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.

Worked example — Regular Tilings Lab

Corner angle of one polygon	90°
q of them at a corner	270°
Left over from 360°	+90°
The test	2 < 4
Vertices − edges + faces	2
Total angular defect	720°
Tiles drawn	6

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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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

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.
  
  $interior angle of a regular p -gon = \frac{180 ( p - 2 )}{p} = \frac{180 \times 2}{4} = 9 0^{\circ}$
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.
  
  $q = 3 : 3 \times 9 0^{\circ} = 27 0^{\circ} ⟹ 36 0^{\circ} - 27 0^{\circ} = 9 0^{\circ} left over$
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.
  
  $(p - 2) (q - 2) = 2 \times 1 = 2 < 4 ⟹ it closes into a solid$
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.
  
  $V = 8, E = 12, F = 6 : V - E + F = 8 - 12 + 6 = 2$
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°.
  
  $8 \times 9 0^{\circ} = 72 0^{\circ} = 4 π (Descartes: the same total for every convex polyhedron)$
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.

$(p - 2) (q - 2) < 4 sphere, = 4 plane, > 4 hyperbolic$

References (6)

Why there is no sixth Platonic solid: Euclid, Elements, Book XIII, Proposition 18 and its closing Remark, which rules out further regular solids on the grounds that six equilateral triangles at a corner "will be equal to four right angles" and so lie flat. The edition by David E. Joyce, Clark University.
The hyperbolic tilings and Escher: H. S. M. Coxeter, "The Non-Euclidean Symmetry of Escher’s Picture ‘Circle Limit III’." Leonardo 12(1), 19–25, 1979.
The nested solids of Kepler: Johannes Kepler, Mysterium Cosmographicum. Tübingen, 1596 — the model that spaced the six known planets by inscribing the five regular solids between their spheres, and which Kepler spent the following decades replacing.
Why the honeycomb is hexagonal: Thomas C. Hales, "The Honeycomb Conjecture." Discrete & Computational Geometry 25, 1–22, 2001 — the proof that the regular hexagonal tiling divides a surface into equal areas with the least total perimeter.
Why so many viruses are icosahedral: D. L. D. Caspar and A. Klug, "Physical Principles in the Construction of Regular Viruses." Cold Spring Harbor Symposia on Quantitative Biology 27, 1–24, 1962.
Hyperbolic space for hierarchies and networks: D. Krioukov, F. Papadopoulos, M. Kitsak, A. Vahdat and M. Boguñá, "Hyperbolic geometry of complex networks." Physical Review E 82, 036106, 2010.

By the 42Math team · Updated 2026-08-11

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

The cube {4,3} - Three squares at a corner come to 270°, which is 90° short of lying flat. That shortfall is what folds the surface, and eight corners × 90° = 720° closes it into a cube.
Icosahedron {3,5} - Five triangles reach 300°, leaving 60°. Twelve vertices × 60° is again 720° — the same total as the cube, and as every convex solid.
Honeycomb {6,3} - Three hexagons meet at exactly 360°. Nothing is left over, so nothing folds, and the tiling runs across the plane forever. Only three regular tilings do this.
Heptagons {7,3} - Three heptagons come to 385.71°, more than the 360° a flat corner has room for. The tiling exists, but only in hyperbolic space — this is the picture Escher borrowed.
Five squares {4,5} - Five squares at every corner: 450°, a full 90° too many. On paper it is impossible; in the disk it is ordinary, and each ring of tiles is larger than the one before it.