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.
 A: I've never seen this question before, so it took some thought to find the right approach. Here's a sketch of an a solution. The hypotheses tell you that the curve is convex, so for each point on the curve there is a unique "opposite" point where the tangent line is parallel; moreover, the unit tangent vectors are indeed opposite at those points.
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 $2R$.
