# Inversion Through P in Euclidean Plane | Fix(Vp)

• Gail Brimeyer
In summary: So, Vp composed of Vp will give us the identity map because the inversion through p will take us back to p.Ok I'm going to write what I came up with, I'm not the best with notation so see if this makes sense to you...Describe Fix(Vp) and proof - I now say that there's only one point fixed in Vp and that's when X=Y=P.Proof: Part 1. Given an element X and and a point P which is X, The inversion of element X through P will result in X therefore Vp(x)=xPart 2. Given a element X and a point P that's not X in E. The inversion through P will produce a point X'
Gail Brimeyer

## Homework Statement

If P is a point in E (euclidean Plane) Then the "Inversion through P" is
Vp = {(x,y)| x,y in E and either,
1. x=y=p, or
2. p is the midpoint of segment xy}What's Fix(Vp)

Show that Vp composed of Vp = The identity

## The Attempt at a Solution

Vp composed of Vp = The identity because Vp is a rigid motion and two rigid motions are equal to the identity?? (not so sure on this one)

I believe is that Fix(Vp) is every point because given point P in E and it's inversion through point P gives us that the inversion is = x=y=p so wouldn't that mean (p,p)?

## The Attempt at a Solution

Last edited:
Gail Brimeyer said:

## Homework Statement

If P is a point in E (euclidean Plane) Then the "Inversion through P" is
Vp = {(x,y)| x,y in E and either,
1. x=y=p, or
2. p is the midpoint of segment xy}
so Vp is the set of all pairs of points satisfying those conditions?

What's Fix(Vp)
What does "Fix" mean? My first thought was "fixed point" but you find fixed points for a function from a set to itself and here V takes points to sets of pairs of points- it can't have any fixed points. And, further, it appears you are applying "Fix" to Vp, not V. Again, what is "Fix"?

Show that Vp composed of Vp = The identity
The way you have defined it, Vp is a set of pairs of points, not a function- you compose functions, not sets.

## The Attempt at a Solution

Vp composed of Vp = The identity because Vp is a rigid motion and two rigid motions are equal to the identity?? (not so sure on this one)

I believe is that Fix(Vp) is every point because given point P in E and it's inversion through point P gives us that the inversion is = x=y=p so wouldn't that mean (p,p)?

HallsofIvy said:
so Vp is the set of all pairs of points satisfying those conditions?
Yes, Vp satisfies those two conditions,
Given a point p, the inversion through p will give either,
1. x=y=p
2. P which is the midpoint of the line segment XY

What does "Fix" mean? My first thought was "fixed point" but you find fixed points for a function from a set to itself and here V takes points to sets of pairs of points- it can't have any fixed points. And, further, it appears you are applying "Fix" to Vp, not V. Again, what is "Fix"?

Let T be a mapping.
A point B in the euclidean plane is called fixed for T if (B,B) is an element of T.

The way you have defined it, Vp is a set of pairs of points, not a function- you compose functions, not sets.
The only way I can help you with this question is to give you an example of another composition

given a point P not lying on a line L, Reflect point P over line L to give you point P' which is the Rl(p) (the reflection of P.)
Then Rl(p) composed of Rl(P) will bring you right back to the beginning point P and therefore
Rl composed of Rl is the identity

Gail Brimeyer said:
If P is a point in E (euclidean Plane) Then the "Inversion through P" is
Vp = {(x,y)| x,y in E and either,
1. x=y=p, or
2. p is the midpoint of segment xy}

OK, you have the formalities here, but you don't seem to be thinking of $$V_p$$ as a function or mapping, which is what you really need. If $$x = p$$, what is $$V_p(x)$$? What about if $$x \neq p$$; what is $$V_p(x)$$, in a rough geometrical description?

It may help you to impose coordinates; in that case it is simpler to start by supposing $$p = (0, 0)$$, and then figure out how to generalize.

Gail Brimeyer said:
Vp composed of Vp = The identity because Vp is a rigid motion and two rigid motions are equal to the identity?? (not so sure on this one)

It is true that $$V_p$$ is a rigid motion of the plane, but it is not true that a rigid motion composed with itself always equals the identity (think of translation by any nonzero distance in any direction, or rotation by say $$\pi/4$$).

Gail Brimeyer said:
I believe is that Fix(Vp) is every point because given point P in E and it's inversion through point P gives us that the inversion is = x=y=p so wouldn't that mean (p,p)?

What is $$\mathop{\mathrm{Fix}}(V_p)$$? It is the set of fixed points of $$V_p$$, that is, $$\mathop{\mathrm{Fix}}(V_p) = \{ x \in E : V_p(x) = x \}$$. You are right to consider whether $$p \in \mathop{\mathrm{Fix}}(V_p)$$ separately from whether $$x \in \mathop{\mathrm{Fix}}(V_p)$$ for $$x \notin p$$, but to build a correct argument, you need to think more carefully of $$V_p$$ as a function, and what it does to each point.

