# 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