Equilateral triangle divided in equilateral triangles
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Let's consider an equilateral triangle which is divided in $100$ equilateral triangles. What is the maximum number of points which are vertices of these triangles such that each straight line that passes through any two of these points is parallel to any edge of the initial triangle? I'm trying to consider the coordinates of every point.
Any ideas? Thank you!
combinatorics
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Let's consider an equilateral triangle which is divided in $100$ equilateral triangles. What is the maximum number of points which are vertices of these triangles such that each straight line that passes through any two of these points is parallel to any edge of the initial triangle? I'm trying to consider the coordinates of every point.
Any ideas? Thank you!
combinatorics
This is an interesting problem but your question has been stated in a way very difficult to understand. In any event, it equivalent to thenon-attacking queens on a triangle
problem for a triangle of side $n = 11$. Nivasch et al. has shown the maximum number is $leftlfloor frac{2n+1}{3}rightrfloor = leftlfloor frac{2(11)+1}{3}rightrfloor = 7$.
– achille hui
Nov 23 at 22:59
Hmm... I didn't notice the original version of this question has a double negative about "no two points that is not parallel to...". So my previous comment didn't apply....
– achille hui
Nov 24 at 3:50
add a comment |
up vote
-2
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favorite
up vote
-2
down vote
favorite
Let's consider an equilateral triangle which is divided in $100$ equilateral triangles. What is the maximum number of points which are vertices of these triangles such that each straight line that passes through any two of these points is parallel to any edge of the initial triangle? I'm trying to consider the coordinates of every point.
Any ideas? Thank you!
combinatorics
Let's consider an equilateral triangle which is divided in $100$ equilateral triangles. What is the maximum number of points which are vertices of these triangles such that each straight line that passes through any two of these points is parallel to any edge of the initial triangle? I'm trying to consider the coordinates of every point.
Any ideas? Thank you!
combinatorics
combinatorics
edited Nov 24 at 3:45
Rócherz
2,7012721
2,7012721
asked Nov 23 at 19:48
rafa
555212
555212
This is an interesting problem but your question has been stated in a way very difficult to understand. In any event, it equivalent to thenon-attacking queens on a triangle
problem for a triangle of side $n = 11$. Nivasch et al. has shown the maximum number is $leftlfloor frac{2n+1}{3}rightrfloor = leftlfloor frac{2(11)+1}{3}rightrfloor = 7$.
– achille hui
Nov 23 at 22:59
Hmm... I didn't notice the original version of this question has a double negative about "no two points that is not parallel to...". So my previous comment didn't apply....
– achille hui
Nov 24 at 3:50
add a comment |
This is an interesting problem but your question has been stated in a way very difficult to understand. In any event, it equivalent to thenon-attacking queens on a triangle
problem for a triangle of side $n = 11$. Nivasch et al. has shown the maximum number is $leftlfloor frac{2n+1}{3}rightrfloor = leftlfloor frac{2(11)+1}{3}rightrfloor = 7$.
– achille hui
Nov 23 at 22:59
Hmm... I didn't notice the original version of this question has a double negative about "no two points that is not parallel to...". So my previous comment didn't apply....
– achille hui
Nov 24 at 3:50
This is an interesting problem but your question has been stated in a way very difficult to understand. In any event, it equivalent to the
non-attacking queens on a triangle
problem for a triangle of side $n = 11$. Nivasch et al. has shown the maximum number is $leftlfloor frac{2n+1}{3}rightrfloor = leftlfloor frac{2(11)+1}{3}rightrfloor = 7$.– achille hui
Nov 23 at 22:59
This is an interesting problem but your question has been stated in a way very difficult to understand. In any event, it equivalent to the
non-attacking queens on a triangle
problem for a triangle of side $n = 11$. Nivasch et al. has shown the maximum number is $leftlfloor frac{2n+1}{3}rightrfloor = leftlfloor frac{2(11)+1}{3}rightrfloor = 7$.– achille hui
Nov 23 at 22:59
Hmm... I didn't notice the original version of this question has a double negative about "no two points that is not parallel to...". So my previous comment didn't apply....
– achille hui
Nov 24 at 3:50
Hmm... I didn't notice the original version of this question has a double negative about "no two points that is not parallel to...". So my previous comment didn't apply....
– achille hui
Nov 24 at 3:50
add a comment |
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This is an interesting problem but your question has been stated in a way very difficult to understand. In any event, it equivalent to the
non-attacking queens on a triangle
problem for a triangle of side $n = 11$. Nivasch et al. has shown the maximum number is $leftlfloor frac{2n+1}{3}rightrfloor = leftlfloor frac{2(11)+1}{3}rightrfloor = 7$.– achille hui
Nov 23 at 22:59
Hmm... I didn't notice the original version of this question has a double negative about "no two points that is not parallel to...". So my previous comment didn't apply....
– achille hui
Nov 24 at 3:50