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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 $[0,2\pi)$?

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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.

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  • $\begingroup$ Doesn't this function have infinite number of critical points? g'(t)=0 for all t in (0,1). $\endgroup$ – User4966 Feb 18 '15 at 6:51
  • $\begingroup$ My mistake. I was thinking about sign changes for some reason. $\endgroup$ – Alec Payne Feb 18 '15 at 7:13
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    $\begingroup$ Note that if the origin is chosen to be the center of mass of C or of the region enclosed by C, then g(t) has at least 4 critical points (see Domokos, Papadopoulos, Ruina, J. Elasticity 36 [1994], 59–66, link.springer.com/article/10.1007%2FBF00042491) $\endgroup$ – Yoav Kallus Feb 18 '15 at 16:07

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