1. What a tessellation is
A covering of the entire plane by one or more shapes — called tiles — such that every point of the plane lies in exactly one tile interior or on a shared edge. Two conditions: no gaps (every point is covered) and no overlaps (tile interiors don't intersect).
That's it. The shapes are allowed to repeat, allowed to rotate, allowed to flip. They just have to fit together perfectly forever, in every direction. A bathroom floor of square tiles is a tessellation. A honeycomb is a tessellation. A jigsaw puzzle is too — but only one with infinite identical copies.
The mathematics lives at the vertices, the points where corners of tiles meet. For the tiles to fit without gap or overlap, the angles meeting at any vertex must add up to exactly one full turn:
$$ \sum (\text{angles at a vertex}) = 360° $$Almost every result in this topic falls out of that single equation. Most of the work is just figuring out which polygons have interior angles that can be combined to make $360°$.
"Tessellation" and "tiling" are interchangeable. "Tessellation" comes from the Latin tessella, the small cube of stone or glass used in Roman mosaics. "Tiling" is the modern, more general mathematical term.
2. Regular tessellations: only three
A regular tessellation uses copies of a single regular polygon — same shape, same size, edge-to-edge. The question "which regular polygons tile?" has a famously short answer: equilateral triangles, squares, and regular hexagons. Nothing else.
To see why, recall the interior angle of a regular $n$-gon:
$$ \theta_n = \frac{(n-2) \cdot 180°}{n} $$For a regular tessellation, some integer number $k$ of these angles has to fit exactly around a vertex:
$$ k \cdot \theta_n = 360° $$Substituting and simplifying:
$$ k \cdot \frac{(n-2) \cdot 180°}{n} = 360° \quad\Longrightarrow\quad k(n-2) = 2n \quad\Longrightarrow\quad k = \frac{2n}{n-2} $$So $k$ must be a positive integer. Plug in values of $n \geq 3$ and see which give whole numbers:
| $n$ (sides) | Shape | Interior angle $\theta_n$ | $k = 2n/(n-2)$ | Tiles? |
|---|---|---|---|---|
| 3 | Triangle | $60°$ | $6$ | Yes — 6 at a vertex |
| 4 | Square | $90°$ | $4$ | Yes — 4 at a vertex |
| 5 | Pentagon | $108°$ | $10/3$ | No — not an integer |
| 6 | Hexagon | $120°$ | $3$ | Yes — 3 at a vertex |
| 7 | Heptagon | $\approx 128.6°$ | $14/5$ | No |
| 8 | Octagon | $135°$ | $8/3$ | No |
| 12 | Dodecagon | $150°$ | $12/5$ | No |
For $n \geq 7$, the interior angle exceeds $120°$, so fewer than 3 copies could possibly fit at a vertex — but you need at least 3 polygons meeting at every vertex to cover the plane. The list ends at the hexagon, and it ends forever.
The three regular tessellations
The notation $\{n, k\}$ — $n$-sided polygon, $k$ of them at each vertex — is called the Schläfli symbol. It encodes everything you need to reconstruct the tiling.
Of the three, the hexagon uses the fewest tiles per vertex (just 3) and has the shortest total edge length per unit of area covered. Bees discovered this long before geometers proved it. The honeycomb conjecture — that the hexagonal grid minimizes perimeter among all unit-area tilings — was stated by Pappus around 300 AD and proved by Thomas Hales in 1999.
3. Why a regular pentagon doesn't tile
The interior angle of a regular pentagon is $108°$. If $k$ of them met at a vertex, we'd need $108k = 360$, i.e. $k = 10/3$. You can't have a fractional number of pentagons.
Geometrically: three pentagons at a vertex cover $3 \cdot 108° = 324°$, leaving a $36°$ gap. Four pentagons would overlap by $72°$. There's no value of $k$ that works.
That gap is fatal: any fourth pentagon you try to drop in will either overlap one of the existing three or leave a different gap somewhere else. The plane will never fully close. This argument generalizes — a regular $n$-gon tiles by itself only when $\frac{360°}{\theta_n}$ is a whole number, which (as the table above showed) happens exclusively for $n = 3, 4, 6$.
