Solution
Lozenges in a Hexagon
Show the problem again
A regular hexagon of side length 3 is cut by three families of parallel lines into 54 unit equilateral triangles. A lozenge is a rhombus formed by two of these unit triangles sharing an edge. In how many ways can the hexagon be tiled exactly by 27 lozenges? Tilings related by a symmetry of the hexagon count as different. Give an integer.
Worked solution
**Answer: 980.** Shade the three orientations of lozenge in three tones and every tiling becomes an isometric drawing of unit cubes stacked in the corner of a 3 × 3 × 3 box, with stacks that never grow taller as you move away from the corner. Reading off the stack heights gives a 3 × 3 array of integers between 0 and 3, weakly decreasing along rows and down columns: a plane partition in a 3 × 3 × 3 box. This correspondence is a bijection. MacMahon's formula counts plane partitions in an a × b × c box as ∏_{i=1..a} ∏_{j=1..b} ∏_{k=1..c} (i + j + k − 1) / (i + j + k − 2). For a = b = c = 3 the product equals 980 (for comparison, side 1 gives 2 and side 2 gives 20). Direct enumeration of all 3 × 3 arrays confirms 980.
Source: Counts plane partitions in a box (P. A. MacMahon, early 20th century); the lozenge-tiling interpretation is classical. Statement written for AxiomIQ.