Let $f:[0,1]\to\mathbb R^k$ be a continuous function with $f(1) = \overrightarrow 0$. Is it true that there always exist $k$ points $0 \le a_1 \le a_2 \le \ldots \le a_k \le 1$ such that $\sum_{i=1}^k (-1)^{k+1} f(a_k) = f(a_1) - f(a_2) + f(a_3) - \ldots = \frac{f(0)}{2}$?

When $k=1$ we have to find one point $a_1 \in [0,1]$ with $f(a_1) = \frac{f(0)+f(1)}{2}$, which is the intermediate value theorem.

When $k=2$ the situation is more complicated. We believe the statement is still true. However we noted that if we are not finding $\frac{f(0)}{2}$ but, say, $.49 f(0)$ instead, there exist counter examples. A counter example looks like a sine function $f(t) = (t, \sin 100t)$, rotated in the 2-d plane a little bit. (This is not the example, but you can imagine changing the constants $.49$ and $100$ a bit makes it work.)

For bigger $k$ we have no idea.

Are there any similar results known before?