Two tiles, fivefold symmetry, and the golden ratio hiding in the counts
Roger Penrose found his tiles in 1974, a decade after Berger’s 20,426. He got the count to six, then to two. There it stayed for almost fifty years — nobody found three, or four, or one, and nobody proved that one was impossible.
The version here is P3: two rhombi with unit sides, a fat one with angles 72° and 108° and a thin one with 36° and 144°. Both are built here by deflation — replacing each tile with smaller ones and rescaling — which is the same machinery the hat proof runs on, in a setting where it is much easier to see.
The rhombi are not the unit of the substitution. Inflate a rhomb by the golden ratio and cut the enlarged copy into rhombi of the original size, and some of them come out cut in half — so the rule has to be written on the halves, and rhombi recovered afterwards by pairing halves across the diagonals they were cut on.
Two rhombi, both with unit sides: the fat one has angles 72° and 108°, the thin one 36° and 144°. Each deflation replaces every triangle with smaller ones and the count multiplies by φ² ≈ 2.618. The fat-to-thin ratio walks towards φ = 1.618034, and the mismatch count — triangles sharing a base that could not be halves of one rhomb — stays at 0.
Try it: Step the deflations up one at a time from zero. The seed is five fat rhombi round a point — 5 × 72° = 360° — or ten thin ones, since 10 × 36° also comes to 360°. Watch the fat-to-thin ratio settle towards 1.618 while the fivefold symmetry of the seed survives at the centre and dissolves further out.
Half a fat rhomb is a golden gnomon, sides 1, 1, φ with a 108° apex. Half a thin rhomb is a golden triangle, sides 1, 1, 1/φ with a 36° apex. Together they are the two Robinson triangles, and everything about the Penrose substitution is a statement about them.
They carry a handedness, and the deflation depends on it — the rule cuts a particular leg, and which leg that is depends on which end of the base you wrote first. Get it backwards and every local check still passes: the children are exact Robinson triangles at exactly 1/φ the size, the areas add up, the edge directions are the usual ten. Only one thing fails, and it is the thing that matters.
Half a fat rhomb is a golden gnomon: sides 1, 1, φ with a 108° apex. Half a thin rhomb is a golden triangle: sides 1, 1, 1/φ with a 36° apex. Only one of a rhomb’s two diagonals gives these — the other cut would leave apexes of 72° or 144°, which no Robinson triangle has.
Try it: Flip between the two labellings. Correct, 50 of the 1,440 triangles are unpaired, and every one is on the rim — the inner half of the patch holds none at all, and their partners simply lie outside. Reversed, 180 are unpaired and 75 of those are in the inner half, the innermost practically at the centre. The count is the weaker signal; where they are is the decisive one, and neither is visible in a picture at the wrong zoom.
A fat rhomb becomes two fat and one thin; a thin one becomes one of each. That is the matrix [[2, 1], [1, 1]] — trace 3, determinant 1, so its eigenvalues are (3 ± √5)/2 and the larger is φ². The number of tiles multiplies by φ² each step and the two kinds settle into the ratio φ.
| deflations | fat, by matrix | thin, by matrix | fat, counted | thin, counted | fat ÷ thin |
|---|---|---|---|---|---|
| 0 | 5 | 0 | 5 | 0 | — |
| 1 | 10 | 5 | 5 | 5 | 1.000000 |
| 2 | 25 | 15 | 20 | 15 | 1.333333 |
| 3 | 65 | 40 | 60 | 35 | 1.714286 |
| 4 | 170 | 105 | 160 | 100 | 1.600000 |
| 5 | 445 | 275 | 425 | 270 | 1.574074 |
| 6 | 1,165 | 720 | 1,135 | 710 | 1.598592 |
| 7 | 3,050 | 1,885 | 3,005 | 1,865 | 1.611260 |
The matrix [[2, 1], [1, 1]] has trace 3 and determinant 1, so its eigenvalues are (3 ± √5)/2 and the larger is φ² = 2.618034. The dominant eigenvector is (φ, 1), which is why the two tiles settle into that proportion — and why a Penrose tiling cannot be periodic, since a repeating patch would force the ratio to be a fraction.
Try it: Read down the last column. This is the same argument that finishes the hat off: the proportion of the two tiles is irrational, and a tiling made by repeating a finite patch would have to make it a fraction.
A crystal cannot have fivefold symmetry. Pentagons do not tile the plane, and the classification of the wallpaper groups allows rotations of order 2, 3, 4 and 6 and nothing else. Yet in 1982 Dan Shechtman photographed a diffraction pattern with tenfold symmetry from an aluminium-manganese alloy, and was told for two years that he was wrong. He got the Nobel Prize in 2011.
A Penrose patch shows how both things can be true. It has no fivefold rotational symmetry as a tiling, but every edge in it runs in one of ten directions 36° apart, and that is what a diffraction pattern sees.
The directions come out as 0°, 36°, 72°, 108°, 144° — five classes, 36° apart, at every size of patch. Not one vertex exceeds 360°, which is the arithmetic form of “no two rhombi overlap”.
Try it: Grow the patch. The direction count stays at five however large it gets — thousands of tiles, five directions. And no vertex ever totals more than 360°, which is the arithmetic form of “nothing overlaps”. The next page explains where the five directions come from: they are the shadow of a lattice in five dimensions.