One shape, thirteen sides, found by an amateur with scissors in 2022
In November 2022 David Smith cut a thirteen-sided shape out of cardboard and could not make it repeat. Here is that shape, built from a grid of kites.
Step through the three layers. On the last, notice that the hat’s eight kites span three hexagons.
A hexagonal tiling. Every hexagon here has edge length 2, so the distance from its centre to an edge midpoint is the square root of 3.
In November 2022 David Smith, an amateur working at his kitchen table in Yorkshire, cut out copies of a thirteen-sided shape and found he could not make them repeat. He had no proof, and he was not a mathematician. He emailed Craig Kaplan. By March 2023 Smith, Joseph Samuel Myers, Kaplan and Chaim Goodman-Strauss had a proof, and the question that had been open since 1961 was closed.
The shape is not exotic. It is eight copies of one quadrilateral glued together, and that quadrilateral is what you get by cutting a hexagon into six pieces. Everything on this page is built from that grid, in exact integer coordinates, so that each claim about the tile can be computed rather than asserted.
Take a hexagonal tiling and draw, in every hexagon, the three lines through opposite edge midpoints. Each hexagon falls into six kites with angles 60°, 90°, 120°, 90° — the hexagon centre, an edge midpoint, a hexagon vertex, the next edge midpoint. The result is a tiling in its own right, the Laves tiling [3.4.6.4], and a shape made of these kites is called a polykite.
Step through the three layers. On the last one, note that the eight highlighted kites do not sit inside a single hexagon: the hat spans three of them, and is not centred on any.
Click kites to build a shape; the panel reports its sides, holes and whether it is the hat. Try building the hat from memory.
Counted up to rotation, reflection and position, there are 1 of 1, 2 of 2, 4 of 3, 10 of 4, 27 of 5, 85 of 6, 262 of 7, 871 of 8 kites. David Smith found the hat among the 871, by hand, in November 2022.
Here is the search Smith was doing, with the scissors replaced by a mouse. Choose kites and the panel below reports what you have made: how many kites, whether it is one connected piece, whether it encloses a hole, how many sides its boundary has after collinear edges merge, and whether it is the hat.
The verdict is not a picture comparison. Every edge of a polykite runs at a multiple of 30° and has a squared length that is an exact integer, so two shapes can be compared by their edge sequences with no tolerance anywhere — and the comparison distinguishes a shape from its mirror image, which will matter in a moment.
Clear the board and try to build the hat from memory. Then consider that there are 871 distinct eight-kite shapes — counted up to rotation, reflection and position, by growing them one kite at a time — and that exactly one of them does what this one does.
Switch between lengths, angles and kites. Every angle is a multiple of 30° and four are reflex; the area is exactly eight kites.
Eight quadrilaterals glued edge to edge would normally have more than thirteen sides. It has thirteen because one pair of kite edges is collinear and merges into the single side of length 2 — 6 of length 1, 6 of length √3, 1 of length 2, and nothing else.
Eight quadrilaterals glued edge to edge would normally leave more than thirteen sides on the outside. The hat has thirteen because one pair of kite edges ends up collinear and merges into a single side of length 2. The other twelve sides come in two classes: six of length 1 and six of length √3, the kite’s two edge lengths.
Switch to the angles. Every one is a multiple of 30°, and four of them exceed 180° — the hat is emphatically not convex, and the notches are what let it interlock in so many different ways. Then switch to the kites and check the area: exactly eight, with no rounding, because the area is a shoelace sum over integer coordinates.
Sweep the turn and starting edge against the mirror image; the match count never reaches thirteen. Switch to “compare with itself” to find the alignment that does.
Each cell shows one edge of the hat above the edge it is being compared with: the length class, then the direction in twelfths of a turn. Exhaustively, no combination of turn and starting edge lines the hat up with its mirror image — which is what makes the hat a chiral tile, and why a tiling of hats needs reflected copies rather than merely happening to contain some.
The hat is chiral: no rotation carries it onto its mirror image. That is easy to say and easy to get wrong by eye, so the demo below settles it by exhaustion. There are twelve directions a polykite edge can take and thirteen edges to start the comparison from, which is 156 ways to line the two shapes up. Every one leaves at least one edge disagreeing.
This matters because a tiling of hats is forced to use both handednesses. The reflected copies are not an artefact of how someone drew the picture; they are unavoidable, and the next page works out exactly how rare they are.
Sweep the turn and the starting edge against the mirror and watch the count stall short of thirteen. Then switch to “compare with itself” and find the alignment that reaches thirteen — which is how you know the test is capable of saying yes.
Colour by metatile and notice how rare T is. Then click a tile and read its chain of supertiles.
Generations stop at 6, which is 242,962 tiles. The substitution keeps going — generation 7 from an H is 1,667,659 — but building that takes about 0.7 s and 25 MB, so the instrument does not offer it.
Hats themselves do not obey a substitution rule. The proof groups them into four clusters — called H, T, P and F — which do, and it is those clusters the instrument draws. An H holds four hats, a T one, and a P and an F two each; every hat in the plane belongs to one of them.
Click any tile and the instrument names the chain of supertiles it belongs to, level by level, all the way out to the seed. That chain is unique, it exists for every tile, and it is the whole argument — which is the subject of the next page.
Colour by metatile and note how much rarer T is than the rest. Then click a tile near the middle and read its chain: a P inside an F inside an H, and so on outward. Every tile has one, and no two tiles at the same position have different ones.
Spot a mistake or have an idea?