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Regular Tilings Lab

Choose the polygon and how many meet at a corner. If their angles fall short of 360° the surface closes into a solid; if they hit it exactly the tiling lies flat; if they overshoot it can only live in hyperbolic space.

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Five, three, and infinitely many — from one inequality

Everything on this page follows from a single comparison. A regular p-gon has corners of 180(p−2)/p degrees; put q of them round a point and you get q·180(p−2)/p. If that is less than 360° the corner has slack and must fold; if it is exactly 360° it lies flat; if it is more, it cannot be built out of flat pieces at all. Rearranged, those three cases are (p−2)(q−2) less than, equal to, or greater than 4 — and the first has exactly five integer solutions, the second exactly three. This is not a modern reformulation. It is the remark Euclid closes Elements XIII with, ruling out a sixth solid on the grounds that six triangles at a corner "would make four right angles" and so lie flat.

On a solid, the curvature is squeezed into the corners

A cube is flat everywhere except at eight points. Walk across a face and nothing curves; walk around a vertex and you find 270° where a flat surface would have given you 360°. The missing 90° is the curvature, and there is nowhere else for it to be. Now add it up: 8 × 90° = 720°. Do the same for the icosahedron and it is 12 × 60° = 720°. For the dodecahedron, 20 × 36° = 720°. The total never moves, for any convex polyhedron whatever, regular or not — Descartes noticed it around 1630. It is the same 720° = 2 × 360° that Euler's V − E + F = 2 is counting, and it is the Gauss–Bonnet theorem from Pythagoras: there the angle excess of a triangle was curvature × area spread smoothly over the surface, and here it is the same quantity concentrated at points.

Kepler built the solar system out of them, and Escher borrowed the other case

The five solids have attracted more theory than they can carry. Kepler, in Mysterium Cosmographicum (1596), nested them inside one another to fix the spacing of the six known planets — a beautiful idea, and wrong; he spent the rest of his career replacing it with the orbits that actually work. The hyperbolic case had to wait until there was a hyperbolic plane to draw it in. Coxeter published a figure of one in the 1950s, Escher saw it, and asked him how it was made; the answer produced the four Circle Limit prints, in which the tiles shrink toward the rim not because they are getting smaller but because the disk is a map of an infinite surface. Set the sides to 7 and the corner count to 3 and you are looking at the same construction.

Twelve pentagons, always

A football is not a design choice. Take any closed cage whose faces are pentagons and hexagons with exactly three meeting at each corner, and count the pentagons: there are twelve. Always — however large the cage, however many hexagons. Three faces per corner gives 3V = 2E, and V − E + F = 2 then collapses to twelve pentagons with the hexagon count cancelling out entirely. That is what it means to read something off a shape without measuring it. A football has 12 pentagons and 20 hexagons; buckminsterfullerene C₆₀ has 12 pentagons and 20 hexagons; a geodesic dome the size of a building still has exactly 12 pentagons hidden among its hexagons. The reason is the 720° on this page. Three hexagons at a corner come to exactly 360°, so they are flat and contribute nothing — every degree of the 720° has to be carried by the pentagons, and at 60° each that takes twelve of them. It is also why so many viruses build icosahedral shells: a closed container assembled from one repeated protein has very few options, and Caspar and Klug worked out in 1962 which ones.

The other two cases are not decorative either

The flat case is not merely possible, it is optimal. Of all the ways to divide a surface into regions of equal area, the hexagonal tiling has the least total perimeter — conjectured for two thousand years and proved by Thomas Hales in 2001. A honeycomb is a wax-economy result, not an aesthetic one. The hyperbolic case earns its keep on the growth you can see in the disk: each ring holds more tiles than the ring before it. A tree branches the same way, and a tree cannot be drawn in the plane without crowding its outer branches together, because the plane only offers area proportional to r². Hyperbolic space offers exponentially more, so hierarchies and scale-free networks fit into it with room to spare — which is why hyperbolic embeddings are now a standard way to model and lay them out. And you can hold one: lettuce, kale and coral ruffle because the growing rim adds length faster than the surface behind it adds area, and a surface with more perimeter than the plane can accommodate has no choice but to buckle.

Problem solved in full

  1. The 90° unused leftover when three squares meet at a corner 5 steps

    Three squares meet at a corner and leave 90° unused. Work out what happens to that leftover — and why it decides whether you get a flat floor, a solid, or neither.

    1. A regular polygon's interior angle follows from its exterior angles summing to a full turn. A square gives 90°, and that single number is the whole input to the question.

      interior angle of a regular p-gon=180(p−2)p=180×24=90∘\text{interior angle of a regular }p\text{-gon}=\frac{180(p-2)}{p}=\frac{180\times2}{4}=90^\circ

    2. Fit three of them round a corner and you use 270°, leaving 90° over. On a flat floor that is a gap, and the shape cannot tile the plane.

      q=3:3×90∘=270∘⟹360∘−270∘=90∘ left overq=3:\quad 3\times90^\circ=270^\circ\quad\Longrightarrow\quad 360^\circ-270^\circ=90^\circ\ \text{left over}

    3. The gap is not waste. Fold the corner up until the edges meet and the leftover angle becomes curvature — the corner is now a vertex of a solid. Whether that is possible at all is decided by one product, and 2 is less than 4.

      (p−2)(q−2)=2×1=2 < 4⟹it closes into a solid(p-2)(q-2)=2\times1=2\ <\ 4\quad\Longrightarrow\quad \text{it closes into a solid}

    4. The solid it closes into is the cube. Count its parts and Euler's relation holds, as it does for every convex polyhedron: vertices minus edges plus faces is 2.

      V=8, E=12, F=6:V−E+F=8−12+6=2V=8,\ E=12,\ F=6:\qquad V-E+F=8-12+6=2

    5. Now total the leftover angle over all eight corners. 720° — and Descartes' theorem says every convex polyhedron gives exactly that, whatever its shape. The cube, the tetrahedron and a 900-faced lump all total 720°.

      8×90∘=720∘=4π(Descartes: the same total for every convex polyhedron)8\times90^\circ=720^\circ=4\pi\quad(\text{Descartes: the same total for every convex polyhedron})

    Answer

    The tool prints 90° per corner, 270° around a vertex, 90° left over, the test as 2 < 4, Euler's 2, and a total defect of 720°. One product decides everything: (p−2)(q−2) less than 4 closes into a sphere, equal to 4 lies flat, greater than 4 will only fit in hyperbolic space. That is why there are exactly five Platonic solids and exactly three regular tilings of the plane — {3,6}, {4,4}, {6,3}, the triangle, the square and the hexagon, and nothing else, ever. Set p = 6, q = 3 and watch the leftover go to zero.

    (p−2)(q−2)<4 sphere, =4 plane, >4 hyperbolic(p-2)(q-2)<4\ \text{sphere},\ =4\ \text{plane},\ >4\ \text{hyperbolic}
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