Equilateral triangle on a grid

[bonanova]

Is it possible to put an equilateral triangle onto a square grid so that all the vertices are in corners?

[phil1882]

no, it is not possible.

consider the coordinates 0,0; a,b; c,d

we have (a-c)^2 +(b-d)^2 = a^2 +b^2 = c^2 +d^2

or a^2 -2ac +c^2 +b^2 -2bd +d^2 = a^2 +b^2 = c^2 + d^2

or 2ac +2bd = a^2 +b^2 = c^2 +d^2

using the fermat identity (a^2 +b^2)*(c^2 +d^2) = (ac +bd)^2 +(ad -bc)^2

let N  = 2*(ac +bd) = a^2 +b^2 = c^2 +d^2; and N/2 = ac +bd

then N^2  = N^2/4 + (ad - bd)^2 or 3*N^2 = 4*(ad - bd)^2

which leaves no integer solution for N; necessary to have a valid answer. (except for 0 of course).

http://math.stackexchange.com/questions/105330/equilateral-triangle-whose-vertices-are-lattice-points

[BMAD]

Can't i connect 3 30-60-90 triangles where their hypotenuses' touch at vertices to make an equilateral? 

Edited August 11, 2015 by BMAD
Forgot spoiler

[bonanova]

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Great observation, and you'd only need two of them actually.

But, as you ask, can you do it?

What is the slope of a 30-60-90 triangle's hypotenuse?

[BMAD]

Hidden Content

Great observation, and you'd only need two of them actually.

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you are right, i do only need two.  Now i know the hypotenuse is irrational assuming the legs are not in my 30-60-90 triangle.  But since my end points in the 30-60-90 make the endpoints  for the equilateral and they themselves are integers.  I would say that it is possible.

[bonanova]

Hidden Content

Great observation, and you'd only need two of them actually.

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Can you give a set of rational coordinates?