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Jul 13, 2020 at 14:49 comment added Joel David Hamkins @AlbertvanderHorst Doesn't Noam's observation (comment on the OP) enable you to pick the right directions? You can tile the whole region between the points before choosing the lines.
Jul 12, 2020 at 8:57 comment added Albert van der Horst @Goucher. Note that one can do an arbitrary line of lenght n by n applications of the ruler. So the arbitrary lines can solve the problem in one step, if you're lucky enough. That makes it seem like your argument is way to pessimistic. On the other hand, it seems that there is a carpenters way, but no mathematicians way to select the directions. Next thing you know the problem is NP complete. Groetjes Albert
Jul 12, 2020 at 8:32 comment added Joel David Hamkins See also this 2008 article (in French) by Xavier Caruso xavier.toonywood.org/papers/publis/troppetit.pdf, which seems to use the same method.
Jul 12, 2020 at 8:30 comment added Joel David Hamkins Thanks for your excellent answer. This is great!
Jul 12, 2020 at 8:28 vote accept Joel David Hamkins
Jul 11, 2020 at 17:30 comment added Adam P. Goucher Note that this naive recursive approach means that, in particular, joining two points separated by a distance of $(1 + \varepsilon)^n$ would take $10^n$ applications of the original straightedge! (It's still polynomial in the length, though, so not too bad.)
Jul 11, 2020 at 17:27 comment added Adam P. Goucher The construction means that you can henceforth proceed as though you have a length-$(1 + \varepsilon)$ straightedge instead of a length-$1$ straightedge. By the same argument, you can proceed as though you have a length-$(1 + \varepsilon)^2$ straightedge (by using the length-$(1 + \varepsilon)$ straightedge to draw each line in the diagram).
Jul 11, 2020 at 17:12 history edited Adam P. Goucher CC BY-SA 4.0
visual demonstration
Jul 11, 2020 at 17:00 history answered Adam P. Goucher CC BY-SA 4.0