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

1. Mar 11, 2017

DaTario

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. Mar 11, 2017

willem2

similar triangles

3. Mar 11, 2017

DaTario

I have also found this one:

4. Mar 11, 2017

Staff: Mentor

That is the same approach in a different way - two similar triangles.

5. Mar 11, 2017

Stephen Tashi

The traditional interpretation of "straight edge" doesn't allow it to be marked as a ruler. I don't know what "unitary segment" means.

6. Mar 11, 2017

Staff: Mentor

You define a given length as 1.

No ruler is used.

7. Mar 11, 2017

DaTario

Ok, but one may also say that this proof is based on the relation between angular coeficient in perpendicular straight lines.

8. Mar 11, 2017

Staff: Mentor

You can also say that this rule follows from the similarity of triangles - both concepts are directly connected.

9. Mar 11, 2017

DaTario

I agree.

10. Mar 14, 2017

DaTario

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?

11. Mar 14, 2017

Staff: Mentor

Seems like it would be fairly simple to prove that the hyperbola y = 1/x is unique in this regard.

12. Mar 14, 2017

DaTario

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´.

13. Mar 14, 2017

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 $|L_2| = \frac 1 {|L_1|}$

14. Mar 15, 2017

DaTario

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).

Last edited: Mar 15, 2017