Posted on June 26, 2026
I'm redoing my kitchen floor using rhombus-shaped tiles composed of two congruent equilateral triangles. How many distinct ways can I tile the outlined region, which consists of 24 equilateral triangles arranged in a regular hexagon?
A regular hexagon of side \(n\) consists of \(6n^2\) equilateral triangles. The kitchen hexagon has side \(n=2\) (since \(6\cdot 4=24\)). Each rhombus covers exactly 2 adjacent triangles, so any tiling uses exactly 12 tiles.
Rhombus tilings of a regular hexagon of side \(n\) are in bijection with plane partitions fitting in an \(n\times n\times n\) box: integer matrices \(\pi\) of size \(n\times n\) with \(0\le \pi_{ij}\le n\) that are weakly decreasing along each row and column.
The key insight is geometric: the three orientations of a rhombus (horizontal, left-leaning, right-leaning) correspond to the top face, left face, and right face of a unit cube in an isometric projection. A valid tiling assembles into a coherent stack of unit cubes fitting inside an \(n\times n\times n\) box — exactly the data of a plane partition. Every tiling looks like a 3D corner-view of stacked cubes.
In 1896, MacMahon derived the exact count of plane partitions in an \(a\times b\times c\) box:
\[ \prod_{i=1}^{a}\prod_{j=1}^{b} \frac{i+j+c-1}{i+j-1}. \]
For a regular hexagon of side \(n\), we set \(a=b=c=n\), giving
\[ M(n) = \prod_{i=1}^{n}\prod_{j=1}^{n} \frac{i+j+n-1}{i+j-1}. \]
With \(n=2\), the product runs over \((i,j)\in\{1,2\}^2\):
\[ M(2) = \frac{3}{1}\cdot\frac{4}{2}\cdot\frac{4}{2}\cdot\frac{5}{3} = 3\cdot 2\cdot 2\cdot\frac{5}{3} = 20. \]
Answer: \(\boxed{20}\) distinct tilings.
How many distinct ways can I tile the patio, which consists of 54 equilateral triangles arranged in a regular hexagon?
The patio hexagon has side \(n=3\) (since \(6\cdot 9=54\)). The product runs over \((i,j)\in\{1,2,3\}^2\):
\[ M(3) = \frac{4}{1}\cdot\frac{5}{2}\cdot\frac{6}{3} \cdot\frac{5}{2}\cdot\frac{6}{3}\cdot\frac{7}{4} \cdot\frac{6}{3}\cdot\frac{7}{4}\cdot\frac{8}{5}. \]
Collecting numerator and denominator:
\[ M(3) = \frac{4\cdot 5\cdot 6\cdot 5\cdot 6\cdot 7\cdot 6\cdot 7\cdot 8}{1\cdot 2\cdot 3\cdot 2\cdot 3\cdot 4\cdot 3\cdot 4\cdot 5} = \frac{313{,}600}{320} = 980. \]
Extra credit answer: \(\boxed{980}\) distinct tilings.
The same formula gives the tiling count for any regular hexagon of side \(n\):
| Side \(n\) | Triangles | Tilings \(M(n)\) |
|---|---|---|
| 1 | 6 | 2 |
| 2 | 24 | 20 |
| 3 | 54 | 980 |
| 4 | 96 | 232,848 |
| 5 | 150 | 700,742,050 |
The count grows rapidly — each additional ring of the hexagon multiplies the tiling count by a combinatorial factor that grows super-polynomially in \(n\).
Adjust \(n\) to see the product of fractions and the resulting tiling count.