Three Regular Tessellations

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Theorem

There exist exactly $3$ regular tessellations of the plane.


Triangles

Equilateral triangles form a regular tessellation:


RegularTriangleTessellation.png


Squares

Squares form a regular tessellation:


RegularSquareTessellation.png


Hexagons

Regular hexagons form a regular tessellation:


RegularHexagonTessellation.png


Proof

Let $m$ be the number of sides of each of the regular polygons that form the regular tessellation.

Let $n$ be the number of those regular polygons which meet at each vertex.

From Internal Angles of Regular Polygon, the internal angles of each polygon measure $\dfrac {\paren {m - 2} 180^\circ} m$.


The sum of the internal angles at a point is equal to $360^\circ$ by Sum of Angles between Straight Lines at Point form Four Right Angles

So:

\(\ds n \paren {\dfrac {\paren {m - 2} 180^\circ} m}\) \(=\) \(\ds 360^\circ\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {m - 2} m\) \(=\) \(\ds \dfrac 2 n\)
\(\ds \leadsto \ \ \) \(\ds 1 - \dfrac 2 m\) \(=\) \(\ds \dfrac 2 n\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 m + \dfrac 1 n\) \(=\) \(\ds \dfrac 1 2\)


But $m$ and $n$ are both greater than $2$.

So:

if $m = 3$, $n = 6$.
if $m = 4$, $n = 4$.
if $m = 5$, $n = \dfrac {10} 3$, which is not an integer.
if $m = 6$, $n = 3$.


Now suppose $m > 6$.

We have:

\(\ds \dfrac 1 m + \dfrac 1 n\) \(=\) \(\ds \dfrac 1 2\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 6 + \dfrac 1 n\) \(>\) \(\ds \dfrac 1 2\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 n\) \(>\) \(\ds \dfrac 1 3\)
\(\ds \leadsto \ \ \) \(\ds 3\) \(>\) \(\ds n\)

But there are no integers between $2$ and $3$, so $m \not > 6$.


There are $3$ possibilities in all.

Therefore all regular tessellations have been accounted for.

$\blacksquare$


Also see


Sources

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