# Interactive Hyperbolic Tiling in the Poincaré Disc

The tiling is made of regular hyperbolic polygons inside a circle $$C_\infty$$. The inside of $$C_\infty$$ is the hyperbolic universe, which is commonly called the Poincaré disc. The circle itself is not included in the universe but can be seen as the circle at infinity.

For a regular hyperbolic polygon, all angles are equal and all sides have the same hyperbolic length. Reflecting a polygon in all its edges and then repeatedly reflecting each newly created polygon in its “free” edges create the tiling. The first reflections are shown as arrows in the figure below. Except for algorithmic reasons, the ordering does not matter. Reflecting any of its neighbours can create every polygon.

## p and q

In the tiling above, $$p$$ is the number of vertices of each polygon, and $$q$$ is the number of polygons adjacent to each vertex.

For a Euclidean tiling of regular polygons, the only possible pair of values are: $$p = 3, q = 6$$ or $$p = 4, q = 4$$ or $$p = 6, q = 3$$. See Geometry - Tessellations and Symmetries for the Euclidean case. In the Euclidean case, the angle sum of a regular polygon with $$p$$ sides is $$(p-2)\cdot \pi$$. Having $$q$$ such polygons meeting at each vertex, following must be true:

\begin{align*} \frac{(p-2)\cdot \pi}{p}\cdot q &= 2\pi \\ \iff \\ (p-2)\cdot q &= 2p \\ \iff \\ p\cdot q-2q-2p+4 & = 4 \\ \iff \\ (p-2)\cdot (q-2) &= 4 \end{align*}

In hyperbolic geometry, the angle sum of a triangle is less than $$\pi$$ and the angle sum of a regular polygon with $$p$$ sides is less than $$(p-2)\cdot\pi$$. Let $$\alpha$$ be the inner angle of a hyperbolic polygon in a pq-tiling, then:

$\frac{(p-2)\cdot \pi}{p}\cdot q > \alpha\cdot q = 2\pi$

For that reason, following relation must hold for any hyperbolic pq-tiling of regular polygons.

$(p-2)\cdot (q-2) > 4$

## Distances

In Euclidean geometry there are only three possibilities when it comes to the values of $$p$$ and $$q$$, but there is no restriction on the size of the polygon. In hyperbolic geometry there are infinitely many pairs of $$p$$ and $$q$$ that can be used for making a tiling of regular polygons, but in any tiling the size of the polygon is uniquely determined by $$p$$ and $$q$$. There is only one hyperbolic size. In the interactive example above, one point can be moved freely, the other point can only be moved along a hyperbolic circle. From a Euclidean point of view, the distance between the points is smaller when the points are close to the edge of the hyperbolic universe. In the hyperbolic universe however, the distance between the two points is constant.

For a given $$p$$ and $$q$$, it is easiest to consider a hyperbolic polygon at the centre of the universe when finding the size of all regular polygons of the tiling.

Let $$d$$ be the Euclidean distance between a vertex of a centered hyperbolic polygon and the centre of the universe $$C_\infty$$. For simplicty, let $$C_\infty$$ be a unit circle in the given Euclidean coordinate system, then $$d$$ can be found through a long series of trigonometric brute force steps.

$d = \sqrt{ \frac{ \cos ( \pi / p + \pi / q ) \cdot \cos (\pi / q) } { \sin(2\pi / q) \cdot \sin( \pi / p) + \cos ( \pi / p + \pi / q) \cdot \cos(\pi / q) } }$

Or if using GeoGebra compatible code:

d = sqrt(cos(pi/p + pi/q)*cos(pi/q) / (sin(2*pi/q) * sin(pi/p) + cos(pi/p + pi/q)* cos(pi/q)))

When $$d$$ has been calculated, any hyperbolic polygon for that tiling can be created by using a hyperbolic compass, as described on the previous page: GeoGebra Constructions in the Poincaré Disc.

For another example of an interactive tiling, where you can load any image and make a hyperbolic tiling of it, see Make Hyperbolic Tilings of Images.