The Angular Kronecker Constant for {1, 2, 3} Is 1/4

With the set {1, 2, 3}, we can compute the angular Kronecker constant using a special metric in RxR, with hexagonal unit balls.  The angular Kronecker constant corresponds to the maximum distance in this metric of any point in the plane from a discrete (additive) subgroup L with two generators.  Our "canonical" generators for this subgroup are P=(2/3, 1/3) and Q=(1, 0).

Related to this geometry is a "Half Domain" to which we can narrow the search for points farthest from L.  Our "canonical" half domain in this case is the rectangle with vertices (0, 0), (0, 1/6), (1, 1/6), and (1, 0).

Related to this half domain is a set T of members of L such that, for all points in the half domain, there is at least one member of T which is among the points of L that are closest.  Here there are four such members of L:   (0, 0), Q= (1, 0), P = (2/3, 1/3),  and Q - P = (1/3, -1/3).

Finally, the farthest points from L within the half domain form a vertical line segment from (1/3, 1/12) TO (1/3, 1/6).  In the two linked pictures, N3.jpg and N3.pdf, this line segment is continued up to (1/3, 1/4) because these points are also farthest from L but are outside the given half domain (the rectangular region); they occur in a "reflection" of the half domain that could also serve as a half domain (specifically, the vertical reflection reflection around the line y = 1/6).

The "worst point" (1/3, 1/12) is interesting to us:  it is equidistant from three distinct members of L, namely  O = (0, 0) ,  P = (2/3, 1/3) and Q - P = (1/3, -1/3).  Notice that two of these three members of L can serve as generators of L.  Also notice that the metric formulas used to realize the distance to (1/3, 1/12) are two horizontal (with opposite orientation) and one vertical.  Note that this "worst point" is NOT the barycenter of the triangle formed by O, P and Q-P.