# Homework Help: P is a point in E

1. Dec 15, 2009

### Gail Brimeyer

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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)?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Dec 15, 2009
2. Dec 16, 2009

### HallsofIvy

so Vp is the set of all pairs of points satisfying those conditions?

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"?

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

3. Dec 16, 2009

### Gail Brimeyer

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

4. Dec 16, 2009

### ystael

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.

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 $$\pi/4$$).

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.

5. Dec 16, 2009

### Gail Brimeyer

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.

6. Dec 17, 2009

### ystael

Most of this is right in essence; I'll show you how to phrase some of it more carefully.
You could just say "by the definition of $$V_p$$, $$V_p(p) = p$$".
This is correct. You might end by observing "and this distance is nonzero, because $$x \neq p$$".
This is correct, but it is simpler to say that since $$V_p(p) = p$$, also $$V_p(V_p(p)) = V_p(p) = p$$.
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.