Total number of equilateral triangles

[Perhaps check it again]

*             *              *             *    

  

       *              *             *             *

 

               *             *             *            *

 

                      *              *

 

              *              *

 

 

 

The 16 points above lie in a plane on an equilateral triangular lattice.

 

Certain sets of three points of the figure correspond to the vertices of equilateral triangles.

 

Suppose you were to form all of the equilateral triangles possible, such that, for any given

 

equilateral triangle, it must have its three vertices coincide with three of the 16 points.

 

 

How many total equilateral triangles can be formed this way in the figure?

 

 

[bonanova]

25.

17 with unit sides; 7 with 2; 1 with 3.

[Perhaps check it again]

bonanova and other users,

 

there are additional equilateral triangles with side lengths different

from the correct (so far) three types you gave, and they have different orientations

from the ones you listed.

 

 

This problem is still open.

Edited August 16, 2014 by Perhaps check it again

[bonanova]

41

The largest triangle that fits has sides of length 13^.5.
Six others types in decreasing order have sides of length 2x3^.5, 3, 7^.5, 2, 3^.5  and 1.
Of each there seem to be 1, 1, 2, 5, 7, 8 and 17, respectively.

Did I miss any?

[bonanova]

Wondering whether a solution has been found.

[witzar]

I get 41.

[witzar]

I get 41.

But I cheeted:

public class TriangleCount {

    final int[][] p = {
        {0, 0},   {2, 0},   {4, 0},   {6, 0},
             {1, 1},   {3, 1},   {5, 1},   {7, 1},
                  {2, 2},   {4, 2},   {6, 2},   {8, 2},
                       {3, 3},   {5, 3},
                  {2, 4},   {4, 4},
    };
    
    void count() {
        int count = 0;
        for (int i = 0; i < p.length; i++)
        for (int j = i + 1; j < p.length; j++)
        for (int k = j + 1; k < p.length; k++) {
            int a = (p[j][0] - p[i][0])*(p[j][0] - p[i][0]) + 3*(p[j][1] - p[i][1])*(p[j][1] - p[i][1]);
            int b = (p[k][0] - p[j][0])*(p[k][0] - p[j][0]) + 3*(p[k][1] - p[j][1])*(p[k][1] - p[j][1]);
            int c = (p[i][0] - p[k][0])*(p[i][0] - p[k][0]) + 3*(p[i][1] - p[k][1])*(p[i][1] - p[k][1]);
            if (a == b && b == c) {
                count++;
                System.out.println("Equilateral triangle #" + count + ": i=" + i + " j="+j + " k=" + k);
            }
        }
    }

    public static void main(String[] args) {
        new TriangleCount().count();
    }
}

Edited September 4, 2014 by witzar

[bonanova]

Looks like that's the answer then. Nice.

 

I'm wondering whether that's the optimal configuration for 16 points to generate ETs.

 

I'm also wondering whether the number of ETs increases proportionately faster then the number of points.

It's easy to see that adding a 17^th point at 6,4 increases the ratio from 2.5625 to 2.70588 (46 ETs.)