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

[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

 

[willem2]

similar triangles

 

[DaTario]

I have also found this one:

 

[mfb]

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

 

[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.

 

[mfb]

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

You define a given length as 1.

No ruler is used.

 

[DaTario]

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

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

 

[mfb]

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

 

[DaTario]

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

I agree.

 

[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?

 

[Mark44]

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?

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

 

[DaTario]

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

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´.

 

[Mark44]

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

 

[DaTario]

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

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).