==> geometry/lattice/equilateral.s <== No. Suppose 2 of the vertices are (a,b) and (c,d), where a,b,c,d are integers. Then the 3rd vertex lies on the line defined by (x,y) = 1/2 (a+c,b+d) + t ((d-b)/(c-a),-1) (t any real number) and since the triangle is equilateral, we must have ||t ((d-b)/(c-a),-1)|| = sqrt(3)/2 ||(c,d)-(a,b)|| which yields t = +/- sqrt(3)/2 (c-a). Thus the 3rd vertex is 1/2 (a+c,b+d) +/- sqrt(3)/2 (d-b,a-c) which must be irrational in at least one coordinate.