Moments of Inertia Associated with the Lozenge Tilings of a Hexagon

Let a, b and c be positive integers and consider a hexagon with side lengths a,b,c,a,b,c whose angles are 120 (see Figure 1). The subject of our interest is lozenge tilings of such a hexagon using lozenges with all sides of length 1 and angles of 60 and 120. Figure 2 shows an example of a lozenge tiling of a hexagon with a = 3, b = 5 and c = 4. We introduce the following oblique angled coordinate system: Its origin is located in one of the two vertices, where sides of length b and c meet, and the axes are induced by those two sides (see Figure 3). The units are chosen such that the side lengths of the considered hexagon are √ 2a, b, c, √ 2a, b, c in this coordinate system. (That is to say, the two triangles in Figure 3 with vertices in the origin form the unit ‘square’.) Let Pa,b,c(x, y) denote the probability that an arbitrary chosen lozenge tiling of the hexagon with side lengths a,b,c,a,b,c contains the horizontal lozenge with lowest vertex (x, y) in the oblique angled coordinate system. Note that S = ((a + b)/2, (a + c)/2) is the centre of the hexagon in question. Consider the probability Pa,b,c(x, y) as if it described the distribution of mass. In [1, Problem 7] Propp suggests to compute the horizontal moment of inertia with respect to S