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I am puzzeling with the following:

Using the beltrami klein disk of hyperbolic geometry (see https://en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model ) (PS not the poincare disk model)

and given a segment inside it

How to construct an equilateral triangle where the segment is one of its sides?

I know you could first translate the endpoints to a Poincare disk points , do the construction the poincare disk and then translate the third point back to the beltrami model.

But is there no simpler way?

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  • $\begingroup$ I don't know how to answer your question, but one thing I am wondering is whether it can be answered if one allows the construction of tangent line given a point on a conic in the projective plane. Allowing such tangents to be constructed, one can construct a perpendicular bisector to a line segment $\ell$ in the Klein model. But so far I cannot extend that to a construction of an equilateral triangle on that line segment. $\endgroup$ – Lee Mosher Jun 6 '15 at 13:35
  • $\begingroup$ Here's the perpendicular bisector construction. Extend $\ell$ to a line $L$. Draw the tangent lines of the circle at infinity at the two ideal endpoints of $L$ and intersect those lines at $P$ in the Mobius band beyond infinity. Draw the two lines $PA$, $PB$ through the endpoints $A,B$ of $\ell$. Draw the intersection points $A',B'$ of $PA,PB$ with the circle at infinity. Draw the tangents to the circle at infinity at the two points $A',B'$ and intersect those two lines at $Q$. The line $\overline{PQ}$, intersected with the disk, is the perpendicular bisector of $\ell=\overline{AB}$. $\endgroup$ – Lee Mosher Jun 6 '15 at 13:40
  • $\begingroup$ the perpendicular bisector is not the problem , but what then? $\endgroup$ – Willemien Jun 6 '15 at 16:04

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