Let C be an oriented plane curve with curvature k>0. Assume that C has at least one point p of self intersection, prove the following questions:
a. There is another point $p_{0}$ such that the tangent line $T_{0}$ at $p_{0}$ is parallel to some tangent at p.
b. The rotation angle of the tangent in the positive arc of C made up by $pp_{0}p$ is >$\pi$
c.the rotation index is >=2.
I think everything is obvious true, but it looks hard to prove it.Any hints? I am beginning studying differential geometry, this question is very strange to me. I have no idea how to use my differential geometry knowledge to prove it. Can someone tell me how to prove them?
Let $\gamma:[0,l]\to\mathbb{R}^2$ be an arc length parametrization of $C$, such that $\gamma(0)=p$. We may assume $\dot{\gamma}(0)=(1,0)$. Let $\theta:[0,l]\to\mathbb{R}$ be the unique continuous function which satisfies$$\theta(0)=0,\quad\dot{\gamma}(t)=(\cos\theta(t),\sin\theta(t)).$$In other words, $\theta$ is the direction of $\gamma$ at time $t$, and whenever the rotation index increases by $1$, the function $\theta$ increases by $2\pi$. By assumption, $\theta$ is a monotonic function.
Since $p$ is a point of self intersection, there exists $a\in(0,l)$ such that $\gamma(a)=p$. Hence,$$\int_0^a\dot{\gamma}(t)dt=0,$$and in particular,$$\int_0^a\sin\theta(t)dt=0.$$Since $\theta(0)=0$, and $\theta$ is monotonic, it follows from the last equation that there exists $b\in(0,a)$ such that $\theta(b)=\pi$, and question a. is solved.
Questin b. follows immediately, since the rotation angle of the tangent between $p$ and $\gamma(b)$ is exactly $\pi$.
To show c., one just needs to repeat the above argument for the restriction $\gamma|_{(a,l)}$. This shows that $\theta(l)>2\pi$, and so the rotation index is $>1$.