Let $f:[0,2\pi]\rightarrow R^2$ be a smooth function such that $f([0,2\pi])$ is a smooth closed simple curve $C$. Suppose $(0,0)$ lies inside the the bounded open region enclosed by $C$ and $f(t)=(x(t),y(t))$. Is it true that $g(t)=x^2(t)+y^2(t)$ has at least 4 critical points in $[0,2\pi)$? 



Number of critical points
My original answer had a silly mistake in it.


*Updated Answer:* There exists a smooth closed simple curve, which encloses the origin, yet whose squared distance to the origin has only 2 critical points in $[0,2\pi)$. 


Let $f(t)=(\frac{1}{2}+\cos{t}, \sin{t})$. This is simply the unit circle shifted to the right by $\frac{1}{2}$, so it is a smooth closed simple curve enclosing the origin. Then, $g(t) = \frac{5}{4} + \cos{t}$, which has critical points at integer multiples of $\pi$. So, $g(t)$ only has 2 critical points in $[0,2\pi)$. We can see this intuitively since the squared distance from $f(t)$ to the origin gets smaller as we go from the point $(\frac{3}{2}, 0)$ to the point $(-\frac{1}{2},0)$ and then increases until we go back to the point $(\frac{3}{2}, 0)$.


It's clear that there does not exist a smooth closed simple curve enclosing the origin where the squared distance to the origin has only one critical point.