4. Semi-regular (Archimedean) tessellations
What if we relax "one shape" to "regular polygons, but more than one kind allowed"? A semi-regular tessellation (also called Archimedean) uses two or more regular polygons, arranged so that every vertex looks the same — the same polygons in the same cyclic order around every vertex.
There are exactly 8. Each is named by the polygons meeting at a vertex, in order around it:
| Vertex configuration | Description | Angle check |
|---|---|---|
| $3.12.12$ | Triangle, dodecagon, dodecagon | $60 + 150 + 150 = 360$ |
| $4.6.12$ | Square, hexagon, dodecagon | $90 + 120 + 150 = 360$ |
| $4.8.8$ | Square, octagon, octagon | $90 + 135 + 135 = 360$ |
| $3.6.3.6$ | Triangle, hexagon, triangle, hexagon | $60 + 120 + 60 + 120 = 360$ |
| $3.4.6.4$ | Triangle, square, hexagon, square | $60 + 90 + 120 + 90 = 360$ |
| $3.3.4.3.4$ | Tri, tri, square, tri, square | $60 + 60 + 90 + 60 + 90 = 360$ |
| $3.3.3.4.4$ | Three triangles, two squares | $60 + 60 + 60 + 90 + 90 = 360$ |
| $3.3.3.3.6$ | Four triangles and a hexagon | $60 \cdot 4 + 120 = 360$ |
Note that not every combination of polygons whose angles sum to $360°$ produces a valid global tiling — the local vertex needs to extend consistently to a covering of the whole plane. The complete list above is what survives that stronger constraint.
Pick a guess like "two octagons and a square." That's $135 + 135 + 90 = 360°$ — it works. That's the $4.8.8$ pattern, the classic "octagon and square" tiling you've seen in bathroom floors and Roman mosaics. Now try "one octagon, one hexagon, and... what fits?" You need $360 - 135 - 120 = 105°$, which isn't the interior angle of any regular polygon. Dead end. The arithmetic is the entire search.
5. Any triangle, any quadrilateral
Drop the "regular" requirement. Surprisingly, the world gets much more permissive.
Every triangle tiles the plane. Take any triangle — scalene, obtuse, however you like. Rotate a copy $180°$ around the midpoint of one of its sides. The original and the rotated copy fit together to form a parallelogram. Parallelograms tile trivially (slide copies in two directions). So every triangle tiles, full stop.
Every quadrilateral tiles the plane — even non-convex ones. The trick is the same in spirit: rotate the quadrilateral $180°$ around the midpoint of one of its edges, and the original-plus-copy is a hexagonal shape with three pairs of parallel sides, which tiles by translation.
Why does the rotation trick work? The angles at each vertex of the original quadrilateral are $\alpha, \beta, \gamma, \delta$ — and they sum to $360°$. When the rotation pairs them up correctly at the meeting vertex, all four angles show up exactly once around that vertex, so they sum to $360°$ and close perfectly. The condition for tiling reduces to an identity that's always true for any quadrilateral.
Once you go to pentagons, things get subtle again. Convex pentagons that tile the plane fall into exactly 15 known types — the last one was discovered in 2015, and Michaël Rao proved the list complete in 2017. Not every convex pentagon tiles. The full story is a beautiful century-long mathematical detective hunt.
6. Escher's recipe: modifying a tile
This is the technique behind M.C. Escher's birds, fish, lizards, and angels — those mesmerizing prints where strange creatures fit together with no gaps. The mathematical core is one sentence: any modification you make to one edge of a tile, you must make the matching opposite modification to the corresponding edge. If translation pairs the edges, copy the modification by translation. If rotation pairs them, rotate it. The total area you remove from one edge is added to its partner; the tile still tiles.
The simplest version uses a square:
- Start with a square — say the one you already know tiles the plane.
- Draw a curve from one corner of the top edge to the other, cutting into the square.
- Take exactly the piece you cut out and translate it to the bottom edge — paste it on as an addition.
- Repeat: cut from one side and paste to the other side.
The shape you've made is no longer a square — it might look like a salamander or a horseman — but it still tiles. Copies fit together because every edge still matches its partner perfectly, just with the same wiggle.
