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I am studying Arlan Ramsay's and Robert Richtmeyer's " Introduction to hyperbolic geometry"

On page 255/6 it gives how to construct an segment with an absolute length of $ \ln (\sqrt{2} +1) $ (via the construction of a Sweikart triangle, a right triangle with two ideal points, $ \ln (\sqrt{2} +1) $ is the length of the altitude from the right angle.)

On page 278 the book mentions that all constructions are also possible with and an finite straightedge (one that can only draw a line trough any two finite points)

This made me wonder:

How to construct an segment with a known length with only a finite straight edge?

Or even:

How to construct an segment with a known length with only a limited straightedge? (a limited straightedge as a straightedge that can only draw lines beteen points that are less than some limited distance away from eachother, like a real the resulting segment will need to be shorter than the straightedge)

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  • $\begingroup$ Do you agree that any straght line of any length can be drawn with a limited straight edge? A straight of limited lenghth can be elongated by taking two points near to its end, and then to put the ruler to these points and go on with the straight. This is not hyp geo, this absolute geo-- independent of the axiom of parallels. $\endgroup$ – zoli Mar 6 '15 at 20:14
  • $\begingroup$ I agree that any straight line of any (finite) length can be drawn with a limited straightedge. (but you cannot draw line between two ideal points) Also I am not convinced that you can draw a straight line between any 2 finite points in the hyperbolic plane with a limited straightedge (and compass). Both constructions given at at math.stackexchange.com/questions/1178135/… won't work in hyperbolic geometry, $\endgroup$ – Willemien Mar 6 '15 at 22:59
  • $\begingroup$ OK. What do you mean by "construct a segment?" (a) Is a straight line given with a point on it? Do you have to construct that segment from the point given on the straight line given? (c) Is the length given a s a number or by a segment somewhere else whose length will have to be copied? Do you have pair of compasses? $\endgroup$ – zoli Mar 7 '15 at 8:12
  • $\begingroup$ Sorry my question is to construct a segment (anywhere) with some fixed absolute length (as hyperbolic geometry has an intrinsic distance scale) you can move and reuse compasses (why do you need two?) $\endgroup$ – Willemien Mar 7 '15 at 8:58
  • $\begingroup$ In English "* the compass" is said like "a pair of compasses." Again: Are you given a number or a segment somewhere to be copied onto a straight line from a given point on it? If you have an irrational number relative to the natural unit length then you cannot create the segment in finite steps. $\endgroup$ – zoli Mar 7 '15 at 9:16
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If $\pi/4$ is the angle of parallelism belonging to the segment to be constructed then I know what to do.

(1) Construct two lines $a$ and $b$ whose angle is $\pi/4$. Let their intersection point be denoted by $P$.

(2) Draw line $b'$, the image of $b$ by reflection in $a$.

(3) Construct the common parallel to $b'$ and $b$.

(4) This common parallel will intersect $a$ in $P'$.

(5) The segment $PP'$ is of the desired length.

Notes:

A. If you don't know how to construct the common parallel then let me know.

B. The construction described above will enable you to construct segments whose length can be associated with an angle that you can construct. (The angle of parallelism belonging to the given length.)

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