | |
$\endgroup$
  • 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 at 3:35
  • 1
    $\begingroup$ If you restrict to a triangular grid with 10 points per side, 29 is optimal. $\endgroup$ – RobPratt Jul 2 at 18:29
  • $\begingroup$ @RobPratt how did you prove this? $\endgroup$ – Dmitry Kamenetsky Jul 3 at 0:56
  • 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 at 1:26
  • $\begingroup$ What is the $\binom{n+1}{2}$? @RobPratt $\endgroup$ – justhalf Jul 3 at 1:37
4
$\begingroup$

My attempt:

enter image description here

There is

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

| improve this answer | |
$\endgroup$
  • $\begingroup$ I think there are more triangles. Have you counted all the sqrt(3) triangles? See the other answer. $\endgroup$ – Dmitry Kamenetsky Jul 1 at 11:51
  • 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 at 12:42
4
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.