Shortest path between two points that stays away from its rotated images

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
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

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).