Yes, there are. For notational purposes, let [a,b] represent a point on the triangular lattice where a is units of movement right (_) and b is movement diagonally up and right (/).
Let's consider the line segment [0,0] to [a,b]. To calculate the length, we will convert each into rectangular coordinates. [0,0] is simply (0,0). We find that [a,b] = (a+1/2b,sqrt(3)/2b) with simple geometry. Using Pythagorean Theorem, the line segment has length:
Therefore, we need to find integers a,b,c such that a^2+ab+b^2=c^2.
Now that I think about it, we could have reached this same conclusion using Law of Cosines. Making a triangle with two of the legs be made entirely of _ and / respectively, results in a triangle with side lengths a and b, and a 120º angle between them. By Law of Cosines, the last side c has the property: c^2=a^2+2abcos(120º)+b^2 = a^2+ab+b^2.
How to find all the solutions to this Diophantine equation is beyond me, but brute force on a calculator confirms that there are indeed solutions: (3,5,7), (7,8,13), (5,16,9), (11,24,31), etc.
This is reminiscent of Euclid's formula for finding Pythagorean triplets, except in this case, we need triplets for a 120º triangle.
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My thoughts on the triangular lattice:
Yes, there are.
For notational purposes, let [a,b] represent a point on the triangular lattice where a is units of movement right (_) and b is movement diagonally up and right (/).
Let's consider the line segment [0,0] to [a,b]. To calculate the length, we will convert each into rectangular coordinates. [0,0] is simply (0,0). We find that [a,b] = (a+1/2b,sqrt(3)/2b) with simple geometry. Using Pythagorean Theorem, the line segment has length:
sqrt((a+1/2b)^2+(sqrt(3)/2b)^2)
=sqrt(a^2+ab+1/4b^2+3/4b^2)
=sqrt(a^2+ab+b^2).
Therefore, we need to find integers a,b,c such that a^2+ab+b^2=c^2.
Now that I think about it, we could have reached this same conclusion using Law of Cosines. Making a triangle with two of the legs be made entirely of _ and / respectively, results in a triangle with side lengths a and b, and a 120º angle between them. By Law of Cosines, the last side c has the property: c^2=a^2+2abcos(120º)+b^2 = a^2+ab+b^2.
How to find all the solutions to this Diophantine equation is beyond me, but brute force on a calculator confirms that there are indeed solutions: (3,5,7), (7,8,13), (5,16,9), (11,24,31), etc.
This is reminiscent of Euclid's formula for finding Pythagorean triplets, except in this case, we need triplets for a 120º triangle.
Interesting stuff.
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