Escher used the rotation variant too — applied to triangles and hexagons — which produces tiles with multi-fold rotational symmetry. Combine the moves freely, swap straight edges for curved ones, and add internal detail, and you get the menagerie that filled his notebooks.
7. The 17 wallpaper groups
So far we've classified tessellations by which polygons they use. There's a deeper, more abstract classification: by the symmetries of the resulting pattern — which combinations of translations, rotations, reflections, and glide reflections leave the pattern looking unchanged.
A remarkable theorem proved in the late 19th century says: there are exactly 17 distinct symmetry types of patterns that repeat in two independent directions across the plane. They are called the wallpaper groups (or "plane crystallographic groups").
Every repeating pattern you have ever seen — a fabric print, a brick wall, a bathroom floor, an Escher tessellation, the floor mosaics of the Alhambra — falls into exactly one of these 17 categories. The square grid is one type. The hexagonal grid is another. Brick-bond is a third. The list is short, complete, and inescapable.
The names use a notation like p1, p2, pm, p3m1, p4mm, p6m, where the letters and numbers encode the allowed rotations (orders 2, 3, 4, or 6 — never 5, never 7+) and reflections. We won't classify them here; just know they exist, that the list ends at 17, and that this number is a property of two-dimensional Euclidean space itself.
The constraint is something called the "crystallographic restriction": any rotation that's a symmetry of a periodic plane pattern must have order 2, 3, 4, or 6 — never 5, never 7, never anything else. (Order $n$ means an $n$-fold rotation, i.e. rotation by $360°/n$.) Combine this with the allowed reflection axes and you get exactly 17 types. Try 5-fold symmetry and you'll find you can't make it repeat periodically — which leads us to the next section.
8. Penrose, aperiodic, and the monotile
Everything so far has been periodic: the pattern repeats by translation. Slide it by some vector and it looks unchanged. The wallpaper groups classify all periodic tilings of the plane.
In the 1970s, the mathematical physicist Roger Penrose found something remarkable: small sets of tiles that force aperiodicity — that cover the plane without gaps but never repeat by translation, no matter how you arrange them. His most famous example uses just two tiles — the kite and the dart, both derived from the golden ratio. Penrose tilings exhibit 5-fold rotational symmetry locally but no global translational symmetry. They look ordered yet never settle into a repeating grid.
This was a stunning discovery, and it acquired a physical correlate when in 1982 Dan Shechtman observed quasicrystals — real metallic alloys whose atomic arrangement was aperiodic in exactly the Penrose sense. He won the 2011 Nobel Prize in Chemistry for it.
For decades a famous open question lingered: is there a single tile that tiles the plane only aperiodically — an "einstein" (German for "one stone")? In March 2023, David Smith, Joseph Samuel Myers, Craig Kaplan, and Chaim Goodman-Strauss announced exactly that: the hat, a 13-sided polygon (plus a closely related spectre) that tiles the plane and only ever tiles it aperiodically. A 50-year-old problem, solved with an answer simple enough to print on a t-shirt.
The "no 5-fold periodic symmetry" rule and the existence of single-tile aperiodicity tell us something deep: the Euclidean plane has both rigid constraints on its periodic patterns and surprising room for ordered-but-non-repeating structure. Tessellations look like a kindergarten topic; pushed hard enough, they bump into Nobel Prize physics and 21st-century mathematics.
9. Common pitfalls
When the problem says "this shape tiles the plane," reflections and rotations of the shape are normally allowed. The shape itself is one tile; copies can appear flipped or rotated. Insisting that every copy be in the same orientation is a much stronger condition that almost no interesting shape satisfies.
Regular tessellations use one regular polygon. Semi-regular (Archimedean) use more than one regular polygon, with identical vertex arrangements. Irregular tessellations use one or more non-regular polygons — like a tiling of arbitrary triangles. These three terms are commonly conflated; keep them distinct.
At a single vertex, you need the interior angles meeting there to total $360°$. But local vertex feasibility doesn't guarantee a tiling of the entire plane — you also need the tiles to extend consistently in every direction. Most local arrangements that pass the angle check do extend, but checking the global condition is what makes (for example) the Archimedean list stop at 8 even though more angle combinations are arithmetically possible.
