# Non-intersecting smooth paths in the plane and the relation to curvature

I'm interested in a problem and have no idea how to approach solving it. Could you please point me in the right direction.

Given 3 smooth paths $\varphi_{1}:[0,1]\to \mathbb{R}^{2}$, $\varphi_{2}:[0,1]\to \mathbb{R}^{2}$ and $\gamma:[0,1]\to \mathbb{R}^{2}$ such that:

1) $\varphi_{1}(1)=\varphi_{2}(1)=\gamma(1)=(x_{0},y_{0})$

2) $\varphi_{1}(0) = (0,y_{1})$, $\varphi_{2}(0) = (0,y_{2})$ and $\gamma(0) = (0,y_{3})$, where $y_{1} < y_{2} < y_{3}$.

3) All three intersect the $x = 0$ line perpendicularly.

4) (Edit) The curvature of each decreases monotonically from $0$ to $1$.

What conditions on the curvature of these three paths ensure that they do not intersect other than at $(x_{0},y_{0})$?

The more specific problem I am interested in is when $(x_{0},y_{0})$ is a point at infinity and all three paths tend to being parallel.

What subject/books concern questions like this? Which theorems would be useful for proving such properties.

P.S. Parameterization of the curves is not important.

Thanks!

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You seem to be assuming that curvature is defined to be nonnegative, which is not the case, see en.wikipedia.org/wiki/Curvature#Signed_curvature – Will Jagy Dec 9 '11 at 21:32

Constant curvature works in this case. Indeed, if you include a point at $\infty,$ this is precisely Hilbert's "field of ends" in $\mathbb H^2,$ any direction (except for straight right) leaving your fixed point $(x_0, y_0)$ defines a unique circular arc with center on the line $x=0$ that also meets the line $x=0$ orthogonally. Straight left gives a line segment.
I see no necessary conditions. Your three curves could separate just after leaving $(x_0, y_0),$ the three then staying in parallel constant width strips but with a meander of arbitrarily high absolute value of curvature ( the connected central path of brown stones in the Rhodes pavement in en.wikipedia.org/wiki/Meander_%28art%29 ), then straighten out again in time to reach $x=0$ orthogonally. – Will Jagy Dec 9 '11 at 0:39