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In the Extended Euclidean Plane, H, let A=(0,0), B=(5,0), and C=(4,0). Show that there exists a point D=(x,0) for some real number x, such that Rx(A,B;C,D)=pi.

I have the formulas for these things. However, I am having trouble applying them correctly.

I have that epsilon= |[A,B] intersect {C,D}| and Rx= (-1)^Epsilon (dE(A,C)/dE (A,D) ) (dE(B,D)/dE (B,C) ) Where dE= Euclidean distance function.
Is dE just the distance between the points?
Any help is appreciated.

lj_growl
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  • I don't clearly catch your notation, but yes, it's just the ordinary (signed) distance of points. So we have $\frac{AC}{CB}=\frac41$, so we need $\frac x{5-x}=\frac{AD}{DB}=\frac4\pi$. – Berci Mar 31 '19 at 00:52
  • So, does writing this show that there is some x? Or must there be some other justification? – lj_growl Mar 31 '19 at 01:14
  • Meanwhile I think I figured it out. I knew a slightly different definition of the cross ratio. In the end, you will have to solve my equation. – Berci Mar 31 '19 at 01:20

1 Answers1

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Yes, $d_E$ here means the usual (Euclidean) distance.

First note that if nonzero vectors $u, v$ are parallel ($v=\lambda u$) then we can divide them: $\frac vu:=\lambda$ which is a signed version of the more generally writable $\frac{|v|}{|u|}$.

To clearly see what happens here, you should prove that the cross ratio as defined in the post, coincides with the following $$\frac{\vec{AC}}{\vec{CB}}\, /\, \frac{\vec{AD}}{\vec{DB}}$$ based on the above notation of parallel vectors.
For that, observe that $\epsilon$ is odd iff exactly one of $C,D$ lies between $A$ and $B$.

Berci
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