# The circle does not intersect the unbounded component of $\mathbb{R}^2 \setminus C$

We get the following problem in our differential geometry class.

Let $C$ be a smooth, non-degenerate simple closed curve traveling counterclockwise. Suppose that the curvature $\kappa$ of C is everywhere positive. Let $R > 0$ be a real number such that for all $p \in C$, we have $R \leq 1/\kappa(p)$. Prove that, for all $p \in C$, the circle whose radius is $R$ and center is $p + R\vec{N}(p)$ does not intersect the unbounded component of $\mathbb{R}^2 \setminus C$.

I can sort of see this intuitively (push a point on the curve inside and with the radius less than the radius of curvature the circle lies inside the curve) but I am not sure how to make this rigorous. Any help is appreciated.

Here's what you want to prove: If $R\le \dfrac1{\kappa(s)}$ for all $s$, then any pair of opposite points are at least $2R$ apart. (This will tell you that the curve cannot come inside any of the circles in your problem.)
Start with an arclength parametrization of your curve. It is a fact (see, e.g., my differential geometry text, p. 27) that there is a smooth function $\theta$ so that the unit tangent vector $\mathbf T(s) = \big(\cos\theta(s),\sin\theta(s)\big)$, and so it is immmediate that $\kappa(s)=\theta'(s)$. Without loss of generality, we may take $\theta=0$ at one of our points (say $s=0$) and $\theta=\pi$ at the other (say $s=s_0$). The vertical distance between the points is then given by $$\int_0^{s_0} \mathbf T(s)\cdot (0,1)\,ds = \int_0^{s_0} \sin\theta(s)\,ds.$$ Use the hypothesis relating $R$ and $\kappa$ to show that this integral is at least $2R$.
• @Federico: The center $C(p)$ of the circle $p+R\vec N(p)$ moves parallel to the tangent vector to the curve at $p$. Thus, for $p,q$ with $q$ and $C(p)$ closest (globally), the tangent vectors at $q$ and at $p$ will end up parallel. This tells us that $p$ and $q$ will be opposite points, hence at least $2R$ apart. – Ted Shifrin Oct 26 '18 at 18:26