A regular pentagon doesn't tile by itself. But there are 15 known irregular convex pentagons that do. And regular pentagons certainly appear in tilings that include other shapes alongside them (just not as a regular or semi-regular tiling). "Doesn't tile" is a precise claim — be sure you know which version you mean.
10. Worked examples
Try each before opening the solution. The point is to see the angle arithmetic become second nature.
Example 1 · Prove the regular hexagon tessellates
Step 1. Compute the interior angle of a regular hexagon. Using the formula $\theta_n = (n-2) \cdot 180° / n$ with $n = 6$:
$$ \theta_6 = \frac{(6-2) \cdot 180°}{6} = \frac{720°}{6} = 120° $$Step 2. Find how many copies fit around a vertex. We need $k \cdot 120° = 360°$, so $k = 3$.
Step 3. Since $k = 3$ is a positive integer, three hexagons fit perfectly around each vertex with no gap or overlap. The pattern extends to the whole plane by translation in two independent directions. ✓
The Schläfli symbol is $\{6, 3\}$.
Example 2 · Show a regular octagon alone doesn't tile
Step 1. Interior angle of a regular octagon:
$$ \theta_8 = \frac{(8-2) \cdot 180°}{8} = \frac{1080°}{8} = 135° $$Step 2. For a vertex to close: $k \cdot 135° = 360°$, so $k = \frac{360}{135} = \frac{8}{3}$.
Step 3. $k$ must be a positive integer. $8/3$ is not. So no whole number of regular octagons can meet at a vertex. ✗
Bonus. Two octagons cover $270°$, leaving a $90°$ gap — exactly the angle of a square. That's why the $4.8.8$ semi-regular tessellation (square + two octagons at every vertex) does work.
Example 3 · Construct an Escher-style tile from a square
Step 1. Start with a unit square with corners at $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$.
Step 2. Modify the top edge. Replace the straight segment from $(0,1)$ to $(1,1)$ with any curve $C$ that starts at $(0,1)$ and ends at $(1,1)$. Some part of $C$ will dip down into the square (removing area) or push up out of it (adding area), or both.
Step 3. Apply the matching modification to the bottom edge. Translate $C$ downward by $1$ unit: it now connects $(0,0)$ to $(1,0)$. Replace the straight bottom edge with this translated curve.
Step 4. Optionally do the same trick to the left and right edges with a different curve $C'$.
Step 5. Verify. Copies of this new tile, translated by the unit vectors $(1, 0)$ and $(0, 1)$, fit together: every right edge meets a left edge with the same shape; every top edge meets a bottom edge with the same shape. No gaps, no overlaps — and the tile is no longer a square. ✓
The total area is preserved: whatever you carve out of one edge, you add back to the opposite edge.
Example 4 · Verify the $4.8.8$ semi-regular tessellation
At each vertex we have a square and two regular octagons.
Step 1. Angles: square $= 90°$, regular octagon $= 135°$.
Step 2. Sum at the vertex: $90° + 135° + 135° = 360°$. ✓
Step 3. Cyclic order is square, octagon, octagon — and crucially, this same pattern can be made to hold at every vertex of the resulting tiling, which is what makes the tessellation Archimedean. (Verifying global consistency requires actually laying out the tiles — but it works, and the resulting pattern is the familiar bathroom-floor "truncated square" tiling.)
Example 5 · Why no semi-regular tiling uses a regular heptagon
Step 1. A regular heptagon ($n = 7$) has interior angle $\theta_7 = \frac{5 \cdot 180°}{7} = \frac{900°}{7} \approx 128.57°$.
Step 2. For any vertex arrangement that includes one heptagon, the remaining angles must sum to $360° - \frac{900°}{7} = \frac{2520° - 900°}{7} = \frac{1620°}{7} \approx 231.43°$.
Step 3. That leftover would have to be expressible as a sum of interior angles of regular polygons. The interior angles of regular $n$-gons are $60°, 90°, 108°, 120°, \dots$ — all rational. But $\frac{1620°}{7}$ has a $7$ in its denominator that no combination of those angles can produce (their denominators are powers of $2, 3, 5$). Contradiction.
So the regular heptagon can't appear in any semi-regular tessellation. The same denominator argument rules out 7, 9, 10, 11, 13, 14, ... and leaves only the polygons that survive in the list of 8.