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  1. Why there are infinitely many regular tessellations of the hyperbolic plane?

  2. Can there be a triangle made up of three straight lines in the hyperbolic plane? I know it's impossible since the angle sum of a triangle in hyperbolic plane is less than $\pi$, but why?

  3. Deduce that there are equilateral triangles with angle $2\pi/n$ for each $n \ge 7$. (this part is easy.)

  4. Also deduce that triangles with angle zero exist, in a certain sense, and that their area is finite. (I know that in the limit as the vertices go to infinity, there are even ideal hyperbolic triangles in which all three angles are 0°, and so the area is finite by $(\pi-A-B-C)R^2$)

  5. Find corresponding results for regular n-gons.

The question may become harder...Thanks in advance!

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    $\begingroup$ For #2: what do you mean by 'made up of three straight lines'? Do you mean 'lines' as opposed to 'line segments'; i.e., triangles whose vertices are all at infinity? Also, once you have the answer to #3 it should trivially give you the answer to #1; and if my reading of #2 is correct, then solving #4 should give you the answer to #2. (Also also, what is the motivation or context for these questions? Are they homework, independent study, or something else?) $\endgroup$ – Steven Stadnicki Jul 15 '13 at 17:56
  • $\begingroup$ @StevenStadnicki, actually my question is why a triangle in the hyperbolic plane can have any angle sum $<\pi$? Why can't it be $\pi$? $\endgroup$ – Ian Jul 15 '13 at 18:12
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    $\begingroup$ Ian: do you understand the angular defect and the way in which the difference between $\pi$ and the angle sum relates to the triangle's area? Once you understand where that formula comes from, that should give you better understanding as to why the sum of angles must be less than $\pi$ (even if only by an arbitrarily small amount). $\endgroup$ – Steven Stadnicki Jul 15 '13 at 18:29
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Infinitely many regular tessellations

  1. Why there are infinitely many regular tessellations of the hyperbolic plane?

In the Euclidean plane, you get a right triangle with corner angles

$$ \alpha=\frac\pi m \qquad \beta=\frac\pi n \qquad \gamma=\frac\pi 2 $$

exactly if $\alpha+\beta+\gamma=\pi$, i.e.

$$\frac1m + \frac1n = \frac12$$

You can then consider this triangle as the $2m$th part of a regular $m$-gon with corner angles $\frac{2\pi}n$, or equivalently as the $2n$th part of a regular $n$-gon with angle $\frac{2\pi}m$. If $m,n\in\mathbb N$ then these polygons will form a regular tessellation. One can check by simple enumeration that there are only very few possible solutions:

$$ \begin{pmatrix}m\\n\end{pmatrix} \in \left\{ \begin{pmatrix}4\\4\end{pmatrix}, \begin{pmatrix}3\\6\end{pmatrix}, \begin{pmatrix}6\\3\end{pmatrix} \right\} $$

So you only have tilings by regular triangles, squares and regular hexagons. Other tessellations by less regular shapes, e.g. isosceles triangles, are possible as well but occur as a supergroup of one of these regular tilings.

In hyperbolic geometry, the game is mostly the same, except for the angle sum condition, which becomes

$$\frac1m + \frac1n < \frac12$$

It is easy to see that there is an infinite number of possible values $m,n\in\mathbb N$ satisfying this inequality.

Straight lines

  1. Can there be a triangle made up of three straight lines in the hyperbolic plane? I know it's impossible since the angle sum of a triangle in hyperbolic plane is less than $\pi$, but why?

Your use of the term “straight line” suggests that you have a particular model in mind. Perhaps it's the Poincaré disk model. An Euclidean straight line which is also a geodesic of the hyperbolic geometry in this model is a line through the center of the circle. So if two straight lines modeling geodesics do intersect, then the point of intersection has to be that center. Since a triangle would require three distinct such points as corners, no such triangle can exist.

In the Poincaré half plane model the situation is similar, except that straight line geodesics there are vertical lines, so the point where they intersect is the point at infinity, which is an ideal point of the hyperbolic geometry.

There are however other models, namely the Beltrami-Klein model, which use Euclidean straight lines to model hyperbolic geodesics. In this model, every hyperbolic triangle will be delimited by straight lines. As a consequence, you can tell that the angle measurement in this model cannot be the same as the apparent Euclidean angles between the lines, since all models of hyperbolic geometry are isomorphic so the angle sum inequality has to hold for all of them.

Equilateral triangles

  1. Deduce that there are equilateral triangles with angle $\frac{2\pi}n$ for each $n\geq7$. (this part is easy.)

Angle sum formula tells you as much.

Zero angles

  1. Also deduce that triangles with angle zero exist, in a certain sense, and that their area is finite. (I know that in the limit as the vertices go to infinity, there are even ideal hyperbolic triangles in which all three angles are $0°$, and so the area is finite by $(\pi−A−B−C)R^2$)

You can also have a single ideal corner point, or two of them. Since these are contained within a completely ideal triangle (of area $\pi$), you know their area to be finite as well.

Regular polygon areas

  1. Find corresponding results for regular $n$-gons.

So you are asking about a bound for the area of a regular polygon, right?

Well, the area is maximal if all corner angles are zero, i.e. all corner points are ideal. Take such an $n$-gon, and divide it into ideal triangles. Sum up their area and you have the area of your regular $n$-gon.

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