Ok I'm going to write what I came up with, I'm not the best with notation so see if this makes sense to you...
Describe Fix(Vp) and proof - I now say that there's only one point fixed in Vp and that's when X=Y=P.
Proof:
Part 1. Given an element X and and a point P which is X, The inversion of element X through P will result in X therefore Vp(x)=x

Part 2. Given a element X and a point P that's not X in E. The inversion through P will produce a point X' where P is the midpoint of XX'. I say that it's impossible to have the inversion through P = P because if X isn't = P then the inversion will result in another element that's equal distance from P that X is.

Vp composed of Vp = Identity
Here's what I came up with, again sorry for the notation.
In the definition of Vp if x=y=p then we know that if x is equal to p then the inversion will give us p', then once again we take the inversion of p' it'll take us back to P. Where P=P'.

Now if X=Y=P, then you have your element x and a point p. the inversion through P will result in a point X' where XP congruent to X'P. because p is the midpoint. So by taking the inversion of X' will result back to point X where p is once again the midpoint.

Most of this is right in essence; I'll show you how to phrase some of it more carefully.
Gail Brimeyer said:
Part 1. Given an element X and and a point P which is X, The inversion of element X through P will result in X therefore Vp(x)=x
You could just say "by the definition of $$V_p$$, $$V_p(p) = p$$".
Gail Brimeyer said:
Part 2. Given a element X and a point P that's not X in E. The inversion through P will produce a point X' where P is the midpoint of XX'. I say that it's impossible to have the inversion through P = P because if X isn't = P then the inversion will result in another element that's equal distance from P that X is.
This is correct. You might end by observing "and this distance is nonzero, because $$x \neq p$$".
Gail Brimeyer said:
Vp composed of Vp = Identity
Here's what I came up with, again sorry for the notation.
In the definition of Vp if x=y=p then we know that if x is equal to p then the inversion will give us p', then once again we take the inversion of p' it'll take us back to P. Where P=P'.
This is correct, but it is simpler to say that since $$V_p(p) = p$$, also $$V_p(V_p(p)) = V_p(p) = p$$.
Gail Brimeyer said:
Now if X=Y=P, then you have your element x and a point p. the inversion through P will result in a point X' where XP congruent to X'P. because p is the midpoint. So by taking the inversion of X' will result back to point X where p is once again the midpoint.
I think you mean to begin with "if $$x \neq p$$". The essence of your argument is correct. $$V_p(x)$$ is the other endpoint of the segment having $$x$$ as one endpoint and $$p$$ as midpoint. $$V_p(V_p(x))$$ is the other endpoint of the segment having $$V_p(x)$$ as one endpoint and $$p$$ as midpoint. But these are one and the same segment; so $$V_p(V_p(x)) = x$$.

Therefore $$V_p(V_p(x)) = x$$ for all $$x\in E$$, whether $$x = p$$ or $$x \neq p$$; that is, $$V_p \circ V_p$$ is the identity map.

## 1. What is "Inversion Through P" in the Euclidean Plane?

"Inversion Through P" is a geometric transformation in the Euclidean Plane where a point P is chosen as the center of inversion. This transformation involves reflecting all points in the plane through the chosen center point, with the distance between each point and the center point being inverted or flipped.

## 2. What is the significance of "Fix(Vp)" in the context of Inversion Through P?

"Fix(Vp)" refers to the set of points in the Euclidean Plane that remain fixed or unchanged under the transformation of Inversion Through P. These points are equidistant from the chosen center point P and form a circle known as the "circle of inversion".

## 3. How does Inversion Through P affect lines and circles in the Euclidean Plane?

Inversion Through P preserves the properties of lines and circles in the Euclidean Plane. Lines that pass through the center point P remain fixed after inversion, while circles not passing through P are inverted into other circles. Additionally, tangent lines to the circle of inversion are flipped into tangent lines to the original circle.

## 4. What is the relationship between Inversion Through P and Mobius transformations?

Mobius transformations are a generalization of Inversion Through P in the Euclidean Plane. They involve transformations of the entire plane, while Inversion Through P only involves transformations within a specific circle. Mobius transformations can also be seen as a combination of Inversion Through P and translation, rotation, and scaling transformations.

## 5. What are some applications of Inversion Through P in the field of mathematics?

Inversion Through P has various applications in fields such as geometry, complex analysis, and number theory. It can be used to prove geometric theorems, solve problems in conformal mapping and geometric optics, and even in cryptography. Additionally, Inversion Through P has connections to other mathematical concepts such as fractals and the Mandelbrot set.

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