Mean value theorem for curves Suppose we have a smooth non constant curve
$$
f(t) = (x(t),y(t))
$$
from $\mathbb{R}\to\mathbb{R}^2$ with $t\in(0,1)$. Suppose we have the limits at $0$ and $1$ of the curve is the same. That is
$$
\lim_{t\to 0} f(t) = \lim_{t\to 1} f(t) = \textbf{a}.
$$
Then is true that there exists a $t_0\in(0,1)$ such that $f’(t_0) = (c,c)$ for some $c>0$?
More generally, is it true that the tangent vector faces every direction at least once? This seems geometrically ok, but I couldn’t find a rigorous proof of this statement.
Thank you in advance.
 A: If $f([0,1])$ is an immersed closed curve (suitably defined --- you need $f'(1) = f'(0)$ --- or, in other words, $f$ should descend to a $C^{1}$  map of $S^{1}$), then the answer is yes.  Otherwise, it's false.
Suppose that $\Gamma = f([0,1])$ is an immersed closed curve.  Precisely, assume that $f(0) = f(1)$, $f'(0) = f'(1)$, and $f'$ is non-vanishing in $[0,1]$.  Up to redefining $f$, we can then assume that $f(t + 1) = f(t)$ and $f'(t + 1) = f'(t)$ and $f$ remains at least $C^{1}$.
Henceforth, given $e \in S^{1}$, denote by $e^{\perp} \in S^{1}$ the vector with $e \cdot e^{\perp} = 0$ and $(e,e^{\perp})$ right-hand oriented.
Pick $e \in S^{1}$.  Define $\ell_{e} : \mathbb{R}^{2} \to \mathbb{R}$ by $\ell_{e}(x) = \langle x,e \rangle$.  Since $\Gamma$ is compact, $\ell_{e}$ achieves its maximum at some point $x_{M} = f(t_{M})$ and its minimum at $x_{m} = f(t_{m})$.  Thus, an advanced calculus exercise shows that $D \ell_{e}(x_{0})$ is parallel to the "normal vector"* $f'(t_{m})^{\perp}$ and $f'(t_{M})^{\perp}$ to $\Gamma$ at $x_{m}$ and $x_{M}$.  This tells us that $f'(t_{M})^{\perp} = \pm e$ and $f'(t_{m})^{-} = \mp e$ since $x_{M}$ is a maximum and $x_{m}$, a minimum.
Since $e \mapsto e^{\perp}$ is a surjective map onto $S^{1}$, it follows that the tangent vector always points in each direction at least once.  (Note we proved that the normal vector always points in each direction at least once.)
A counter-example if $\Gamma$ is not immersed: take $f$ so that it traces out a square.  (This can be done smoothly by asking that $f$ stops and takes a break at each corner point.)  The normal vector (where defined) only takes on four values.  (If it's the coordinate square, then these four are $(1,0)$, $(-1,0)$, $(0,1)$, and $(0,-1)$ so you won't see $(1/\sqrt{2},1/\sqrt{2})$.)  The problem is $f'$ vanishes.

*

*"Normal vector" in quotes since $\Gamma$ is immersed.  For example, if $\Gamma$ is a figure-eight, there is no well-defined normal vector at the intersection point between the two loops.  However, $f'(t)$ and $f'(t)^{\perp}$ are always well-defined and the non-vanishing condition on $f'$ ensures that, locally in $t$, we only see one of the possibilities.

A: This answer is motivated by what Peter wrote in his answer. I realize now that when I was thinking about this question, I was thinking of something along the lines of intermediate value theorem, but didn't know how to apply it for a curve in $\mathbb{R}^2$.
Suppose that $f'(t)$ be a continuous non zero function of $t$. Choose a unit vector $e_1$ and let $e_2$ be perpendicular to $e_1$. Then it follows that $t\mapsto \langle e_2, f'(t)\rangle$ is a continuous function of $t$. Now if $\langle e_2, f'(t)\rangle|_{t=0} < 0$ and $\langle e_2, f'(t)\rangle|_{t=1} > 0$, we can apply the intermediate value theorem for the function $t\mapsto \langle e_2, f'(t)\rangle$ and get the required result.
