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Total number of equilateral triangles

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*             *              *             *    

  

       *              *             *             *

 

               *             *             *            *

 

                      *              *

 

              *              *

 

 

 

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?

 

 

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7 answers to this question

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Posted · Report post

25.

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

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Posted (edited) · Report post

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 by Perhaps check it again
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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?

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Posted · Report post

Wondering whether a solution has been found.

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I get 41.

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Posted (edited) · Report post

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 by witzar
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Posted · Report post

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 17th point at 6,4 increases the ratio from 2.5625 to 2.70588 (46 ETs.)

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