In the last post I have presented a hyperbolic kaleidoscope with two- and three-fold rotational symmetries. Could we have other rotational symmetries?
Yes, we simply move the vertical lines! To get an n-fold rotational symmetry the circle has to intersect a straight line at an angle of π/n. The distance between the line and the center of the unit circle is then cos(π/n). We can freely choose the rotational symmetry at the left and at the right of the unit circle as long as the sum of the angles of the kaleidoscopic triangle is less than π.
Here is an example with 2- and 4-fold rotational symmetry:
And here comes a symmetric kaleidoscope with two 3-fold rotational symmetries: