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In the plane equipped with an orthonormal basis, let us consider the two points $A$ and $B$ whose coordinates are $(-2,0)$ and $(1,1)$, respectively. Is there a path from $A$ to $B$ (i.e. a continuous map $\gamma : [0,1] \to {\mathbb R}^2$, with $\gamma(0)=A, \gamma(1)=B$) such that the sets $I,\rho (I), \rho^2 (I)$ and $\rho ^3(I)$ are all disjoint, where $\rho$ is the rotation whose center is the origin and whose angle is $\frac{\pi}{2}$, and $I$ is the image of the path : $I=\gamma([0,1])$.

Update 14:30 Now that I know that such paths do exist (see carlop’s answer), I ask : what is the minimal length of a path meeting those constraints ?

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I could draw such on my paper, probably some arc of an appropriate circle would do that. –  Berci Dec 1 '12 at 13:11
    
It's probably not possible to attain the greatest lower bound on path length, only to come within epsilon –  hardmath Dec 1 '12 at 13:34
    
@hardmath : indeed, the “optimal” path probably intersects its rotated image at a few “extremal” points. But I’m curious to know what this path looks like ? –  Ewan Delanoy Dec 1 '12 at 13:56
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The piecewise-linear path $A \rightarrow \rho(B) \rightarrow B$ has length $2 + \sqrt{2}$, and if the middle point is perturbed slightly upward, the images $I$, $\rho(I)$, etc. are disjoint. So we can approach this length as closely as we wish, but not attain it. –  hardmath Dec 1 '12 at 13:58

1 Answer 1

up vote 3 down vote accepted

Take a spiral centered in the origin that pass for the two point...

In general every function such that the distance from the origin is increasing works fine for every angle greater than $0$ (not only for the right angle).

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