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I How to obtain the inverse (reciprocal) of a line segment?

  1. Mar 11, 2017 #1
    Hi All,

    Which are the ways one can geometrically obtain, given a line segment AB with length x and an unitary segment OC, a line segment with length 1/x ?

    Straight edge and compass are allowed (also some auxilliary curve).

    Best wishes,

    DaTario
     
  2. jcsd
  3. Mar 11, 2017 #2
    similar.png
    similar triangles
     
  4. Mar 11, 2017 #3
    I have also found this one:

    invert method old 1.jpg
     
  5. Mar 11, 2017 #4

    mfb

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    Staff: Mentor

    That is the same approach in a different way - two similar triangles.
     
  6. Mar 11, 2017 #5

    Stephen Tashi

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    The traditional interpretation of "straight edge" doesn't allow it to be marked as a ruler. I don't know what "unitary segment" means.
     
  7. Mar 11, 2017 #6

    mfb

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    You define a given length as 1.

    No ruler is used.
     
  8. Mar 11, 2017 #7
    Ok, but one may also say that this proof is based on the relation between angular coeficient in perpendicular straight lines.
     
  9. Mar 11, 2017 #8

    mfb

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    You can also say that this rule follows from the similarity of triangles - both concepts are directly connected.
     
  10. Mar 11, 2017 #9
    I agree.
     
  11. Mar 14, 2017 #10
    But, apart from the hyperbola y = 1/x, which is rather obvious, is there any other auxiliary curve that can be used to invert the line segment?
     
  12. Mar 14, 2017 #11

    Mark44

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    Seems like it would be fairly simple to prove that the hyperbola y = 1/x is unique in this regard.
     
  13. Mar 14, 2017 #12
    In some sense, the circle also can be understood as an auxiliary curve to this construction. See figure below, from point P through point T we find point P´.
    figure forum circle inverter.jpg
     
  14. Mar 14, 2017 #13

    Mark44

    Staff: Mentor

    If you're talking about the relationship between, for instance, the sine and cosecant, those all reduce to reciprocal relationships.I.e., ##\csc(x) = \frac 1 {\sin(x)}##. In other words, the same as y = 1/x.

    I'm assuming that by inverting a line segment, you mean finding another segment so that the product of the lengths of the two segments is 1, then the basic relationship is ##|L_1| |L_2| = 1##, or ##|L_2| = \frac 1 {|L_1|}##
     
  15. Mar 15, 2017 #14
    Yes it is precisely so. If we graph the functions ##\csc(x)## and ##\frac 1 {\sin(x)}## it yields a method for inverting a point. Not practical, however. Pick a point O so as to obtain the line segment OP. Then, the line segment OP´naturally appears (figure below). func inverter forum.jpg
     
    Last edited: Mar 15, 2017
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